Basic Mathematics & Measurements
1
Section I Single Correct Option
1, K = ∴
1 mv2 2
p2 = ∆m ∆v ∆K = +2 K max m v
∴ Maximum error = 2% + 2 (3%) In the estimate of, kinetic energy ( K ) = 8% Option (c) is correct. m m 2. d = = 3 V l ∴ ∆m ∆l ∆d × 100 = × 100 + 3 × 100 d max m l = 4% + 3(3%) = 13% Option (d) is correct. F F 3. p = = A L2 ∴ Permissible error in pressure ( p) = 4% + 2 (2%) = 8% Option (a) is correct. 4. ⇒
p1V1 = p2 V2 V p2 = p1 1 V2 = p1 =
V1 V1 − 10% of V1
p1 90%
p1 9
∴ Percentage increase in pressure =
100 9
= 111 . Option (a) is correct. p2 5. K = 2m ∴ Error in the measurement of kinetic energy (K) = 2 × 100% = 200% Option (d) is correct. 6. 3400 = 3.400 × 103 ∴ Number of significant figures = 2 Option (d) is correct. 7. A = 3.124 m × 3.002 m = 9.378 / 248 m 2 = 9.378 m 2 Option (a) is correct. GM 8. g= 2 R constant = R2 1 2 K = Iω 2 1 2 = MR2 ω 2 5 = constant × R2
Basic Mathematics & Measurements ∴ Decrease in R (radius) by 2% world increase g by 4% and decrease K (rotational kinetic energy) by 4%.
16. Q = I α e
= 12 × 6 × 2.45 = 176.4 = 1764 . × 102 cm 3 2 = 2 × 10 cm 2 Option (b) is correct. P 11. I = 4 πr2 i.e., Ir2 = constant i.e., if r is increased by 2% the intensity will decrease by 4%. Option (d) is correct. 12. Option (b) is correct. 4 13. V = πr 3 3 ∆V ∆r =3 V r = 3(1%) = 3% Option (c) is correct. 14. a 3 = 6a2 (given) ∴ a=6 ⇒ V = 63 = 216 m 3 Option (b) is correct. l 15. g = 4 π2 2 T ∆l ∆T ∆g ∴ × 100 = × 100 + 2 × 100 T l g max 1 mm 015 . = × 100 + 2 × 10 2003 100 m = 01 . % + 01 . % = 0.2 % Option (a) is correct.
(given)
Q = It
9. Heat (H) = 22 Rt
10. V = lbt
| 3
We know that
Option (b) is correct. ∴ Maximum error in measuring heat (H) = 2 (2%) + 1% + 1% = 6% Option (b) is correct.
tI − (∆ V ) ε 0β
−
∴
t=α e
tI (∆V ) ε 0β
[α ] = [ t ] tI and [β ] = ( ∆V ) ε 0 ⇒
⇒
or
β = tI = I 1 α ( ∆V ) ε 0 t ( ∆V ) ε 0 1 1 = [Resistance] ε 0 1 1 β = α [ML2 T –3A –2 ] [M–1L–3T 4 A 2 ] 1 = –1 = [velocity] [L T] 1 = [µ 0 ε 0 ]1/ 2
Option (a) is correct. pq 17. a = 2 3 r s 1 ∆p 1 ∆q ∆a × 100 = × 100 + × 100 a max 2 p 2 q ∆r ∆s +2 × 100 + 3 × 100 r s 1 1 = (1%) + (3%) +2 (0.5%) + 3(0.33%) 2 2 = 0.5% + 1.5% +1% + 1% = 4% Option (c) is correct. 18. Least count of main scale 2 mm = = 0.5 mm 4 least count of main scale Least count = 50 = 0.1 mm Zero error = − 30 × 0.01 mm = −0.3 mm (–ive sign, zero of circular scale is lying above observed reading of plate thick) = 2 MSR + 20 CSR
4 | Mechanics-1 = (2 × 05 . mm) + (20 × 0.01 mm) = 1 mm + 0.2 mm = 1.2 mm. Plate thickness (corrected reading)
= observed reading − zero error = 1.2 mm + 0.3 mm = 1.5 mm Option (d) is correct.
More than One Correct Options
1. Maximum percentage error in x : = (% error in a) + 2 (% error in b) +3 (% error in c) = 15% Assertion and Reason
1. Least count of screw gauge Pitch = Number of divisions of circular scale Less the value of pitch, less will be least count of screw gauge leading to len uncertainty that is more accuracy in the measurement.
Thus, assertion is true. From the above relation we conclude that least count of screw gauge is inversely proportional to the number of divisions of circular scale. Thus reason is false. Option (c) is correct.
Match the Columns
1. (a) F =
GM1M2 r2
∴ GM1M2 = Fr2 i. e., [GM1M2 ] = [MLT –2 ][L2 ] = [ML3T –2 ] ∴ (a) → (q) 3 RT 3 pV 3 work (b) = = M n nM 3 RT ML2 T −2 ∴ = = [L2 T −2 ] M M
qvB sin θ = qB = v2 sin2 θ ∴
2
F2 –1 2 2 –2 q2 B2 = [LT ] = [L T ]
∴ (c) → ( r) (d)
g=
GM e R2e
GM e = gRe Re
∴ (b) → (r) F2 F (c) 2 2 = qB qB
2
⇒
GM e –2 R = [LT ][L] e = [L2 T –2 ]
∴ (d) → ( r)
2
Units & Dimensions Vectors Section I
Single Correct Option
Force Area [MLT −2 ] [ p] = [L2 ]
1. Pressure ( p) =
∴
= [ML−1T −2 ] Option (d) is correct. 2.
∴
∴
W = I 2 Rt [ML2 T −2 ] [ML2 T −2 A −2 ] [R] = …(i) = [T ] [A 2 T] dI and W = Vq V =L dt W dt L= q dI [L] =
[ML2 T −2 ][T] = [ML2 T −2 A −2 ] 2 [A T]
Using Eq. (i) [ L] [T ] L [T ] = R [R] =
i.e.,
Option (c) is correct. 3.
∴
F = 6π ηav [F ] [ η] = [ av ] [ML T −2 ] = [L L T −1 ]
= [ML−1 T −1 ] Option (d) is correct.
4.
φ = Li [φ ] = [ L ][ i] ∴ = [ML2 T −2 A −2 ] [ A ] = [ML2 T −2 A −1 ] Option (a) is correct.
5. Linear impulse ( I ) = F ⋅ ∆t
[I] = [ML T −2 ] [T] = [ML T −1 ] Option (c) is correct. mm 6. F = G 1 2 2 d [ F ][ d2 ] [ML T −2 ] [L2 ] [G] = = [ m1 m2 ] [M2 ] = [M−1 L3 T −2 ] Option (c) is correct. µ i ⋅i 7. F= 0 1 2 4π d [ML T −2 ][L] ∴ [ µ 0] = = [ML2 T −2 A −2 ] [A 2 ] Option (c) is correct. [L] 8. [ k] = = [T] [L T −1 ] Option (c) is correct. [ML T −2 ] 9. [ a ] = = [ML T −3 ] [T] [ b] =
[ML T −2 ] = [ML T −4 ] [ T2]
Option (c) is correct.
6 | Mechanics-1 10. E = hν
11. [Energy] = [ML2 T −2 ]
q = CV and V = iR ∴ q = iCR i t = iCR ⇒ [CR] = [ t ] = [M0 L0 T A 0 ] Option (a) is correct. 1 q1 q2 18. F = 4 πε 0 r2
= [M][L T −1 ] 2 ∴ [Mass] = [ Ev −2 ] Option (c) is correct. 1 Energy 12. ε 0 E2 = Energy density = 2 Volume 2 [ML T −2 ] 1 2 ε E = ∴ 0 3 2 [L ]
∴ Unit of ε 0 = Newton-metre2 /coulomb2 . Option (b) is correct. nh 19. Angular momentum ( J ) = 2π I = Σ mr2 h 2π J / n mvr ∴ = = 2 I Σmr Σ mr2
17.
[ML2 T −2 ] ∴ [ h] = = [ML2 T −1 ] [T −1 ] nh Angular momentum ( J ) = 2π [ J ] = [ h] = [ML2 T −1 ] Option (b) is correct.
−1 h = [L T ] = [T −1 ] I L
= [ML−1 T −2 ] Option (b) is correct. [ a ] = [T 2 ]
13.
[T 2 ] [ b] = [L ][ML−1 T −2 ] ∴
a = [M T −2 ] b
Option (b) is correct. dv 14. Velocity gradient = dx [L T −1 ] [Velocity gradient] = [L] = [T −1 ] = [M0 L0 T −1 ] Option (a) is correct. [Force] = [ML T −2 ] [F] [Mass] = [L T −2 ]
15.
∴
= [F L−1 T 2 ] Option (a) is correct. 16. Coefficient of friction (µ)
= ∴
Limitting frictional force Normal force
[µ] = [M0 L0 T 0 ]
Option (b) is correct.
= Frequency Option (a) is correct. b 20. v = at + t+c b t +
[ c] = [ T ] = [v ] c
[ b] = [L T −1 ] [T] = [L] [ at ] = [v ] = [L T −1 ] [a] = [L T −2 ] ⇒ Option (a) is correct. 2π 21. y = A sin ( c t − x) λ 2 π x 2π = A sin c t − λ λ 2πx = θ (angle) λ ∴ [ x ] = [ λ ] = [L ] Further, y = A sin θ ∴ [ A ] = [ y ] = [L ] Option (a) is correct. or
22.
[ X ] = [M−1 L−3 T 3 A 2 ] [T A 2 ] = [ML2 T −2 ]
Units & Dimensions Vectors 2
[ t] [ i ] [ Work ]
=
[QW = i2 Rt]
∴ X is resistance. →
^
F = 2 ^i − 3 ^j + 4 k
23.
→
^
or
→
24.
0.25 + 0.64 + c2 = 1 c2 = 1 − 0.89 c = 0.11 Option (b) is correct. or or
→
→
→
→
→ →
→ →
A2 + 2 A ⋅ B + B2 = C2 → →
A ⋅B = 0 →
or
cos θ = 0 π θ= 2
→
→
→
→
→ →
→
122 + 52 + 2 P⋅ Q = 132 → →
→
→ →
P⋅ Q = 0 →
→ →
→
33. Option (b) is correct.
→
→
∴
→ →
27. Work ( = F⋅ s ) is a scalar quantity.
or
Option (d) is correct.
or
→
28. Speed = |v|
or
Option (d) is correct. →
→
29. |A |= 3, |B|= 5 and angle between A and B →
→
A ⋅ B = |A ||B|cos 60°
→
→
→
→
34. P + Q + R = 0
Option (d) is correct.
is 60°.
π 2
Option (b) is correct.
|A |= |B|
→
→
∴ Angle between P and Q =
A2 − B2 = 0 A=±B
→ →
→ →
→ →
A ⋅ A + B⋅ A − B⋅ B − A ⋅ B = 0
∴
→
(P + Q) ⋅ (P + Q) = R⋅ R
26. ( A + B) ⋅ ( A − B) = 0
→
→
P2 + Q2 + 2 P⋅ Q = R2
→
→ →
→ →
|A ||B|cos θ = 0
→
→ →
∴ Angle between A and B = 90°
→ →
→ →
32. P + Q = R
A⋅ B = 0
→
→ →
A ⋅ A + 2 A ⋅ B + B⋅ B = C⋅ C
or
→
→
→ →
→
→ →
Option (d) is correct. →
A2 + B2 + 2 A ⋅ B = A2 + B2 − 2 A ⋅ B
→
→ →
31. Magnetic field intensity.
( A + B) ⋅ ( A + B) = ( A − B) ⋅ ( A − B)
i.e.,
→
Option (d) is correct.
→
→
→
or
25. |A + B|= |A − B| →
→
or
^
→
→
or
(0.5)2 + (0.8)2 + ( c)2 = 1
→
→
( A + B) ⋅ ( A + B) = C⋅ C
or
τ = 17 ^i − 6 ^j − 13 k
→
→
∴
^ ^ ^ i j k → → → τ = r × F = 3 2 3 2 −3 4 →
= 7.5 Option (b) is correct. 30. A + B = C
^
^
r =3i+2j+3k
∴
7
1 = (3) (5) 2
→
P+Q= −R →
→
→
→
→
→
(P + Q) ⋅ (P + Q) = ( − R) ⋅ ( − R) → →
→ →
→ →
→ →
P⋅ P + Q⋅ Q + 2 P⋅ Q = R⋅ R → →
P2 + Q2 + 2 P⋅ Q = R2
Let Q2 = P2 and R = P 2 Thus, Eq. (i) takes the form P2 + P2 + 2 PQ cos θ = 2 P2 or 2 PQcos θ = 0
…(i)
8 | Mechanics-1 or or
( − Q sin θ) Q sin θ − P + Q cos θ P − Q cos θ = ( − Q sin θ) Q sin θ 1− ⋅ ( P + Q cos θ) ( P − Q cos θ) 2 PQ sin θ = 2 P + Q2 cos 2θ
cos θ = 0 θ = 90° →
→
∴ Angle between P and Q is 90° →
→
→
→
→
P+Q+R=0 →
P+R= −Q
∴ →
→
→
→
→
(P + R) ⋅ (P + R ) = ( − Q) ⋅ ( − Q)
or or or or or
P2 + R2 + 2 PR cos φ = Q2 2 PR cos φ = Q2 − P2 − R2 2 PR cos φ = − R2 2P cos φ = − R 2 P cos φ = − P 2 1 cos φ = − 2
or
→
→
or
→
→
→
P − Q will vary from 0 to π. Option (b) is correct. 36. R2 = P2 + Q2 + 2 PQ cos θ
for R = P = Q P2 = P2 + P2 + 2 PP cos θ 1 or cos θ = − 2 or θ = 120° Option (b) is correct.
φ = 135°
∴
→
∴ Angle between P and R is 135°.
→ →
W = F⋅ s
37. 135° → R
= (3 ^i + 4 ^j) ⋅ (3 ^i + 4 ^j) → P 90°
= 25 J Option (b) is correct.
→ Q
135°
→ →
→
→
→ –Q
→ → +Q θ P φ
→ P
→ φ' P – Q→
tan φ = →
^
^
^
^
^
Q sin θ P + Q cos θ →
→
Angle φ′ between P − Q and P Q sin ( π + θ) tan φ′ = P + Q cos ( π + θ) − Q sin θ = P − Q cos θ tan φ − tan φ′ tan[ φ + ( −φ′ )] = 1 + tan φ tan φ′
= a2 − 2a − 3
→
35. Angle ( φ) between P + Q and P − Q → Q
^
38. P⋅ Q = ( a i + a j + 3 k) ⋅ ( a i − 2 j − k)
Option (a) is correct. →
→
This implies that angle between P + Q and
→
→ → →
For P ⊥ Q, P⋅ Q = 0 i.e., a2 − 2a − 3 = 0 or ( a − 3)( a + 1) = 0 ⇒ a=3 Other value is − ive. Option (d) is correct. 39. If a vector makes angles α, β and γ with
the co-ordinate axes, then cos2 α + cos2 β + cos2 γ = 1 2
2
3 9 6 Now, = , = 7 49 7 9 and + 49 ∴ Option (a) is correct. →
^
40. A = 4 i − 3 j
∴
→
→
B = 8 ^i + 8 $j
and →
2
36 2 4 , = 49 7 49 36 4 + =1 49 49
→
A + B = C = 12 ^i + 5 ^j
Units & Dimensions Vectors →
^
C=
∴ B=2N Option (c) is correct.
12 ^i + 5 ^j C = |C| 122 + 52
→
θ=
C = − i + 2$j + 3 k →
∴ Vectors A , B, and C will be coplanar if their scalar triple product is zero i.e., →
→
→
(A × C ) ⋅ B = 0 $ i$ $j k ^ 2 3 −2 ⋅ (5 ^i + n ^j + k) = 0 −1 2 3 or
^
^
→
→
→
→
→
→
→
→
=a ×a +b×a −a ×b−b×b →
→
→
→
=0−a × b−a × b−0 →
→
→
→
= − 2(a × b) = 2( b × a ) Option (a) is correct. 44.
→
^
→
^
A = 3 ^i + 4 ^j + 5 k
2 + 32 ⋅ 2
47. R2 = (3 P)2 + (2 P)2 + 2 × 3 P × 2 P cos θ
43. (a + b) × (a − b) →
2+3 2
Option (a) is correct.
42. Option (a) is correct. →
=
→
(13 ^i − 4 ^j + 7 k) ⋅ (5 ^i + n ^j + k) = 0
→
→
Component of A along ^i + ^j → 5 C= (2 ^i + 3 ^j ) 26 → 5 |C|= 2
or 65 − 4 n + 7 = 0 or n = 18 Option (a) is correct. →
A ⋅B →
|A ||B| 5 = 13 ⋅ 2 5 = 26
^
→ →
B = ^i + ^j → →
^
^
^
→
and
^ → ^ ^ 41. A = 2 i + 3 $j − 2 k, B = 5 i + n$j + k, →
^
46. Angle between A = 2 i + 3 j
12 ^ 5 ^ = i+ j 13 13 Option (b) is correct. →
or …(i) R2 = 13 P2 + 12 P2 cos θ Further (2R)2 = (6 P)2 + (2 P)2 + 2 × 6 P × 2 P cos θ or 4 R2 = 40 P2 + 24 P2 cos θ …(ii) Dividing Eq. (ii) by Eq. (i), 10 P2 + 6 P2 cos θ = 13 P2 + 12 P2 cos θ or 6 cos θ = − P or θ = 120° Option (b) is correct. Q sin α 48. tan θ = P + Q cos α
B = 3 ^i + 4 ^j − 5 k → →
cos θ =
( A ⋅ B) →
=0 ⇒ θ = 90° Option (c) is correct. 45.
Q = 2P
→
|A ||B| 9 + 16 − 25 = 2 3 + 42 + 52
A + B=7 A − B=3
9
R θ = 90° P
As ∴ i.e.,
θ = 90°, tan α = ∞ P + Q cos α = 0 P Q P =− 2P
cos α = −
10 | Mechanics-1 1 2 α = 120°
or
=−
∴
− P × 16 = − 96
∴ Option (a) is correct.
Option (a) is correct. → →
49. A ⋅ B = 0
→
→
→ →
51. |A × B|= 3 ( A ⋅ B)
→ A → → B ×C
→
→
→
tan θ = 3 θ = 60° →
→
→
→ → →
→
→
→
→
|A × B|= [ A2 + B2 + AB] 1/ 2
A ⋅C = 0 →
…(ii)
A⊥C
⇒
→
= A2 + B2 + 2 AB cos 60° = A2 + B2 + AB
…(i)
A⊥B
⇒
→
∴|A × B|2 = |A |2 × |B|2 + 2|A ||B|cos θ
→ C →
→
|A ||B|sin θ = 3 |A | |B|cos θ
⇒ ⇒ ⇒
→ B
P = + 6N Q = 10 N
→
→
→
52. C is perpendicular to both A and B → → → C=A×B
From Eq. (i) and Eq. (ii), we conclude that → A is perpendicular to the plane containing → → B and C. →
This implies that A is perpendicular to B ⋅ C. Option (c) is correct. or
2
2
or or
P + Q + 2 PQ cos α = 8
( P + Q)2 + 2 PQ (cos α − 1) = 64
^
2 PQ(cos α − 1) = − 192
Using Eq. (ii) and Eq. (i), P ( − P) − PQ = − 96 or
^
^
− P ( P + Q) = − 96
^
^
^
⇒
− 8 + 12 + 8α = 0
∴
α=−
Option (c) is correct. →
→
→
→
P+Q+R= 0
54.
γ
P + Q cos α = 0 Q cos α = − P
→
(A × B ) ⋅ A = 0
53. (2 i + 3 j + 8 k) ⋅ ( −4 i + 4 j + α k) = 0
(16)2 + 2 PQ (cos α − 1) = 64
PQ cos α − PQ = − 96 …(i) Q sin α (as tan θ = =∞ P + Q cos α θ = 90°)
→
Option (d) is correct.
or
∴
C⋅ A = 0 →
or
2
or P2 + Q2 + 2 PQ + 2 PQ cos α − 2 PQ = 64 or
→ →
∴
P2 + Q2 + 2 PQ cos α = R2
50.
→ B
→ A
→ →
→ Q
…(ii)
→
If,
→
→ P
β α
→
|P| |Q| |R| = = sin α sin β sin γ
→ R
1 2
Units & Dimensions Vectors →
→
→
56. Using answer to questions no. 35, as angle
OA + AB = OB
55.
→
→
→
→
between A + B and A − B is 90°
B (–2,6,4)
A2 + B2 cos 2θ = 0
Z
O
or
A2 = − B2 cos 2θ
or
π A2 = − B2 cos 2 2
or
A2 = − B2 cos π
or
A2 = B2
⇒
A=B
Y A (0, 3, –1)
X →
→
→
AB = OB − OA
∴
11
^
^
^
= ( −2 i + 6 ^j + 4 k) − (0 ^i + 3 ^j − k)
Qθ = π 2
Option (a) is correct.
^
= − 2 ^i + 3 ^j + 5 k Option (c) is correct. Match the Columns →
→
→ →
→
1. (a) |A × B|= |A ⋅ B| →
or
→
→
→
→ →
tan θ = ± 1 π 3π ⇒ θ= , 2 4 Thus, (a) → (r) (s). →
→
(b) A × B = B × A →
or →
→
→
→
→
→
→ →
→
→
→
→ →
→ →
→ →
→ →
4 A⋅ B = 0 →
→
A⊥B →
→
→
or or or
→
→
→ →
→ →
→ →
→ →
→ →
→ →
→ →
2 A ⋅ B = 0 (Q A2 + B2 = C2 ) →
→
A⊥B
Thus, (d) → (q).
Section II Subjective Questions
1. 2 × 1011 N/m 2 =
(2 × 1011)(105 dyne) (104 cm 2 )
= 2 × 1012 dyne/cm 2 (72)(10−5 N) 2. 72 dyne/cm = = 0072 . N/m (10−2 m ) 3.
[ a ] = [ y] = [ L]
→ →
A2 + 2 A ⋅ B + B2 = C2
⇒
→
→
A ⋅ A + B⋅ A + B⋅ A + B⋅ B = C⋅ C
or
|A + B|2 = |A − B|2
→
( A + B) ⋅ ( A + B) = C⋅ C
→
→
→
(d) A + B = C
(c) |A × B|= |A × B| or
→
Thus, (c) → (q).
or sin θ = − sin θ or 2 sin θ = 0 ⇒ θ = 0 rad Thus, (b) → (p). →
→ →
⇒
|A ||B|sin θ = − |A ||B|sin θ
→
→
→ →
or
A × B = − (B × A )
or
→
= A ⋅ A − B⋅ A − A ⋅ B + B⋅ B
(given) →
→
→ →
or
→
→
or A ⋅ A + B⋅ A + B⋅ A + B⋅ B
|A ||B|sin θ = ± |A ||B|cos θ
→
→
⇒ ( A + B) ⋅ ( A + B) = ( A − B) ⋅ ( A − B)
[ωt ] = [M0L0 T 0 ] ∴ [ω ] = [ T −1 ] [θ ] = [M0L0 T 0 ] E J 4. h = = = J-s ν per sec [ h] = [ML2 T –2 ][T] = [ML2 T –1 ] 5. [ b] = [ x2 ] = [ L2 ]
12 | Mechanics-1 b = [ p] at
→ → → 14. A + B = R (say) = 3 i$ + $j
L2 b ∴ [a] = = = [M–1T 2 ] 2 –3 tp TML T 1 2 1 6. St = ut + at − u( t − 1) + a( t − 1)2 2 2 1 1 2 = u.1 − a(1) + at(1) = u + at − a 2 2 a = u + (2t − 1) 2 Here t in second. Hence the given equation seems to be dimensionally incorrect. But it is correct because 1 is hidden. 7. LHS is dimensionless. While RHS has the dimensions [L−1 ]. 8. LHS is dimensionless. Hence n = 0. 9. Just write the dimension of different physical quantities. 10. E = km x n y a z . Here k = a dimensionless constant ∴ [ E ] = [ m ]x [ n] y [ a ] z 2 –2 ∴ [ML T ] = [M]x [T –1 ] y [L] z ∴ x = 1, y = 2 and z = 2 x y z
11. F = km v r
(k = a dimensionless constant) x [ F ] = [ m ] [v ] y [ r] z ∴ [MLT –2 ] = [M]x [LT –1 ] y [L] z Solving we get, x = 1, y = 2 and z = − 1 kmv2 ∴ F= r
→
→ → A − B = S (say) = i$ + 5$j →
→ →
R⋅S 3+5 8 4 cos θ = = = = RS 9 + 1 1 + 25 260 65 4 ∴ θ = cos −1 65 15. Their dot product should be zero. $ should be 16. Ratio of coefficients of $i, $j and k same. 17. No solution is required. →
–2 x
→
$ =R R R → → → $ 20. A × B = C (say) = − 3 i$ + 8 $j + 2k
→
→
→
→
Now A ⋅ C = −6 + 8 − 2 = 0 →
→
∴ A⊥C →
→
B⋅C = 0+8 −8 = 0 ∴
→
→
B⊥C →
A ⋅B 6−2+8 13. cos θ = = AB 9 + 1 + 4 4 + 4 + 16 12 3 = = 336 21
→
21. Area of parallelogram = |A × B| → R
22. → B
B sin θ ⇒ → A
A B cos θ
z
∴ [ML ] = [MLT ] [L] [T] Equating the powers we get, x = 1, y = − 4, z = 2 ∴ [ d] = [FL–4 T 2 ] Similarly other parts can be solved.
→
A ⋅B B
→ → → $ 19. A + B = R = 5$i − $j + k
θ y
→
→
18. Component of A along B = A cos θ =
12. (a) [ d] = [ F ]x [ L ] y [ T ] z –3
→
Angle between R and S
B sin θ = R =
B 2
θ = 30°
∴
→
→
∴ Angle between A and B = 180 − θ = 150° → → → 23. A + B = ( 4 $i + 6$j ) + B = 10 $i + 9 $j
∴
→
B = (6$i + 3 $j)
Units & Dimensions Vectors 24.
3P
25. → C
A
4P
→ B
B
C
P
→ A
Applying sine law, we have a b = sin (180 − A) sin (180 − B) c = sin (180 − C) a b c or = = sin A sin B sin C
2P
→
R = P $i + 2 P $j − 3 P $i − 4 P $j = ( − 2 P i$ − 2 P $j ) 26.
R2 = P2 + Q2 + 2 PQ cos θ S2 = P2 + Q2 − 2 PQ cos θ ∴ R2 + S2 = 2 ( P2 + Q2 )
13
3
Motion in One Dimension Introductory Exercise 3.1
1. Suppose
a particle is moving with constant velocity v (along the axis of x). Displacement of particle in time t1 = vt1 Displacement of particle in time t2 = vt2 ∴ Displacement of the particle in the time interval ∆t ( = t2 − t1) = vt2 − vt1 = v ( t2 − t1) ∴ Average velocity in the time interval v ( t2 − t1) ∆t ′ = ( t2 − t1) =v Now, as the particle is moving with constant velocity (i.e., with constant speed in a given direction) its velocity and speed at any instant will obviously be v. Ans. True.
2. As the stone would be free to acceleration
under earth’s gravity it acceleration will be g. 3. A second hand takes 1 min i.e., 60s to
complete one rotation (i.e., rotation by an angle of 2π rad). ∴ Angular speed of second hand 2π rad = 60 s π = rad s −1 30 Linear speed of its tip = radius × angular speed
= 2.0 cm ×
π rad s −1 30
π cms −1 15 As the tip would be moving with constant speed. π Average speed = cms −1 15 In 15 s the second hand would rotate through 90° i.e., the displacement of its tip will be r 2. ∴ Modulus of average velocity of the tip of second hand in 15 s. r 2 = 15 2 2 = cms −1 15 =
4. (a) Yes. By changing direction of motion,
there will be change in velocity and so acceleration. (b) (i) No. In curved path there will always be acceleration. (As explained in the previous answer no. 3) (ii) Yes. In projectile motion the path of the particle is a curved one while acceleration of the particle remains constant. (iii) Yes. In curved path the acceleration will always be there. Even if the path is circular with constant speed the direction of the acceleration of the particle would every time be changing.
Motion in One Dimension Circumference Speed 2π × 4 cm = = 8π = 25.13 s 1 cm / s
4v 2 2πr v 2 2 v2 = π r 2 2 (1)2 = π 4
5. (a) Time speed =
(b) As particle is moving with constant speed of 1 cm/s, its average speed in any time interval will be 1 cm/s. r 2 |Average velocity|= T/4 =
4r 2 2 2 speed = π 2πr speed
| 15
=
= 0.23 cm s −2 6. Distance = Speed × time
D1 = v1t1 D2 = v2 t2
2 2 cms −1 π = 0.9 cms −1 v 2 |Average acceleration|= T/4
Average speed =
=
=
D1 + D2 v1t1 + v2 t2 = t1 + t2 t1 + t2 ( 4 × 2) + (6 × 3) 2+3
= 5.2 ms −1
(where v = speed)
Introductory Exercise 3.2 1. Acceleration (due to gravity).
1 1 = ut + at2 − u ( t − 1) + a( t − 1)2 2 2 st = u + at –
1 a 2
Displacement Velocity
Therefore, the given dimensionally incorrect.
Acceleration
equation
v = t 3/ 4 ds = t 3/ 4 dt
4.
1 2. st = u + at − a is physically correct as it 2 gives the displacement of the particle in tth second (or any time unit). st = Displacement in t seconds − displacement in ( t − 1) seconds
∴
s=
∫t
(given) …(i) 3/ 4
dt =
3 +1 t4
3 +1 4
+c
4 7 /4 t +c 7 i.e., s ∝ t7 /4 Differentiating Eq. (i) w.r.t. time t, 3 d2 s 3 4 − 1 = t dt2 4 ⇒ a ∝ t −1/ 4 or
s=
5. Displacement (s) of the particle
is
3. Yes. When a particle executing simple
harmonic motion returns from maximum amplitude position to its mean position the value of its acceleration decreases while speed increases.
s = ( 40 × 6) +
1 ( − 10) 62 2
= 240 − 180 = 60 m (in the upward direction) Distance covered ( D) by the particle Time to attain maximum height
16 | Mechanics-1 40 = 4s < 6 s 10 It implies that particle has come back after attaining maximum height (h) given by u2 h= 2g = ∴
( 40)2 = 80 m 2 × 10
D = 80 + (80 − 60) = 100 m
v = 40 − 10t dx ∴ = 40 − 10t dt or dx = ( 40 − 10t) dt or x = ∫ ( 40 − 10t) dt or x = 40t − 5t2 + c As at t = 0 the value of x is zero. c=0 ∴ x = 40t − 5t2 For x to be 60 m. 60 = 40t − 5t2 or t2 − 8 t + 12 = 0 ∴ t = 2 s or 6 s Displacement in time t 7. Average velocity = t 1 2 ut + at 2 = t 1 = u + at 2 6.
8.
∴
v2 = v1 + at at = v2 − v1 Displacement in time t t 1 2 v1t + at 2 = t 1 = v1 + at 2 v2 − v1 = v1 + 2 v1 + v2 = 2
Ans. True.
∴
=
125 = 0 ⋅ t +
9.
1 2 gt 2
t = 25 s 125 m Average velocity = 5s
(downwards)
= 25 m/s
(downwards)
⇒
2
…(i
v = 10 + 5t − t
10.
) dv = 5 − 2t dt At t =2s a=5−2×2 = 1 m/s2 From Eq. (i), dx = 10 + 5t − t2 dt ∴ x = ∫ (10 + 5t − t2 ) dt 5t2 t 3 or x = 10t + − +c 2 3 As, at t = 0 the value of x is zero c=0 5 t3 ∴ x = 10t + t2 − 2 3 Thus, at t = 3 s 5 33 x = (10 × 3) + (3)2 − 2 3 = 30 + 22.5 − 9 = 43.5 m a=
∴
→
^
11. u = 2 i m/s →
a = (2 cos 60° ^i + 2 sin 60° ^j ) m/s2 = (1 ^i + 3 ^j) m/s2 y
Average velocity =
a = 2 m/s2 60° u = 2 m/s →
→
→
v=u+at = 2 ^i + (1 ^i + 3 ^j) 2 = 4 ^i + 2 3 ^j
x
Motion in One Dimension →
Part II Yes. As explained below.
|v|= 42 + 12 = 2 7 m/s → → 1→ s = u t + a t2 2 1 ^ = (2 i) 2 + (1 ^i + 3 ^j) 22 2
→
v = 2 ^i + 2t ^j implies that initial velocity of
the particle is 2 ^i m/s2 and the acceleration is 2$j m/s2 → 1 ∴ s (at t = 1 s) = (2 ^i × 1) + (2 ^j) 12 2
= 4 ^i + 2 ^i + 2 3 ^j
= (2 ^i + ^j) m
= 6 ^i + 2 3 ^j
13. x = 2t and y = t2
→
|s |= 36 + 12
∴
→
v = (2 ^i + 2t ^j)m/s
…(i)
→
dv = 2 ^j dt →
a = 2 ^j m/s2
From Eq. (i), →
ds = (2 ^i + 2t ^j ) dt s = ∫ (2 ^i + 2t ^j ) dt
2
or, x2 = 4 y (The above is the equation to trajectory) x = 2t → dx ∴ = 2 i.e., v x = 2 ^i dt y = t2 → dy ∴ = 2t i.e., v y + 2t ^j dt →
Thus,
→
∴
x y = 2
∴
=4 3m 12. Part I
| 17
→
→
v = vx + v y = (2 ^i + 2t ^j) m/s
→
s = 2t ^i + t2 ^j + c
→
Taking initial displacement to be zero. →
s (at t = 1 s) = (2 ^i + ^j) m
→
a=
dv = 2 ^j m/s2 dt
Introductory Exercise 3.3 1. At t = t1 v
s
α
t2 β
β
t1 0
θ t1
t2
φ
v = tan θ As θ < 90°, vt1 is + ive. At t = t2 vt 2 = tan φ As φ > 90°, vt 2 is − ive. Corresponding v-t graph will be
t
t
Acceleration at t = t1 : at1 = tan α As α < 90°, a t1 is + ive constant. Acceleration at t = t2 at 2 = tan β As β < 90°, at 2 is + ive constant. 2. Let the particle strike ground at time t
velocity of particle when it touches ground
18 | Mechanics-1 1 mg2 t2 2 i.e., KE ∝ t2 . While going up the velocity will get − ive but the KE will remain. KE will reduce to zero at time 2 t when the particle reaches its initial position.
Corresponding velocity-time will be
would be gt. KE of particle will be
Velocity (m/s)
4 KE
4.
t =
KE =
2t
2h g
time
8
t (s)
h 2 = tan θ = ( t − 2) (2 − 1) h = 2 ( t − 2)
⇒ 2
a (m/s )
1 1 2h mg2 t2 = mg2 2 2 g
= mgh
2
(t–2)
θ
3. Speed of ball (just before making first
collision with floor) = 2gh = 2 × 10 × 80 = 40 m/s Time taken to reach ground 2 × 80 2h = = =4s g 10 Speed of ball (just after first collision with floor) 40 = = 20 m/s 2 Time to attain maximum height −20 t= = 2s −10 ∴ Time for the return journey to floor = 2 s. Speed (m/s)
1
2 θ
t h
Particle will attain its initial velocity i.e., net increase in velocity of the particle will be zero when, area under a-t graph = 0 (1 + 2) × 2 ( − h) ( t − 2) + =0 2 2 or 3 − ( t − 2)2 = 0 or ( t − 2)2 = 3 or t−2= ± 3 or t=2± 3 Ans . At time t = 2 + 3 s (t = 2 − 3 not possible).
4
8
t (s)
Introductory Exercise 3.4 1. Relative acceleration of A w.r.t. B
a AB = ( + g) − ( + g) = 0 2. Velocity of A w.r.t. B = v A − vB
∴ Relative displacement (i.e., distance between A and B) would be 1 s = (v A − vB ) t + a AB t2 2
Motion in One Dimension or
sin 30° sin θ sin (180° − 30° + θ) = = 150 t 20t 500 × 103 1 sin θ = 300 20 1 θ = sin −1 15 15 sin (30° + θ) 1 Now = 3 300 t 500 × 10 224 5000 or = sin 30° cos θ + cos 30° sin θ 3t
s = (v A − vB ) t tan θ = (vA–vB)
s
θ
θ
0
t
3. In figure, u = speed of boat
v = speed of river flow B
C Actual path of boat
Boat sailing direction
or
u d = 400 m
⇒
A v
Time to cross river AB BC AC = = = 2 u v u + v2 400 m = 10 m/s = 40 s v BC = AB u 2 m/s = × ( 400 m ) 10 m/s = 80 m 4. Let C be the point along which pilot
should head the plane. B
30°
5 00
km
North
θ
Drifting due to wind
Speed = 150 m/s
A Wind/speed = 20 m/s
Apply sine formula in ∆ ABC
| 19
East
1
5000 1 224 3 1 = + ⋅ 3t 2 15 2 15 5000 = 0.5577 3t 5000 t= 3 × 0.5577 = 2989 s = 50 min
5.
10 m A
B
a A = 1 m/s2 , aB = 2 m/s2 v A = 3 m/s, vB = 1 m/s Acceleration of A w.r.t. B = 1 − 2 = − 1 m/s2 Velocity of A w.r.t. B = 3 − 1 = 2 m/s Initial displacement of A w.r.t. B = − 10 m At time relative displacement of A w.r.t. B 1 s = − 10 + 2 t + ( −1) t2 2 or s = − 10 + 2t − 0.5t2 For s to be minimum ds =0 dt or 2 − (0.5 × 2t) = 0 i.e., t = 2s ∴ smin = − 10 + (2 × 2) − 0.5 × (2)2 = − 10 + 4 − 2 = −8m Minimum distance between A and B = 8 m.
20 | Mechanics-1 AIEEE Corner Subjective Questions (Level 1) 1. (a) D = v1t1 + v2 t2
km km 1 = 60 × 1 h + 80 × h h h 2 = 100 km (b) Average speed 100 km = = 66.67 km/h. 1.5 h 2. (a) Displacement in first two seconds
= ( 4) (2) +
1 (6) 22 2
= 20 m
20 m ∴ Average velocity = = 10 m/s 2s (b) Displacement in first four seconds 1 = ( 4) ( 4) + (6) 42 = 64 m 2 ∴ Displacement in the time interval t = 2 s to t = 4 s = 64 − 20 = 44 m 44 m ∴ Average velocity = = 22 m/s. 2s
(b) Average velocity = =
Net displacement Time taken
− 60 m 6.1 s
= − 9.8 m/s (downwards) = 9.8 m/s Σ vt 4. Average velocity = Σt vt0 + 2 vt0 + 3vT 2.5v = t0 + t0 + T ∴ or
5t0 + 2.5T = 3 t0 + 3 T T = 4 t0
5. (a) Average acceleration
Final velocity − initial velocity ( 4 + 8) s 0−0 = 12 = 0 m/s2 v (c) a = tan θ = max 4 =
v vmax
3. Let the particle takes t time to reach
ground. +20 m/s
θ
h
4s
– 60 m
∴
− 64 = 20 t +
5t2 − 20t − 64 = 0 t = 6.1 s If the particle goes h meter above tower before coming down 0 = (20)2 + 2 ( −10) h ⇒ h = 20 m Total distance moved (a) Average speed = Time taken (20 + 20 + 60) = = 16.4 m/s 6.1 i.e.,
12s time
vmax (Q a = 4 m/s2 ) 4 i.e., vmax = 16 m/s Displacement of particle in 12 seconds = Area under v-t graph v = 12 × max 2 12 × 16 = 2 = 96 m Average velocity Displacement (96 m) at 12 s = Time (12 s)
∴ 1 ( − 10) t2 2
8s
4=
= + 8 m/s
Motion in One Dimension (b) As the particle did not return back distance travelled in 12 s = Displacement at 12 s ∴ Average speed = 8 m/s. 21 6. (a) Radius ( R) of circle = m 22 ∴ Circumference of circle = 2πR 22 21 m =2× × 7 22 = 6m Speed (v) of particle = 1 m/s ∴ Distance moved by particle in 2 s = 2 m Thus, angle through which the particle moved 2 2π = × 2π = = 120° 6 3 Magnitude of Average velocity Magnitude of displacement = Time ( = 2 s)
7. Position vector at t = 0 s →
r1 = (1 ^i + 2 ^j) m Position vector at t = 4 s →
r2 = (6 ^i + 4 ^j) m (a) Displacement from t = 0 s to t = 4 s →
= (6 ^i + 4 ^j) − (1 ^i + 2 ^j) = (5 ^i + 2 ^j) m ∴ Average velocity =
(b) Average acceleration Final velocity − Initial velocity = 4s
120° O
AB 2R sin 60° = 2 2 3 =R 2 21 3 21 3 m/s = = 22 2 44 (b) Magnitude of average acceleration → → vB − v A = 2s 120° 2 v sin 2 = 2 =
[Q
→
→
|v A |= |v B |= v = 1 m/s) = v sin 60° 3 m/s2 = 2
(5 ^i + 2 ^j) m 4s
= (125 . ^i + 05 . ^j) m/s
=
(2 ^i + 10 ^j) − ( 4 ^i + 6 ^j) 4
=
−2 ^i + 4 ^j 4
A(t = 0 s)
(t = 2s) → vB
→
= ( r 2 − r1 )
→ vA
B
| 21
= ( −05 . ^i + ^j) m/s2 (c) We cannot find the average speed as the actual path followed by the particle is not known. Uniform acceleration (a) One dimensional motion 8. If at time t the vertical displacement
between A and B is 10 m 1 2 1 gt − g( t − 1)2 = 10 2 2 or or or
t2 − ( t − 1)2 = 2 t2 − ( t2 − 2t + 1) = 2 2t = 3 t = 1.5 s
9. The two bodies will meet if
Displacement of Displacement of first after attaining = second before attaining highest highest point point
22 | Mechanics-1 v0 t − or or
or
10.
⇒ T O
1 2 1 gt = v0 ( t − t0 ) − g( t − t0 )2 2 2 1 2 0 = − v0 t0 − g( t0 − 2t0 t) 2 1 2 gt0 t = gt0 + v0 t0 2 1 gt0 + v0 t= 2 g t0 v0 = + 2 g 1 5 = gt2 2 t = 1s 5m A 25 m
H W
∴ or or t=
2t0 ± ( −2t0 )2 − 41( − t20 )
2 2t0 ± 2t0 2 = 2 = t0 + t0 2 (− absurd) = (1 + 2) t0 = 2141 . t0
Time interval 1s
ive
t second for A
12. (t–1) second for B
⇒
v=0 A
B
5 m/s H
For A H = 0. t + H=
1 2 gt 2
1 2 gt 2
15m
…(i)
1 …(ii) g( t − 1)2 2 1 2 1 or gt − g( t − 1)2 = 25 2 2 [Substituting value of H from Eq. (i)] 1 g[ t2 − ( t − 1)2 ] = 25 2 t2 − ( t2 − 2t + 1) = 5 2t − 1 = 5 ⇒ t=3s Substituting t = 3 s in Eq. (i) 1 H = × 10 × 32 = 45 m 2 1 2 11. s = at0 Forward motion 2 H − 25 =
Backward motion
u
(a) For H 02 = u2 + 2( −10) H i.e., 20H = 325 or H = 1625 . m
For B
v = at0
being
52 = u2 + 2 ( −10)15 u2 = 325
R
or
sign
From the begining of the motion the point mass will return to the initial position after time 3.141t0 .
B u=0
E
1 ( − a) t2 2 1 1 − at20 = ( at0 ) t − at2 2 2 2 2 − t0 = 2t0 t − t 2 t − 2t0 t − t20 = 0 − s = ( at0 ) t +
(b) For t 0 = 325 + ( −10) t 325 t= 10 = 1. 8 s
∴
13.
At rest
a x
u
a
15 m/s
60 m 6.0 s
(a) …(i) 152 = u2 + 2a × 60 and …(ii) 15 = u + a × 6 Substituting the value of 6a from Eq. (ii) in Eq. (i) 225 = u2 + 20 (15 − u)
Motion in One Dimension 2
i.e.,
u − 20 u + 75 = 0 (u − 15)(u − 5) = 0 ∴ u = 5 m/s (15 m/s being not possible) (b) Using Eq. (ii) 5 a = m / s2 3 (c) u2 = 02 + 2ax u2 (5)2 i.e., x= = 2a 2 × 5 3 = 7.5 m 1 (d) s = at2 2 1 5 = × × t2 2 3 5 = t2 6 t(s )
0
1
3
v (m/s)
0
5/6
7.5
6
Starts A
…(iii) 9
12
30 15 t (s)
12
Differentiating Eq. (iii) w.r.t. time t 5 v= t 3 t(s )
0
3
6
9
12
v (m/s)
0
5
10
15
20
v (m/s) 20 15 10 5 6
9
12
Stops B B
s2
Journey A to P vmax = 0 + xt1 and v2max = 2xs1 v ⇒ t1 = max x
60
3
t2
4 min
30 67.5 120
9
t1 4 km
90
6
a = –y
s1
120 s (m)
3
a=+x
14.
| 23
t (s)
…(i) …(ii)
Journey P to B 0 = vmax + ( − y) t2 2 and vmax = 2 ys2 v ⇒ t2 = max y vmax vmax ∴ + = t1 + t2 = 4 x y 1 1 vmax + = 4 x y From Eq. (ii) and Eq. (iv) v2 v2 s1 + s2 = max + max 2x 2y v2max 1 1 or 4= + 2 x y or
…(iii) …(iv)
…(v)
…(vi)
Dividing Eq. (vi) by Eq. (v) vmax = 2 Substituting the value of vmax in Eq. (v) 1 1 (Proved) + =2 x y 15. Let acceleration of the particle be a using 2m 4 0 t = 10 s t = 6 s v=0
t=0s
+ive x-axis
v = u + at 0 = u + a6 u ∴ a=− 6 (a) At t = 10 s, s = − 2 m 1 −2 = u × 10 + a × 102 2 or −2 = ( −6a) 10 + 50a or −10a = − 2 or a = 0.2 m/s 2
24 | Mechanics-1 →
(b) v(at t = 10 s) = u + a 10 = − 6a + 10a = 4a = 08 . m/s
→
u = 0 m/s
→
a 2 = − 4 ^j m/s2 (t = 2 s to t = 4 s) t2 = 2 s (a) Velocity →
→
→
→
a=
16.
→
= 0 + (2 ^i) 2
F 10 N north = m 2 kg
= 4 ^i
2
= 5 m/s , north ^
= 5 j m/s
→
→
→
v2 = (8 ^i − 8 ^j ) m/s
or
^
= 10 i m/s
(b) Co-ordinate of particle → → → 1 s 1 = s 0 + u t1 + → a 1 t12 2
→
using v = u + a t →
v = 10 ^i + (5 ^j × 2) ^
→
= 4 ^i + (2 ^i − 4 ^j ) 2
u = 10 m/s, east
→
→
v2 = v1 + (a 1 + a 2 )t2
2
→
→
→
v1 = u + a 1 t1
(c) Two or three dimensional motion
= (2 ^i + 4 ^j) + (0)(2) +
^
= 10 i + 10 j
1 ^ 2 (2 i) 2 2
= 2 ^i + 4 ^j + 4 ^i
→
|v|= 10 2 m/s
= 6 ^i + 4 ^j
→
v = 10 2, north-east → → 1→ using s = u t + a t2 2 1 ^ = (10 i × 2) + (5 ^j)22 2
1 ( a1 + a2 ) t22 2 1 = 6 ^i + 4 ^j + ( 4 ^i) 2 + (2 ^i − 4 ^j) 22 2
→
→
→
s 2 = s 1 + v1 t2 +
= 6 ^i + 4 ^j + 8 ^i + 4 ^i − 8 ^j
= (20 ^i + 10 ^j) m
= 18 ^i − 4 ^j
→
|s|= 202 + 102
Co-ordinate of the particle [18 m , − 4 m ]
= 10 5 m 20 cotθ = =2 10 θ = cot−1 2
→
^
→
^
→
18. u = (2 i − 4 j) m/s, s 0 = 0 m →
a = ( 4 ^i + ^j) m/s2
North
(a) Velocity →
θ East
→
s = 10 5 m at cot−1(2) from east to north.
→
^
^
17. s 0 = (2 i + 4 j) m →
→
= (2 ^i − 4 ^j) + ( 4 ^i + ^j)2
^j O ^ i
→
v=u+at
→ s
a 1 = 2 ^i m/s2 (t = 0 s to t = 2 s) t1 = 2 s
= (10 ^i − 2 ^j ) m/s (b) Co-ordinates of the particle → → → 1→ s = s 0 + u t + a t2 2 → 1 ^ = 0 + (2 i − 4 ^j) + ( 4 ^i + $j )22 2
Motion in One Dimension = 10 ^i − 2 ^j
=
∴ Co-ordinates of particle would be [10 m , − 2 m ] u = 8 ^j m/s
→
|a av|= 52 + 202
→
a = ( 4 ^i + 2 ^j) m/s2
→
= 206 . m/s 20 tan θ = 5 i.e., θ = tan1( 4) Non uniform acceleration.
→
s0 = 0 →
s = 29 ^i + n ^j
→
→
→
(a) s = s 0 + u t +
1→ 2 at 2
21. x = 2 + t2 + 2t 3
1 ^ ( 4 i + 2 ^j) t2 2
→
29 ^i + n ^j = 0 + (8 ^j) t +
(a) At t = 0, x = 2 m (b)
29 ^i + n ^j = 8 ^j t + 2 ^i t2 + ^j t2 Comparing the coefficients of ^i and ^j 29 = 2t2 n = 8 t + t2 29 t= = 3.807 s 2
and ⇒
Substituting value of t in Eq. (ii) n = 8 × 3.807 + (3.807)2 = 44.95 29 (b) Speed at t = s 2 →
→
…(i) …(ii)
d2 x = 2 + 12t dt2 ( a) t = 2 s = 2 + (12 × 2) = 26 m/s2 dv 22. a= −v dx = − (10) × 3 = − 30 m/s2 23. s = t 3 − 9 t2 − 15t
ds = 3 t2 − 18 t − 15 dt dv ie, a= = 6t − 18 dt Acceleration (a) in the interval 0 ≤ t ≤ 10 s will be maximum at t = 10 s a(at t = 10 s) = (6 × 10) − 18 = 42 m/s2
∴
→
= 8 ^j + ( 4 ^i + 2 ^j)(3.807) = (2 × 3.807) ^i + {( 4 × 3.807) + 8 } ^j = (7.614 ^i + 23.228 ^j) m/s →
speed = |v|= (7.614)2 + (23.228)2 = 24.44 m / s 20. s 0 = 5 i m →
→
Average velocity v av ^
=
^
→
→
s − s0 = 002 . ^
(51 . i + 0.4 j) − (5 i) 002 .
v=
a = 3 − 2t dv = 3 − 2t dt
24.
^
at t = 002 . s, s = (51 . ^i + 0.4 ^j) m
dx = 2t + 6t2 dt dx = 0 m/s dt t = 0 s
(c)
v=u+at
→
01 . ^i + 0.4 ^j 002 .
= 5 ^i + 20 ^j m/s
→
19.
| 25
or
∫ dv = ∫ (3 − 2t) dt
or
v = 3 t − t2 + c
or
v = 3 t − t2 + v0 [as at t = 0, v = v0 ]
or or
(3 t − t2 + v0 ) dt 3 t2 t 3 s= − + v0 t 2 3
∫ ds = ∫
26 | Mechanics-1 (a) Displacement at = Displacement at t = 0s t = 5s 3.52 53 0= − + v0 5 2 3 s ⇒ v0 = 5/6 = 0833 . (b) v = 3 t − t2 + v0 Velocity at t ( = 500 . s) = 3.5 − 52 + 0833 . = 15 − 25 + 833 = − 9.167 m/s 25. v = 3 t2 − 6t 2
∫ ds = ∫3(3 t
− 6t) dt s = t − 3 t2 + c s = t 3 − 3 t2 = t2 ( t − 3) (3.5)2 (3.5 − 3) (a) Average velocity = 3.5 = 3.5 × 05 . = 175 . m/s (b) Distance covered
∴ or or
Now, at t = 2 s, v = 6 m/s 62 ∴ = ( 4 × 2) + C 2 Thus, C = 10. v2 i.e., …(ii) = 4 t + 10 2 ∴ at t = 3 s, v2 = ( 4 × 3) + 10 2 v2 = 44 v = 44 = 2 11 m/s Substituting above found value of v in Eq. (i), 2 11 a = 4 2 i.e., a= 11 = 0.603 m/s 2 27. According to question, the velocity of the
particle varies as shown in figure. v 20 m/s
v
O
1 2
3 3.5
t
–3
=
2
2
3.5
∫ 0 (6t − 3 t ) dt + ∫ 2
(3 t2 − 6r) dt
= [3 t2 − t 3 ]20 + [ t 3 − 3 t2 ] 23.5 = 3(2)2 − (2) 3 + (3.5) 3 − 3(3.5)2 − (2) 3 + 3(2)2 = 12 − 8 + 42875 . − 3675 . − 8 + 12 = 14.125 m 14.125 (c) Average speed = 3.5 = 4.036 m/s 4 26. v= a i.e., …(i) va = 4 or ∫ v dv2 = ∫ 4 dt v or = 4t + C 2
30 m/s s
v s + =1 20 30 or 3v + 2s = 60 or …(i) 3v = 60 − 2s Differentiating above equation w.r.t. time t dv ds 3⋅ = − 2⋅ dt dt or 3a = − 2v 60 − 2 s or 3 a = − 2 ⋅ [using Eq. (i)] 3 2 or a = − (60 − 2 s) 9 2 [at s = 15 m] = − (60 − 2 × 15) 9 20 m/s2 =− 3
∴
Motion in One Dimension 1 S1 = gt 2 2 1 h = g 2 g h = 2 ∴Assertion is true. Reason is also true as assertion is based on this.
| 39
12. Please see answer to question no. 2 of objective questions (level 1). um T2 = ( mg + f )( mg − f ) Now, if masses of the bodies are different, the value of T2 (time to reach earth) will be different if f (air resistance) is same. Thus, Assertion is false further Reason is correct as explained above.
Objective Questions (Level 2) Single Correct Option 1. Applying sine formula in ∆ AOB
∠ B′ AO = 180° − (90° + α ) − (θ − 37° )
A
= 90° − α − θ + 37° = 127° − (α + Q )
v→ Nm
→ vr
Comparing Eqs. (i) and (ii) sin (α − 37° ) sin [127° − (α + θ )] = cos 37° sin (θ − 37° ) sin α cos 37° − cos α sin 37° cos 37° sin 127° cos (α + θ ) + cos 127° sin (α + θ ) = sin θ cos 37° − cos θ sin 37°
37°
α
B –D
m
→
O
vm
→
37°
AO OB = sin (96° + 37° ) sin (α − 37° ) vr vm = cos 37° sin (α − 37° )
…(i)
Applying sine formula in A' → vr
O
→
vm
θ
37°
B' 37°
∆ AOB′ AO OB′ = sin (θ − 37° ) sin [(α + θ )] vr vm …(ii) = sin (θ − 37° ) sin [127° − (α + θ )]
sin α − cos α tan 37° 4 3 cos (α + θ ) + sin (α + θ ) 5 5 = 4 3 sin θ − cos θ 5 5 4 cos (α + θ ) + 3 sin (α + θ ) 3 or sin α − cos α = 4 sin θ − 3 cos θ 4 3 or sin α − cos α 4 4 [cos α cos θ − sin α sin θ ] + 3 [sin α cos θ + cos α sin θ ] = 4 sin θ − 3 cos θ 3 or sin α − cos α 4 4 [cos α − sin α tan θ ] + 3 [sin α + cos α tan θ ] = 4 tan θ − 3 3 or sin α − cos α 4 7 7 4 [cos α − sin α ] + 3 [sin α + cos α ] 8 8 = 7 4⋅ − 3 8 3 or sin α − cos α 4 4 [ 8 cos α − 7 sin α ] + 3 [ 8 sin α + 7 cos α ] = 4
40 | Mechanics-1 or 4 sin α − 3 cos α = 32 cos α − 28 sin α + 24 sin α + 21 cos α √50 7 α 1
or 32 cos α + 21 cos α + 3 cos α = 28 sin α − 24 sin α + 4 sin α or 56 cos α = 8 sin α 7 tan α = 1 From Eq. (i) vr vm = cos 37° sin (α − 37° ) vm cos 37° vr = ∴ sin α cos 37° − cos α sin 37° vm cos 37° = cos α cos 37° [tan α − tan 37° ] vm = 3 cos α 7 − 4 4 vm = ⋅ 25 cos α 4 5 = × × 50 25 1 = 32 Option (b) is correct. dv 2. = − 4v + 8 dt a = − 4v + 8 At the time the body acquires terminal speed its acceleration (a) must be zero. Thus − 4v + 8 = 0 v = 2 m/s Terminal speed = 2 m/s. Option (b) is correct. 3. Particles will collide if → v1 → |v1 →
→
^
− −
→ → v2 r = 2 → → v 2| |r2 ^
− −
^
→ r1 → r1| ^
→
→
v1 − v 2
∴
→ |v1
−
→ v 2|
→ → → ^ r2 − r1 = r2 = 30 i → → |r2 − r1| = 30 → → r2 − r1 ^ =i ∴ → → |r2 − r1|
^
^
→
→
|v1 − v 2| = 5 2
→
→
→
→
v1 − v 2
≠
|v1 − v 2|
∴Option (c) is correct. 4. Displacement along y-axis at time t Y v0
P a = 2t u=0
X
a =2t dvy =2t dt or ∫ dvy =2 ∫ 2t dt or vy = t + C1 At t = 0, vy = 0 (given) ∴C1 = 0 vy = t 2 ∴ dy = t2 ⇒ dt 2 or ∫ dy = ∫ t + C2
t3 + C2 2 At t = 0, Y = 0 (given) t3 Y= ∴ 3 Displacement along x-axis at time t : or
or
y=
x = v0t x t= v0
Substituting above value of t in Eq. (i) ^
^
Y=
v1 − v 2 = (5 i + 10 j + 5 k ) − (10 i + 5 j + 5 k ) = −5i + 5 j
1 ^ 1 ^ i+ j 2 2
=−
1 x 3 v0
3
Option (a) is correct. 5. t = αx 2 + βx Differentiating w.r.t. time t
…(i)
Motion in One Dimension
or or ∴
dx dx 1 = α ⋅ 2x + β⋅ dt dt v (β + 2α ⋅ x ) = 1 v = (β + 2α ⋅ x )−1 dv 1 dx =− ⋅ 2α ⋅ dt dt (β + 2α x )2 2α a=− ⋅v (β + 2α x )2 2α =− ⋅v (β + 2α x )2
7. Force = − kx
∴ Acceleration = − ⇒ or
v
∴ Thus
6. f = a − bx dv = a − bx dx or ∫ v dv2 = ∫ (a − bx )2dx v bx or = a⋅x − +C 2 2 As at x = 0 car is at rest C = 0 v2 bx 2 = ax − ∴ 2 2 For v to be maximum. d bx 2 ax − =0 dx 2
…(i)
a in Eq. (i), b 2 vmax a b a 2 = a ⋅ − 2 b 2 b 2 a = 2b a vmax = b
Car will come to rest when bx 2 ax − =0 2 2a x= b ∴Distance between two stations = Option (a) is correct.
k a3 m 3
v2 kx 2 ka 3 =− + 2 m ×3 m ×3 2 ka 3 3m
Option (d) is correct. 8. vx = 4 + 4t
Substituting x =
∴
C=
v=
a − bx = 0 a x= b
or
v2 k x3 =− +C 2 m 3
∴Velocity at x = 0
v
i.e.,
kx 2 dx m
Now at x = a, v = 0
= − (2α v2 ) ⋅ v = − 2α v3 ∴Retardation = 2α v3 Option (a) is correct. ∴
k2 2 x 2m
dv kx 2 =− dx m
∫ v dv = ∫ −
or
| 41
2
2a b
and
vy = 4 t
dx dy = 4 + 4t and = 4t dt dt or ∫ dx = ∫ (4 + 4t ) dt
∴
and ∫ dy = ∫ 4t dt 2 or x = 4t + 2t + C1 C1 = 1 ∴ Thus …(i) x = 4t + 2t 2 + 1 and y = 2t 2 + C2 At t =0x =1 and at t = 0, y = 2 C2 = 0 ∴ Thus …(ii) y = 2t 2 + 2 Substituting value of t in Eq. (i) in terms of y using Eq. (ii) we won’t get relationship as mentioned in option (a), (b) or (c). ∴Option (d) is correct. 1 9. a = gt 2 2 = 4 × tan 120° = 4 × [ − tan 60° ] = − 4 3 m/s 2 ∴Magnitude of acceleration = 4 3 ms −2. Option (a) is correct.
42 | Mechanics-1 10.
1 2 gt 2 1 x1 + x2 = g (2t )2 2 1 1 x2 = g ⋅ 4t 2 − gt 2 ∴ 2 2 3 2 x2 = gt 2 x2 − x1 = gt 2 x2 − x1 ⇒ t= g
or
14. Distance =
1 αβ 2 t 2 α + β
0=
1 1 × 4 2 t 2 1 + 4
∴ ⇒
t = 500 s
= 22.36 s Option (a) is correct. 15. Let BC = t′
y2 + x 2 = l2 dy dx 2y⋅ + 2x ⋅ =0 dt dt dy x dx =− dt y dt
∴
v (m/s) 24
x dx x tan 30° dt = − 2 3 m/s =−
A
O
12. Maximum separation (xmax ) between the police and the thief will be at time t (shown in figure) Police
108 km/h 90 km/h
v Thief
t 7s
time (t)
⇒ ∴ Let OA = t AB = 56 − t ∴ 1 1032 = [56 + (56 − t )] × 24 2 t = 20 s ⇒ 24 m / s ∴maximum acceleration = 20 s = 1.2 m/s 2
−1
16. From ∆ OAB P
(4 – )m 2t 0.
13. If meeting time is t 2v sin θ = u sin 30° 3 3 i.e., sin θ = 4
Option (b) is correct.
2t 0.
v 90 kmh = a 5 ms −2 90 1000 s = × 5 3600 i.e., t =7s 1 –1 xmax = [90 kmh × 7 s ] − × 5 s × 90 kmh –1 2 90 × 1000 9 = × m 3600 2 = 112.5 m Option (a) is correct. t −2 =
C t (s)
B
24 =4 t′ t′ = 6 s OB = 50 s
∴
Option (c) is correct.
2s
3 4
Option (c) is correct.
Option (a) is correct. 11.
θ = sin −1
or
x1 =
– 0.2t Q
cos AOB =
60° B
(3 – 0.2t) m
O
OA 2 + OB2 − AB2 2 ⋅ OA ⋅ OB
Motion in One Dimension 2
2
2
OA + OB − AB = OA ⋅ OB AB2 = OA 2 + OB2 − OA ⋅ OB = (4 − 0.2t )2 + (3 − 0.2t )2 + (4 − 0.2t )(3 − 0.2t ) 2 2 or AB = 16 + 0.04t − 1.6t + 9 + 0.04t 2 − 1.2t + 12 − 1.4t + 0.04t 2 2 2 or AB = 0.12t − 4.2t + 37 For AB to be minimum 0.24t − 4.2 = 0 or t = 17.5 s 2 ∴ ( AB)min = 0.12(17.5)2 − 4.2 × (17.5) + 37 = 36.75 − 73.5 + 37 ( AB)2min = 0.75 m 2 = 7500 cm 2 = 50 3 cm Option (d) is correct. ∴ i.e.,
17. Velocity of ball w.r.t. elevator = 15 m/s Acceleration of ball w.r.t. elevator Max. height attained by ball
+
H 15 m/s = vBE
5 m/s2
2m
10 m/s
2
Now v = u + 2as 02 = (25)2 + 2( − 10) H ∴ or H = 31.25 m Maximum height by ball as measured from ground = 31.25 + 2 + 50 = 83.25 m Option (c) is correct. 19. Displacement of ball w.r.t. ground during its flight = H = 31.25 m
Option (d) is correct. 20. Displacement of ball w.r.t. floor of elevator at time t 1 s = 15 t + ( −15) t 2 + 2 2 s will be maximum, when ds =0 dt i.e., 15 − 15t = 0 i.e., t = 1s 1 ∴smax = (15 × 1) + ( − 10)(1)2 + 2 2 = 12 m Option (a) is correct. 21. Let the particles meet at time t i.e., displacement of the particles are equal and which is possible when vB
B'
v 50 m
A uA θ C P
P'
θ
B uB
= ( − 10) − ( + 5) = − 15 m/s 2 Final displacement of ball w.r.t. elevator = −2m 1 − 2 = 15 t + ( − 15) t 2 ∴ 2 i.e., 15t 2 − 30t − 4 = 0 30 + 900 + 4 × 4 × 15 t= ∴ 2 × 15 = 2.13 s Option (a) is correct. 18. Velocity of ball w.r.t. ground (vBE ) = (15 + 10) m/s = 25 m/s
| 43
2
vA O
4
A' R Q time (s)
Area of ∆ ACB = Area of ∆ A′ B′ C [Area OBCA′ θO being common] or Area of ∆ APC = Area A′ P′ C 1 1 AP × PC = A′ P′ × P′ C 2 2 or AP × PC = A′ P′ × P′ C or ( PC tan θ ) PC = ( P′ C tan θ ) P′ C or PC = P′ C′
44 | Mechanics-1 23. u A = 6 ms −1, u B = 12 ms −1 and at t = 4 s common velocity = 8 ms −1 To find velocity of A at t = 10 s vA − 8 u A − 8 = 10 − 4 4 vA − 8 6 − 8 or = 6 4 vA = 5 ms −1 ∴
OR = PC + P′ C′ = 2 PC = 8 s. ∴ Option (c) is correct. 22. Area of ∆ A′ B′ O = Area of ∆ ABO 1 1 (vB − vA ) ⋅ 4 = (u A − u B ) ⋅ 4 ∴ 2 2 i.e., vB − vA = u A − u B = 5 − 15 = − 10 vA − vB = 10 ms −1 ⇒ Option (b) is correct.
Option (d) is correct.
More than One Correct Options a = − α v1/ 2 dv i.e., = − α v1/ 2 dt or ∫ v−1/ 2 dv = − α ∫ dt 2 v1/ 2 = − α t + C1 Now, at t = 0, v = v0 C1 = 2 v10/ 2 ∴ 2 v1/ 2 = − α t + 2 v10/ 2 ⇒ i.e., the particle will stop at 2 v1/ 2 t= 0 α Option (a) is correct. Option (b) is incorrect. From Eq. (i), dv v = − α v1/ 2 dt
1. Q
or
1/ 2
∫v
dv = − α ∫ dx
2 3/ 2 v = − α + C2 3 2 As at x = 0, v = v0, C2 = v03/ 2 3 2 3/ 2 2 v = − αx + v03/ 2 ∴ 3 3 or
i.e., when the particle stops 2 3/ 2 x= v0 3α Option (d) is correct. Option (c) is incorrect. 2. a = − 0.5 t (m/s 2)
or
dv t =− dt 2 1 dv = − t dt ∫ 2∫
…(i)
v=−
t2 + C1 4
At t = 0, v = 16 m/s t2 + 16 4 From above relation v is zero at t = 8s Options (a) is correct. From relation (i), ds t2 =− + 16 dt 4 t2 ∴ ds = ∫ ∫ − 4 + 16 dt v=−
∴
i.e.,
s=− s=−
…(i)
t3 + 16t + C2 12
t3 + 16t 12
[As at t = 0, s = 0]
At t = 4 s s = 58.67 m Option (b) is correct. The particle returns back at t = 8 s From relation (ii) 83 s8 = − + (16 × 8) = 85.33 m 12 103 s10 = − + (16 × 10) = 76.66 m 12 Distance travelled in 10 s = s8 + ( s8 − s10 ) = (2 × 85.33 ) − 76.66 = 94 m Option (c) is correct. Velocity of particle at t = 10 s
Motion in One Dimension 2
(10) + 16 4 = − 25 + 16 = − 9 m/s ∴Speed of particle at t = 10 s is 9 m/s. Option (d) is correct. v10 = −
→
3. |v|is scalar. ∴Option (a) is incorrect.
Position of A relative to B : →
→
→
S AB = S A − S B ^
→
v = (Let) Rest
B
t1 Rest
East X (km)
South → vA
= − 20 j kmh −1
→ vB
= (40 cos 37° ) i + (40 sin 37° ) j kmh −1
^
^
^
^
= (32 i + 24 j) kmh −1 → → → vAB = vA − vB ^
^
= ( − 20 j) − (32 i + 24 j) ^
^
= − 32 i − 44 j Option (a) is correct. Option (c) is incorrect. At any time →
^
^
^
SA = 3 i + 4 j + ( − 20 j) t →
^
v' = 2v
2a1
t3 – a2
Average velocity = v2
S2
20 km/h
^
Rest
Average velocity = v1
Case II.
A (3, 4)
^
– a2
v = a1t1 v = a 2t2 1 1 s1 = a1t12 + a 2t22 2 2
d v^ ^ → =a≠a dt ∴Option (d) is incorrect.
37°
t2
a1 S1
∴
4
^
t1
|v|
h m/ 0k
^
Option (b) is correct. Option (d) is incorrect.
is v^ (unit vector)
4. North Y (km)
^
^
→ →
^
= (3 − 32t ) i + (4 − 44t ) j
(by definition).
=a
dt ∴Option (b) is correct. v2 is scalar. ∴Option (c) is incorrect. v
^
= (3 i + 4 j − 20t j) − (32t i + 24t j)
5. Case I.
→
dv
| 45
^
S B = (32 i + 24 j) t
v′ = 2a1t1 = 2v 2v = a 2t3 2 ( a 2t2 ) = a 2t3 t3 = 2t2 ⇒ 1 1 s2 = (2a1 ) t12 + a 2t32 2 2 1 2 = a1t1 + a 2(2 t2 )2 2 = a1t12 + 2 a 2t22 2s2 = a1t12 + a 2t22 s1 > 2s1 ∴ 4s1 = 2a1t12 + 2a 2t22 s2 < 4s1 ∴ Combining Eqs. (i) and (ii), 2s1 < s2 < 4s1 Option (d) is correct. s1 In Case I. vav = t1 + t2 1 1 a1t12 + a 2t22 2 2 i.e., v1 = t1 + t2
…(i) …(ii)
46 | Mechanics-1 ( a1t1 ) t1 + ( a 2t2 ) t2 2 ( t1 + t2 ) vt1 + vt2 v = = 2 ( t1 + t2 ) 2 =
a=−
8.
5a = − s + 30 dv 5 ⋅v = − s + 30 ds
In Case II. vav = i.e.,
s2 t1 + t3
a1t12 + 2a 2t22 t1 + t3 ( a1t1 ) t1 + a 2t2 (2 t2 ) = t1 + 2 t2 vt + v (2 t2 ) = 1 t1 + 2t2
v2 =
= v = 2v1 Option (a) is correct. Option (b) is incorrect. 6. If the particle’s initial velocity is + ive and has some constant − ive acceleration the particle will stop somewhere and then return back to have zero displacement at same time t ( > 0). Also if particle’s initial velocity is − ive and has some constant + ive acceleration the particle will stop somewhere and then return back to have zero displacement at same time t ( > 0). ∴Options (b) and (c) are correct. and options (a) and (d) are incorrect. 7. F = α t ⇒
ma = α t α or …(i) a= t m ∴Graph between a (acceleration) and time ( t ) will be as curve 1. ∴Option (a) is correct. From equation dv α = ⋅t dt m α ∴ ∫ dv = m ∫ t dt α t2 i.e., v= ⋅ + c m 2 ∴Graph between velocity (v) and time ( t ) will be as curve 2. Option (b) is correct.
6 s+6 30
or
or or
∫ 5v dv = ∫ ( − s + 30) ds 5v2 s2 =− + 30s + c1 2 2 5 2 S2 v =− + 30s 2 2
…(i)
[C1 = 0 as at s = 0 particle is at rest] Substituting s = 10 m in Eq. (i), 5 2 (10)2 v =− + 300 2 2 i.e.,
v = 10 m/s
∴Option (b) is correct. From Eq. (i) v to be maximum − s + 30 = 0 s = 30 2
∴ or
5 2 (30) vmax = − + (30 × 30) 2 2 5 2 vmax = 450 2
or vmax = 180 m/s Option (c) is correct. 9. If particle’s path is (i) straight with backward motion (ii) not straight somewhere. Distance moved will be greater than the modulus of displacement ∴
→
|v av | < vav
Option (a) is correct. If particle returns to its initial position, the → value of v av will be zero while its average speed (vav ) will not be zero. ∴Option (c) is correct. 1 10. If u = 0, v = at and s = at 2 2 ∴v-t graph will be as shown in (a). s-t graph will be as shown in (d).
Motion in One Dimension t = t1 + t2 + t3 2l l 2l = 1 + 2 + 3 v v v l1 l2 l3 l1 l3 = + + + + v v v v v l 1 = + [ l1 + l2 ] v v l 1 1 1 = + α t12 + β t22 v v 2 2
11. Swimmer will reach point B B b
→ u
v cos θ
θ v sin θ
u
if v sin θ = u i.e., v>u Option (c) is correct. Time to cross the river b b t= = 2 v cos θ v − u2
1 α t12 1 β t22 l + v + 2 v 2 v2 2 v 1 α t12 1 β t22 l = + v + 2 2 v 2 β 2t22 2 α t1 l v 1 1 = + + v 2 α β =
[as v sin θ = u]
Option (b) is correct. b v (cos θ )max b = v Option (a) is correct. tmin =
12. At time T, particle’s velocity change from − –ive to + ive. Option (a) is correct. As slope v-t is same throughout, particle’s acceleration is constant. Option (b) is correct. As net area under the curve is zero. vT − vT Anet = 2 + 2 , displacement of particle at time 2T is zero. Option (c) is correct. →
Initial speed = |v| = v →
Final speed = |v| = v Option (d) is correct. 13. t1
Velocity O Accel :
v
(α) l1
t2
l2
v (− β)
t3
O
Option (b) is correct. For t to be minimum. d l v 1 1 + + =0 dv v 2 α β − or or
v = βt3
1 2 1 αt1 , l2 = vt2, l3 = βt32 2 2 1 1 = vt1 = vt3 2 2
l1 =
l 1 1 1 + + =0 v2 2 α β l α+β = 2αβ v2 2 lα β v= α+β
Option (d) is correct. 14. x = t 2 and y = t 3 − 2 t dx dy and = 2t = 3 t2 − 2 dt dt dx dy At t = 0, = 0 but = −2 dt dt ∴at t = 0 particle is moving parallel to y-axis. Option (a) is correct. Option (d) is incorrect. At t = 0,
l3
l
v = αt1
| 47
and At t = 0, i.e.,
→
^
v = −2 j
d 2x =2 dt 2 d2 y = 6t dt 2 →
^
a =2i →
→
v⊥a
Option (b) is correct.
48 | Mechanics-1 At t =
2 : 3 dy 2 dx 2 but = 3 − 2 =2 3 dt dt 3
Speed (v) at
Y
= a ( S1 + S1 ) = a (25a t12 ) v = 5a t1 ∴ or v = 10 m/s Option (a) is correct.
t2
Time t to reach P from T 10 = 2 + at 8 t= ⇒ a
∴Particle is moving parallel to x-axis. Option (c) is correct. 15. Speed 2 m/s X Starting point
S1
14 m/s P (Mid point) of X and Y
t1 S2
S1 + S2 2
∴
S1 + S2 : 2 S + S2 v2 = 2a 1 2
…(i)
Time t′ to reach Y from P 14 = 10 + at′ 4 t′ = ∴ a Comparing Eq. (i) and Eq. (ii), t = 2t′ Option (c) is correct.
at1 = 2 and at2 = 14 t2 = 7t1 1 1 S1 + S2 = at12 + at22 2 2 1 2 = a ( t1 + t22 ) 2 1 = a [ t12 + ( 7t1 )2 ] 2 = 25a t12
…(ii)
Match the Columns 1. (a)
(b) a (acceleration)
a (acceleration) + –
+ –
t
v (velocity) + –
+ –
t Speed increasing
(a) (a) → (r) (s).
v (velocity)
v (velocity) t
v (velocity) + – + –
Speed decreasing
t
t Speed decreasing
Speed increasing
(b) (b) → (r) (s). (c) s = kt 2 v=
ds = 2kt dt
Motion in One Dimension ∴v must be increasing with time. ⇒ speed must also be increasing with time. (c) → (p). (d) Slope is + ive and decreasing. ∴Velocity must be decreasing with time. ⇒ speed must also be decreasing with time. (d) → (q). →
^
2. (a) v = − 2 i
→
∴Initial speed = |v| = 2 →
^
a = −4 j
→
Velocity (ω ) at time t → → → ω =v + at ^
^
= − 2 i − 4t j Speed at time t = |ω| = ( − 2)2 + ( − 4 t )2 At t > 0 →
|ω| > |v| Speed increasing. (a) → (p). ∴ →
^
(b) v = 2 i
→
|v| = 2 →
^
→
→
→
^
^
^
ω =2i + 2ti + 2t j →
2
|ω| = (2 + 2 t ) + (2 t )
∴
→
As t > 0,
→
|ω| > |v| (a) → (p).
→
→
^
^
(c) v = − 2 i and a = + 2 i →
→
→
Using ω = v + a t →
^
^
ω = −2i + 2ti →
or
^
ω = (2 t − 2) i →
|ω|= |2 t − 2|
∴
→
At t = 0 s
Speed = |ω| = 2 m/s
t =1s
|ω| = 0 m/s
t =2s
|ω| = 2 m/s
t =3 s
|ω| = 4 m/s
→ → →
^
→
→
^
^
a = −2i + 2 j
→
→
ω =v + at ^
^
^
^
^
=2i −2ti + 2t j = (2 − 2 t ) i + 2 t j →
|ω| = (2 − 2 t )2 + (2 t )2 →
At t = 0 1 t= 2
|ω| = 2
t =1
|ω| = 2
t =2
|ω| = 20
→
|ω| = 2 →
→
3. (a) From A to B, v is increasing and area is + ive. (a) → (p). ∴ (b) From B to C, v is increasing and area is + ive. (b) → (p). ∴ (c) From C to D, v is decreasing while area is + ive. (c) → (q). ∴ (d) From D to E, v is − ive and |v|is increasing, area is −ive. (d) → (r). ∴
Using ω = v + a t ∴
→
(d) v = 2 i and
→
^
a =2i + 2 j
→
∴(c) → (s).
∴(d) → (s).
→
→
| 49
2
4. (a) Displacement = Area 10 × 4 = = 20 unit 2 20 = + 5 unit ∴ Average velocity = + 4 i.e., (a) → (r). 1 ×5 (b) Displacement = + 20 − = + 17.5 unit 2 + 17.5 unit ∴ Average velocity = 3 = 5.83 unit (b) → (s). ∴ 2 × ( − 10) (c) Displacement = + 20 + 2 = + 10 unit + 10 ∴ Average velocity = 6 i.e., (c) → (s).
50 | Mechanics-1 (d) Rate of change of velocity at t = 4 s 0 − 10 → a= = −5 4 −2
i.e.,
∴Rate of change of speed at t = 4 s would be 5. i.e., (d) → (r). 5. (a) x = − 20 + 5t 2 0 = − 20 + 5t 2 t =2s
⇒ i.e.,
(a) → (r).
(b) x = − 20 + 5t
2
dx = 10 t dt d 2x = 10 dt 2 Velocity will be numerically acceleration at t = 1 s. Thus
equal
to
(b) → (q).
t =1s Y = 4 − 4t + t 2 dy = − 4 + 2t dt = − 4 + (2 × 1) = − 2 unit
∴(a) → (q). (b) Y = 4 − 4t + t 2 0 = 4 − 4t + t 2 [For particle to cross x-axis] i.e., t =2s x = 1 − 2t + t 2 dy = − 2 + 2t dt [at t = 2 s] = − 2 + (2 × 2) = + 2 unit ∴(b) → (p). dx (c) = − 2 unit dt t = 0 s
(c) At t = 0 s
x = − 20 m
dy and = − 4 unit dt t = 0 s
t =1s
x = − 15 m
∴ Initial velocity of particle
t =2s
x =0m
t =3 s
x = + 25 m
Particle is always moving along + ive x-axis. (c) → (s). dx (d) v = = 10 t is zero at t = 0 s. dt (d) → (p). ∴ ∴
6. (a) x = 1 − 2t + t 2 0 = 1 − 2t + t 2 [for particle to cross y-axis]
[at t = 1 s]
= ( − 2 )2 + ( − 4 )2 i.e., d 2x (d) 2 = 2 dt
= 2 5 unit (c) → (s).
d2 y =2 dt 2 ∴ Initial acceleration of particle = 2 2 unit i.e., (d) → (s). and
Motion in One Dimension
| 27
Graphy 28. OA : slope is + ive and increasing.
30. Case I
s C B
v
θ
D
A
s
m = tan θ
⇒
t
s0 t
v0 a
t
O
∴ velocity is + ive and acceleration is + ive. AB : slope is + ive and constant ∴ velocity is +ive and acceleration is zero. BC : sope is + ive and decreases. ∴ velocity is + ive and increasing. CD : slope is –ive and increasing ∴ velocity is − ive and acceleration is − ive. 29. v v M2
M1
θ t
O
θ O
⇒
v = mt − v0 (st-line) ∴
∫ ds = ∫ (mt2 − v0 ) dt
t − v0 t + s0 2 1 or s = s0 − v0 t + at2 2 dv Further, a = = tan θ dt i.e.,
(Parabola)
s= m⋅
(Q m = a)
As for 0° ≤ θ < 90°
t
tan θ is + ive, a is + ive.
v M4 θ
a = tan θ
Case II
M3 v t
s
θ
⇒ s0 t
t
In M1 and M 3 : 0° ≤ θ < 90° ∴ slope is + ive i.e., acceleration is + ive. In M2 and M4 : 90° < θ ≤ 180° ∴ slope is − ive i.e., acceleration is − ive. (a) M1 : Magnitude of velocity is increasing. M2 : Magnitude of velocity is decreasing. M 3 : Magnitude of velocity is increasing. M4 : Magnitude of velocity is decreasing. Ans : M1 and M 3. (b) P → M1 Q → M2 R → M3 S → M4
a
⇒ t
v = k (constant) (st line time-axis) ⇒ ∫ ds = ∫ k dt s = kt + s0 dv Further, a= =0 dt
parallel
to
28 | Mechanics-1 Case III.
Time-acceleration graph s
v v0
time
s0 ⇒ θ t
t
0-2s 2-4s 4-6s 6-8s 8 - 10 s
slope
acceleration 5 m/s 2 0 m/s 2 − 5 m/s 2 − 5 m/s 2 + 5 m/s 2
5 zero −5 −5 +5
a a ⇒
t
5
a = tan θ
v = v0 + mt ∴ ∫ ds = ∫ (v0 + mt2) dt t i.e., s = v0 t + m + s0 2 1 or s = s0 + v0 t − at2 2 dv Further, a = = tan θ dt As for 90° < θ ≤ 180° tan θ is − ive, a is − ive.
2
(Q m = − a)
Time-displacement graph Time
at 0 s 2s 4s 6s 8s 10 s
Area Initial Net under the Displacement displacement graph 0m 10 m 30 m 40 m 30 m 20 m
− 10 m − 10 m − 10 m − 10 m − 10 m − 10 m
− 10 m 0m + 20 m + 30 m + 20 m + 10 m
4
6
8
31. a = slope of v- t graph. S = area under v − t graph. Corresponding graphs are drawn in the answer sheet. s Net area A1 − A2 32. Average velocity = = = t Time t d A1 + A2 Average speed = = t t 33. Average acceleration v f − vi −10 − 20 = = = −5 m / s2 time 6 v f − vi = Area under a- t graph. 34. (a) Acceleration = slope of v- t graph. 35. (b) rf − ri = s = area under v- t graph. (c) Equations are written in answer sheet.
s
Relative Motion
+ 30 m
35. (a) Acceleration of 1 w.r.t. 2 + 20 m + 10 m
2 – 10 m
10 t
4
6
8
10 t
= ( − g) − ( − g) = 0 m/s2 (b) Initial velocity of 2 w.r.t. 1 = ( +20) − ( −5) = 25 m/s (c) Initial velocity of 1 w.r.t. 2 = ( − 5) − ( + 20) = − 25 m/s
Motion in One Dimension
= − 25 m/s (d) Initial relative displacement of 2 w.r.t. 1 →
S 0 = − 20 m → → → 1→ Using S = S 0 + u rel t + a rel t2 , 2 as at time t (= time of collision of the particles) the relative displacement of 2 → → w.r.t. 1 will be zero (i.e., S = O) → 1 O = ( − 20) + ( − 25) ⋅ 2
t = 0.85 s
⇒
36. Let length of escalator L (= 15 m)
walking speed of man =
L 90
L Speed of escalator = 60 Time taken by man walking on a moving L escalator = L L + 90 60 = 36 s Required time has been found without using the actual length of the escalator. →
37. S 0 = Initial displacement of elevator w.r.t.
ball = + 10 m. Relative velocity of elevator w.r.t. ball = (2) − (18) = − 16 m/s Accelerator of elevator w.r.t. ball = (0) − ( − 10) = 10 m/s2
38. 3.5 m/s2
time = 0 2.2 m/s2
time =ts
T
T
A A x 60 m
1 (a) (2.2) t2 = 60 2
…(i)
t = 7.39 s 1 2 (b) (3.5) t = 60 + x 2 ⇒
x = 35.5 m (c) At the time of overtaking Speed of automobile = (3.5) (7.39) = 25.85 m/s Speed of truck = (2.2) (7.39) = 16.25 m/s 39. Let, acceleration of lift = a (upward)
∴ acceleration of thrown body w.r.t. lift = ( − g) − ( + a) If time of flight is t, using vrel = urel + arel t
10 m
12 m
2 m/s Elevator 5m
→
→
Using S = S 0 + urel t +
1 a t2 2 rel
…(ii)
Dividing Eq. (ii) by Eq. (i), 60 + x 3.5 = 60 2.2 x 1.3 or = 60 2.2
= − ( g + a)
18 m/s
| 29
1 0 = ( + 10) + ( −16) t + ( +10) t2 2 or 5t2 − 16t − 10 = 0 t = 3.65 s Position of elevator when it meets ball = 5 + (2 × 3.65) = 12.30 m level
1 ∴ Velocity of 1 w.r.t. 2 at time t = s 2 1 = ( − 25) + (0) 2
( − v) = ( + u) + { − ( a + g)} t ⇒ ( a + g) t = 2 u 2u or a= −g t
30 | Mechanics-1 A
40. 20 m
2 4 = sin ( 45° − θ) sin 45°
B OAB = OA = 20 m
1 θ = sin −1 − 45° 2 2
2√2 m/s 45° O
42. Let pilot heads point R to reach point Q 2
2
OB = (OA) + ( AB)
R
= (20)2 + (20)2 = 20 2 m Speed along OB is 2 2 m/s ∴ Time taken to reach B =
200 km/h
RQ PR 200 t = 500 t
sin θ =
41. In ∆ OPQ, B River flow
P Q
θ
⇒
45° O
or →
OP = |v br |
East
P South
= 10 s A
1000 km
θ
Wind
20 2 m 2 2 ms −1
North Q
or
→
OP = |v r | PQ OP = sin ∠ POQ sin ∠ OPQ
θ = sin −1 (0.4) ( PR)2 = ( RQ)2 + ( PQ)2 (500 t)2 − (200 t)2 = (1000)2 1000 t= (500)2 − (200)2 10 10 h = = 25 − 4 21
Objective Questions (Level 1) Single Correct Option
mg + f 0=u− T1 m um T1 = mg + f
1. As the packet is detached from rising
balloon its acceleration will be g in the downward direction. Option (b) is correct. 2. While going up : Stops
Using v2 = u2 + 2as mg + f 02 = u2 − 2 s m i.e.,
S
S=
While coming down
u
1 mg − f 2 T2 2 m mu2 1 mg − f 2 = T2 2 ( mg + f ) 2 m s=
f (Air resistance) mg
mu2 2( mg + f )
or
…(i)
Motion in One Dimension ⇒
T2 =
um ( mg + f ) ( mg − f )
…(ii)
8. For first one second
2=
Using Eq. (i) and Eq. (ii) T2 = T1
mg + f mg − f
2π rad s −1 60 Speed of the tip of seconds hand π [Q v = rω and v= × 1 cm/s 30 r = 1 cm] As in 15s the seconds hand rotates through 90°, the change in velocity of its tip in 15 s will be π 2 =v 2= cm s −1 30 Option (d) is correct. 5+5 4. Average speed = = 40 ms −1 5 5 + 30 60 =
Option (c) is correct. 5. Relative velocity of boat w.r.t. water
= (3 ^i + 4 ^j) − ( − 3 ^i − 4 ^j) = 6 ^i + 8 ^j Option (b) is correct. (18 × 11) + ( 42 × v) 6. 21 = 18 + 42 ⇒ v = 25.29 m/s 8 t3 7. x = 32t − 3 dx …(i) v= = 32 − 8 t2 dt ∴ Particle is at rest when 32 − 8 t2 = 0 i.e., t = 2s Differentiating Eq. (i) w.r.t. time t dv a= = − 16 t dt ∴ a at time t = 2 s (when particle is at rest) = − (16) × (2) = − 32 m/s2 Option (b) is correct.
1 a × 12 2
a = 4 m/s 2
⇒
Velocity at the end of next second v = ( 4) × (2)
⇒ T2 > T1 Option (c) is correct. 3. Angular speed (ω) of seconds hand
| 31
= 8 m/s Option (b) is correct. 9. x = − 3 t + t 3
∴ displacement at time t (= 1 s) = − 3(1) + (1) 3 = − 2 m and displacement at time t (= 3 s) = − 3(3) + (3) 3 = 18 m And as such displacement in the time interval (t = 1 s to t = 3 s) = (18 m ) − ( − 2 m ) = + 20 m Option (c) is correct. 10. Acceleration a = bt
∴
dv = bt dt
or
∫ dv = ∫ bt dt
1 2 bt + C 2 Now, at t = 0, v = v0 ∴ C = v0 1 i.e., v = bt2 + v0 2 ds 1 or = v0 + bt2 dt 2 1 2 or ∫ ds = ∫ v0 + 2 bt dt 1 or s = v0 t + bt 3 + k 6 At t = 0, s = 0, ∴ k=0 1 ⇒ s = v0 t + bt 3 6 or
v=
Option (a) is correct.
32 | Mechanics-1 11.
1 g (2t2 + 6) m (given) 2
v2 = 2 ^i + 4 ^j Option (b) is correct.
3rd drop 1 gt2 2
time = t
∴
2nd drop
u2 =3 v20
or u = v0 3 Option (a) is correct.
1 g (2t)2 2
time = t
13. v20 = 2 gh and u2 = 2 g (3 h)
14. vx = 8 t − 2
dx = 8t − 2 dt or ∫ dx = ∫ (82 t − 2) dt or x = 4 t − 2t + k1 Now at t = 2, x = 14 ∴ 14 = 4 ⋅ 22 − (2) 2 + k1 i.e., k1 = 2 Thus, x = 4 t2 − 2t + 2 Further, vy = 2 dy i.e., =2 dt or ∫ dy = ∫ 2 dt or y = 2t + k2 Now, at t = 2, y = 4 ∴ k2 = 0 Thus, y = 2t y Substituting t = in Eq. (i), 2 2 y y x = 4 − 2 + 2 2 2 ∴
1st drop
Ground
1 2 5 gt = m 2 4 ∴ Height of 2nd drop from ground 5 = 5m − m 4 = 3.75 m Option (c) is correct. ∴
12. [at time = 1] 0 m/s
20 m/s
S
A
B
[At time = t − 1] 1 s = × 10t2 2 1 × 10 ( t − 1)2 2 t2 = 4 ( t − 1) + ( t2 − 2t + 1) 2t − 3 = 0 3 t= s 2 2 1 3 s = × 10 × 2 2 = 20 ( t − 1) +
or or or ∴
= 11.25 m Option (c) is correct. t3 ^ ∴ v = t ^i + j 2 Thus, velocity of particle at time t (= 2 s) will be
or x = y2 − y + 2 Option (a) is correct. 15. x = 5t and y = 2t2 + t
dx dy = 5 and = 4t + 1 dt dt dy dy / dt Now, = dx dx / dt 4t + 1 tan 45° = 5 ⇒ t =1s Option (b) is correct.
∴
16. y = 8 t − 5t2 and x = 6t
∴
dy dx = 8 − 10t and =6 dt dt
…(i)
…(ii)
Motion in One Dimension At t = 0
v
20.
dy dx = 8 and =6 dt dt ∴ Velocity of projection 2
vmax
2
dx dy = + = 62 + 82 dt dt
∴
2 g
Acceleration = In time interval t2
O
t =T 3
H W
H/9
vmax = 2a t2 v = 2 ⋅ max t1
i.e., Now ∴
R Ground
t=T
vmax t2
∴
H – H = 8H 9 9
E
vmax =a t1
Retardation =
H 9 Top of tower
T
t
t2
In time interval t1
= 10 ms −1 Option (c) is correct. 2H 17. T= g T = 3
t1
Option (c) is correct. 18. Distance of farthest corner from one
corner = a + a + a = 3a 3a ∴ Time taken = u Option (a) is correct. 19. Time-velocity
graph of the time-acceleration graph will be
given
t1 = 2t2 t1 + t2 = t 2t2 + t2 = t t or t2 = 3 Depth of shaft = Displacement of lift 1 = vmax ( t1 + t2 ) 2 1 = vmax (2t2 + t2 ) 2 3 = vmax t2 2 3 = (2at2 ) t2 2 = 3 at22 2 t = 3 a 3
at2 3 Option (b) is correct. 1 1 21. s = 2ut + at2 = ut + 2at2 2 2 =
v 8 (m/s)
4
8
12 t (s)
∴ Height of lift above the starting point when it comes to rest = Area under t-v graph 4 + 12 = × 8 = 64 m 2 Option (b) is correct.
P
a 2u
P
time = 0 Q
time = t
u 2a
Q s
⇒
ut −
1 2 at = 0 2
| 33
34 | Mechanics-1 1 at = 0 2 2u t= a 2 2u 1 2u s = 2u + a a 2 a
or
u−
or ∴
4u2 2u2 + a a 6u2 = a Option (a) is correct.
v (m/s) 10 A1 O
1
A2 2
t
– 10
=
22. u = vertical speed (w.r.t. cart) of the
particle. Actual path followed by the particle
u
∴ Displacement of particle at time (t = 2) = ( + 5 m ) + ( −5 m ) = 0 m. i.e., the particle crosses its initial position at t = 2 s. Option (b) is correct. 24. As downward direction is considered to be
t=T u = Vertical speed (w.r.t. cart of the particle)
t=0 30 m/s 80 m
80 m 8 = s 30 m / s 3 1 For particle 0 = uT + ( − g) T2 2 1 2 i.e., gT = uT 2 2u 8 T= = g 3 2u 8 i.e., = 10 3 40 or u= ms −1 3 Option (c) is correct. 1 23. A1 = ( +10) ( +1) = + 5 m 2 1 A2 = ( −10)( +1) = − 5 m 2 For cart T =
f (Air resistance)
u=0
S
+ ive, velocity of ball at t = 0 will be − v0 . Thus, option (a) and (c) are incorrect. Now, as the ball will have +ive acceleration throughout its motion option (b) is also incorrect. ∴ Correct option is (d). 25. Let acceleration of lift = a (upwards)
Displacement of ball in time t = Displacement of lift in time t 1 1 v0 t − gt2 = at2 2 2 gt at or v0 − = 2 2 2 v0 − gt ∴ a= t Option (a) is correct. 26. a = − 0.2 v2
∴ or
dv = − 0.2 v2 dt −2 ∫ v dv = − 0.2 ∫ dt
v −2 + 1 = − 0.2t + C −2+1 1 or − = − 0.2t + C v Now, at t = 0, v = 10 m/s (given) 1 ∴ − =0+C 10 1 i.e., C=− 10 or
Motion in One Dimension 1 1 Thus, − = − 0.2t − v 10 Q For velocity v at time t (= 2 s) 1 1 − = − (0.2 × 2) − v 10 1 4 1 − =− − v 10 10 1 1 = v 2 ⇒ v = + 2 m/s. Option (a) is correct. 27. For displacement ( S1) of train 1 before
coming to rest 02 = (10)2 + 2 ( −2) S1 i.e., S1 = 25 m For displacement ( S2 ) of train 2 before coming to rest 02 = (20)2 + 2 ( −1) S2 i.e., S2 = 200 m ∴ Smin = S1 + S2 = 225 m Option (b) is correct. 28. Let the balls collide after time t the first
ball is shot. ∴ displacement (S) of ball 1 at time t = displacement ( S) of ball 2 at time ( t − 2) 1 1 40t + ( − g) t2 = 40 ( t − 2) + ( − g) ( t − 2)2 2 2 1 2 2 or g[ t − ( t − 2) ] = 80 2 or ( t + t − 2) ( t − t + 2) = 16 or 2t − 2 = 8 or t = 5s 1 S = 40 × 5 + ( −10) 52 2 = 200 − 125 = 75 m Option (b) is correct. 29.
i.e.,
and
0 = u − gT u = gT 1 H = uT − gT2 2 1 2 h = ut − gt 2
…(i) …(ii) …(iii)
| 35
Substracting Eq. (ii) from Eq. (iii), 1 h − H = u ( t − T) + g ( T2 − t2 ) 2 1 = gT ( t − T) − g ( t − T) ( t + T) 2 1 = ( t − T) g T − ( t + T) 2 2 T − t − T = ( t − T) g 2 1 = − g ( t − T)2 2 1 i.e., h = H − g ( t − T)2 2 Option (d) is correct. t2 30. x= 2 dx ∴ vx = =t dt x2 ( t2 / 2)2 t4 y= = = 2 2 8 dy t 3 ∴ vy = = dt 2 31.
x = t+3 i.e.,
x = t2 + 6t + 3 dx = 2t + 6 dt i.e., v∝t Option (b) is correct. 32. Aeroplane’s velocity at time t (= 20 s)
= Area under curve ( t-a graph) 1 = (20 × 5) − × 10 × 2 2 = 90 m/s Option (c) is correct. 33. v = 5 1 + 5
At s = 0 (i.e., initially), velocity of particle = 5 m/s Option (b) is correct. Differentiating equation (i) w.r.t. s 1 − dv 1 = 5 (1 + 5) 2 ds 2 1 1 − dv 5 v⋅ = (1 + 5) 2 ⋅ 5 (1 + 5) 2 ds 2 = 12.5 m/s 2
…(i)
36 | Mechanics-1 Option (b) is correct.
= 12.5 − 16.0 + 10 = 6.5 m/s Option (b) is correct.
34. See answer to question no. 2.
mg − f mg + f
Time of ascent = Time of descent
37. u2 = 2 gH
mg − ma mg + ma
= =
g−a g+a
=
10 − 2 10 + 2 =
H=
⇒
H 2 2 u = 100 + gH
102 = u2 − 2 g ⇒
Stops
2 3
Option (b) is correct.
H 2
N θ
10 m/s
H
35.
10 t
u2 2g
River flow
W
u
E S
Thus,
5t
sin θ =
5t 10 t
=
1 2
H=
100 + gH 2g
or gH = 100 ⇒ H = 10 m Option (b) is correct.
⇒ θ = 30° Option (b) is correct. 36. Fnet = 3 t2 − 32
manet = 3 t2 − 32 3 t2 − 32 or anet = 10 2 or anet = 0.3 t − 3.2 dv or = 0.3 t2 − 3.2 dt 2 or ∫ dv = ∫ (30.3 t − 3.2) dt 0.3 t or v= − (3.2) t + k 3 or v = 0.1t 3 − (3.2) t + k Now, at t = 0, v = 10 m/s ∴ 10 = k Thus, v = 0.1 t 3 − (3.2) t + 10 ∴ at t = 5 s v = 0.1 (5) 3 − (3.2) (5) + 10 i.e.,
38.
10 m/s
15 m u t=0 2
10 = u2 − 2 g × 15 100 = v2 − 2 × 10 × 15 i.e., u = 20 m/s Now, v = u − gt = 20 − 10 × 3 = 20 − 10 ⋅ (3) = − 10 m/s, downward Option (d) is correct.
Motion in One Dimension
| 37
JEE Corner Assertion and Reason
Displacement t0 1 v0 t0 v =2 = 0 t0 2
→
1. As acceleration (a ) is just opposite to
=
→
velocity ( v), the particle will move along a straight line. a
∴ Assertion is true. Reason is also true as proved. 4. Acceleration (a) will be zero only when v
∴ Assertion is false (motion being one dimensional). →
→
Reason is true as v and a do not depend upon time. Option (d) is correct. 2. Displacement-time graph is parabolic only
when the slope of the straight line velocity-time graph is not zero i.e., acceleration is not zero. If acceleration is zero. s = ut i.e., the displacement-time graph will be a straight line. ∴ Assertion is wrong. Reason : v = u + at ds i.e., = u + at dt or ∫ ds = ∫ (u1 + at) dt s = ut + at2 + k 2 If at t = 0, s = 0 The value of k will be zero. 1 ∴ s = ut + at2 2 Thus, reason is true. 3. Displacement in time t0
= Area under v-t graph 1 = v0 t0 2 ∴ average velocity in time interval t0
velocity does not change with time dv a =v⋅ ds ds dv = ⋅ dt ds dv = dt dv ∴ a = 0 if =0 dt or v is constant with time. Thus, Assertion is wrong. dv Reason is correct as a is equal to which dt is instantaneous acceleration. 5. If acceleration is in the opposite direction
to the velocity of the particle the speed will decrease. u = 10 m/s a = − 2t m/s 2 dv = − 2t dt ∫ dv = ∫ − 22 t dt t v = − 2⋅ + k 2 At t = 0 s, velocity = + 10 m/s ∴ k = + 10 m/s ∴ v = − t2 + 10 Time (t)
v m/s
speed (m/s)
15 25 35
9 6 1
9 6 1
∴ Assertion is true.
38 | Mechanics-1 Reason is false as when acceleration is positive the speed will increase. da 6. (given) = 2 (ms −3) dt [This implies that reason is true] ∫ da = ∫ 2 dt a = 2t + C If at t = 0, a = 0 we have C=0 ∴ a = 2t dv or = 2t dt ∴ Assertion is false. Option (d) is correct. 7. If initially particle velocity is − ive and
acceleration is uniform and + ive, the particle will return to its initial position after a certain time interval. In this time interval the average velocity will be zero as net displacement will be zero. ∴ Assertion is false. For average velocity to zero, the particle must return to its initial position (as discussed above) and for this velocity can’t remain constant. ∴ Reason is true. Option (d) is correct. 8. From O to A velocity of the particle is
increasing while from A to B it is decreasing without change in direction as for the velocity to change its direction the slope of s-t graph must be negative. ∴ Assertion is false. If the slope of s-t graph is + ive the velocity of the particle will be + ive while if it is − ive the velocity of the particle will also be − ive. ∴ Reason is true. 2
9. S1 = 2t − 4 t
and S2 = − 2t + 4 t2 ⇒ displacement of particle 2 w.r.t. 1. S21 = S2 − S1 = ( − 2t + 4 t2 ) − (2t − 4 t2 ) = − 4 t + 8 t2
Time t Relative Displacement 0s 1s 2s 3s
0m 4m 24 m 60 m
As relative displacement is increasing the relative velocity would also be increasing. ∴ Assertion is false. Reason is true. 10. If v = u + ( − a) t [acceleration being made
− ive] Velocity (v) of the particle will be zero u at t= . a u Thus, for t < , v is + ive a i.e., the acceleration can change its direction without change in direction of velocity. ∴ Assertion is true. If ∆V changes sign say from + ive to − ive, ∆V the acceleration which equals will also ∆t change sign from + ive to − ive. ∴ Reason is true but it is not the correct explanation of the assertion. 11. At time t when the two are at the same
height. A S1
u=0 a=+g
A
B
time t
S2 u = √gh B
S1 + S2 = h 1 gt2 + ut − 1 gt2 = h 2 2 ut = h gh ⋅ t = h t=
h g
4
Projectile Motion Introductory Exercise 4.1
1. A particle projected at any angle with
horizontal will always move in a plane and thus projectile motion is a 2-dimensional motion. The statement is thus false.
= 80 − ∴
= 402 + 202
2. At high speed the projectile may go to a
place where acceleration due to gravity has some different value and as such the motion may not be uniform accelerated. The statement is thus true.
i.e.,
3. See article 4.1.
= 802 + 602
As horizontal acceleration would be zero. vy u
B
sx
v
φ
vx sy
α A
= 20 5 m/s v y 20 tan φ = = vx 40 1 φ = tan −1 2 s = s2x + s2y
4. u = 40 2 m/s, θ = 45°
uy
1 × 10 × 22 = 60 m 2 v = v2x + v2y
g
ux
vx = ux = u cos θ = 40 m/s sx = ux t = (u cos θ) t = 80 m A : position of particle at time = 0. B : position of particle at time = t. As vertical acceleration would be − g v y = u y − gt = u sin θ = gt = 40 − 20 = 20 m/s 1 s y = u y ⋅ t − gt2 2
i.e., 5. s y = u y t +
= 100 m sy tan α = sx 60 = 80 3 α = tan −1 4 1 ( − g) t2 2
or
s y = (u sin θ) t −
or
15 = 20t −
or i.e.,
1 2 gt 2
1 2 gt 2 t2 = 4 t + 3 = 0 t = 1 s and 3 s
6. See figure to the answer to question no. 4.
u = 40 m/s, θ = 60° ∴ ux = 40 cos 60° = 20 m/s
52 | Mechanics-1 Thus, vx = ux = 20 m/s As φ = 45° tan φ =
→
vy vx
v = v2x + v2y = vx 2
Thus,
= 20 2 m/s
r = 2.4 ^i m
i.e.,
x = 2.4 m
11. vx = ux = 10 m/s v u = 20 m/s
φ
vx
60° ux = 20 cos 60°
u2 sin 2α / g 2 u sin α / g
2 u2 sin α cos α g × g 2 u sin α
8. Change in velocity t=T u sin α α
α u cos α
x = 0m →
= ucos α + u sin α u
r = 0 ^i m
i.e., At t = 0.8 s,
After attaining highest point − 20 = 40 3 − 10t i.e., −2=2 3 − t or t = 2 ( 3 + 1) s Range 7. Average velocity = Time of flight
=
→
∴
Before reaching highest point v y = u y + ( − g) t ∴ 20 = 40 sin 60° − 10 t or 2=2 3 − t ⇒ t = 2 ( 3 − 1) s
=
^
y-coordinate will be zero when 4 t − 5t2 = 0 4 i.e., t = 0 s, s 5 t = 0 belongs to the initial point of projection of the particle.
=1
y y = vx = 20 m/s
∴
^
10. r = [3 t i + ( 4 t − 5t2 ) j] m
– u sin α u
= ( − u sin α) − ( + u sin α) = − 2usin α = 2usin α (downward) 9. Formulae for R, T and H max will be same
if the projection point and the point where the particle lands are same and lie on a horizontal line.
= 10 m/s vx = ux = 10 m/s vx = v cos φ v ⇒ cos φ = x v 10 = v 10 20 [as, v = (given)] = 10 2 =1 i.e., φ = 0° ∴ Speed will be half of its initial value at the highest point where φ = 0°. Thus,
u sin θ g 20 sin 60° = = 3s 10
t=
Projectile Motion
53
Introductory Exercise 4.2 1. Time of flight
10 3 × 3 2
= vx vy
α
= 5 m/s
vnet
u
3. Let the particle collide at time t. v β
u α = 60° β = 30° u = 10 m/s2 g = 10 m/s2
T=
1
θ h
10 m
2u sin (α − β) g cos β
x1
(2 × 10 sin(60° − 30° ) = 10 cos 30° 2 s = 3 Using,
2
x1 = (u cos θ) t x2 = v t d = x2 − x1 = (v + u cos θ) t = [10 + 10 2 cos 45° ] t = 20 t 1 Using equation, s = ut + at2 2 For vertical motion of particle 1 : 1 h − 10 = (u sin θ) t + ( − g) t2 2 1 i.e., …(i) h = 10 + (u sin θ) t − gt2 2 1 or h − 10 + 10 t − gt2 2 For the vertical motion of particle 2 : 1 20 − h = gt2 2 1 i.e., …(ii) h = 20 − gt2 2 Comparing Eqs. (i) and (ii), 1 1 10 + 10 t − g t2 = 20 − g t2 2 2 ⇒ t = 1s ∴ d = 20 m and ∴
v = u + at vx = ux = u cos 60° 1 = 10 × = 5 m/s 2 v y = u y + ( − g) T = u sin 60° − gT 3 2 = 10 − 10 2 3 3 − 4 = 10 2 3 5 m/s =− 3 vnet = v2x + v2y −5 = 52 + 3 =5 1+ =
2
1 3
10 m/s 3
2. Component of velocity perpendicular to
u = 10 m/s v = 5 2 m/s
4.
plane = vnet cos β =
10 × cos 30° 3
x2 d
v u φ
θ A
B d
54 | Mechanics-1 θ = 30° φ = 45° d = 15 m Let the particles meet (or are in the same vertical time t). ∴ d = (u cos θ) t + (v cos φ) t ⇒ 15 = (10 cos 30° + 5 2 cos 45° ) t or 15 = (5 3 + 5) t 3 or s t= 3 +1
u = 40 m/s α = 60° β = 30° g = 10 m/s 2 φ=β vx = ux = u cos α vx = v cos φ = v cos β u cos α = v cos β u cos α 40 cos 60° v= = cos β cos 30° 1 40 2 = 3 2 40 m/s = 3
∴
or ⇒
= 1.009 s Now, let us find time of flight of A and B 2 u sin θ TA = g = 1s As TA < t, particle A will touch ground before the expected time t of collision. Ans : NO. ∴
7. (a) At time t, vertical displacement of A
= Vertical displacement of B
5. For range to be maximum
u α
vA
=
20
/s m
time t
a = – 10 m/s2
β = 30° = π 6
A
π β α= + 4 2 π/6 π = + 4 2 π = 3 = 60°
1 2 1 gt = vB t − gt2 2 2 v A sin θ = vB
(v A sin θ) t −
→
v A u α
φ
A
∴ (b) x = (v A cos θ) t
= (20 cos 30° ) × =5 3 m
β
vB vA 10 1 = = 20 2 θ = 30°
sin θ =
6. At point A velocity ( v) of the particle will
be parallel to the inclined plane.
B
x
i.e.,
vB = 10 m/s
1 2
Projectile Motion
55
AIEEE Corner Subjective Questions (Level 1) 1. (a) R =
u2 sin 2θ (20 2)2 sin 90° = g 10
or ∴
= 80 m 2
H=
2
2
2
u sin θ (20 2) sin 45° = 2g 10
= 40 m 2 u sin θ 2 (20 2) sin 45° T= = g 10 =4s →
^
→
^
(b) u = (20 i + 20 j) m/s and a = − 10 ^j m/s2 →
→
T=
T2 − 4 T − 8 = 0 − ( −4) + ( −4)2 − 4(1)( −8) 2 (1)
Leaving − ive sign which is not positive. 4 + 48 4 + 4 3 = = 2 2 = (2 + 2 3 ) s = 5.46 s R = 20 × 5.46 = 109.2 m 1 (b) s = ut + at2 2
→
−v=u+at
+
20 m/s
= (20 ^i + 20 ^j) + ( − 10 ^j) 1(at t = 1 s) – 40 m
= (20 ^i − 10 ^j) m/s
a = – 10 m/s2
(c) Time of flight
T
2 u sin θ T= g 1 2 × 20 2 × 2 = 10 =4s (c) ∴ Velocity of particle at the time of collision with ground. = (20 ^i + 20 ^j) + ( − 10 ^j ) 4
R
1 ( −10) T2 2 T =2 2s = 2.83 s R = 2.83 × 20 = 56.6 m
( −40) = 0 × T + ⇒
(c) s = ut +
1 2 at 2
= (20 ^i − 20 ^j ) m/s 2. (a) s = ut +
1 2 at 2
20 √2 m/s
– 40 m
20 m/s 20√2 m/s +
– 40 m
R
45°
1 ( −10) T2 2 5T2 + 20T − 40 = 0 T2 + 4 T − 8 = 0
( −40) = ( −20) T + or or
a = – 10 m/s2 T R
∴ or
T
1 ( −10) T2 2 5T2 − 20T − 40 = 0
( −40) = ( + 20) T +
∴
T= =
− 4 + ( 4)2 − 4 (1) ( −8) 2 (1) −4+ 4 3 = − 2 + 2 3 = 1.46 s 2
56 | Mechanics-1 R = 20 × 1.46 = 29.2 m
∴
100 1 40 × 2 5 m/s = 2
v=
∴
3. (a) Change in velocity (vc )
= Change in vertical velocity (as horizontal velocity does not change). = (u sin θ − gt) − (u sin θ) = − gt = − (10 × 3) m/s = − 30 m/s = 30 m/s (downward)
5. Horizontal compnent of velocity at
P = Horizontal component of velocity at O v sin θ P 60 m/s
→
(b) u = (20 ^i + 20 ^j) m/s →
a = − 10 ^j m/s 2
= 60 ^i + 15 ^j v av =
60 ^i + 15 ^j 3
= 20 ^i + 5 ^j →
⇒ |v av |= (20)2 + (5)2 = 425 = 20.6 m/s u Player
t=0
v
R t=T
50 m
60 cos 60°
vT = 50 −
t=0
Differentiating above equation w.r.t. time t dy dx dx …(i) ∴ =b − 2cx dt dt dt dy dx (at x = 0) ⇒ =b dt dt 2
2
⇒
60°
O –
6. Y = bx − cx2
4. R + vT = 50
Coach
+
v cos θ = 60 cos 60° v cos 45° = 60 cos 60° 60 cos 60° ⇒ v= cos 45° = 30 2 m/s For point P : v sin 45° = 60 sin 60° + ( −10) t1 3 or 30 = 60 − 10 t1 2 t1 = 3 ( 3 − 1) = 2.20 s For point Q : − v sin 45° = 60 sin 60° + ( −10) t2 ∴ t2 = 3 ( 3 + 1) = 8.20 s
= 60 ^i + 60 ^j − 45 ^j
→
Q θ θ t = t1 v cos θ v a = –10m/s2 – v sin θ
∴ ⇒
∴ Displacement at time t (= 3 s) → 1→ = u t + a t2 2 1 ^ = (20 i + 20 ^j) 3 + ( −10 ^j) 32 2
∴
v
2
u sin θ g
2 u sin θ (20)2 v× = 50 − g 10 2 × 20 × sin 45 ° v× = 10 10
dx dy v= + dt dt 2
2
dx dx = + b2 dt dt = 1 + b2
dx dt
2
…(ii)
Projectile Motion Differentiating Eq. (i) w.r.t. time t 2 d2 y d2 x d2 x dx = b ⋅ − 2 c − 2cx 2 …(iii) 2 2 dt dt dt dt Acceleration of particle 2
d2 x d2 x − a = 2 + 2 dt dt d2 x =0 dt2 d2 x d2 y Substituting and = 0 = − a in dt2 dt2 Eq. (iii) 2 dx − a = − 2c dt
∴
2
(6.1) = + 2 ( −10) (9.1) ⇒ u y = 37.21 + 182 = 14.8 Angle of projection uy θ = tan −1 ux uy = tan −1 ux 14.8 = tan −1 7.6 = 62.82° u2 sin 2θ Range = g
dx a = dt 2c
i.e.,
=
dx Substituting above value of in Eq. (ii) dt a v= (1 + b2 ) 2c ^
=
→ v
^
uy
u
B +h
A
(u2x + u2y ) sin 2θ g (v2x
u2y )
+
sin 2θ
g
[(7.6)2 + (14.8)2 ] sin 125.64 ° = 10 = 22.5 m (c) Magnitude of velocity just before the ball hits ground
7. (a) Velocity at point A = 7.6 i + 6.1 j vy
57
u2y
1
= [( −u y )2 + (ux )2 ]2 9.1 m
a = – 10 m/s2
1
= [( −14.8)2 + (6.1)2 ]2
θ O
φ
∴ Initial vertical velocity at a = 6.1 ^j m/s. Final vertical velocity at B (highest point) = 0 m/s 2 2 Using v = u + 2as (Between A and B) 02 = (6.1)2 + 2 ( −10) ( + h) (6.1)2 ⇒ h= 20 = 1.86 m ∴ Maximum height attained by ball = 9.1 + 1. 86 = 10.96 m (b) Let magnitude of vertical velocity at O (point of projection) = u y Using v2 = u2 + 2as (Between O and A)
= 16 ms
[ux = vx ]
−1
−1 u y
(d) φ = θ = tan ux 14.8 = tan −1 7.6 = tan −1 (1.95) 8. As
initial vertical velocities of both particles will be zero and both fall under same acceleration (g), at anytime t, the vertical displacement of both will be same
u1
A θ
time = t
12
u2
+h
a=+g time = t
B (90°− θ)
58 | Mechanics-1 i.e., both will always remain in the same horizontal line as shown in figure. At time t : Vertical velocity of A = Vertical velocity of B = 0 + ( + g) t = gt At A : Vertical velocity of particle 1 tan θ = Horizontal velocity of particle 1 gt or …(i) tan θ = u1
Y y + = tan α x R−x
or or
tan θ + tan φ = tan α (Students to remember this formula) 3 3 ⇒ tan α = + 6 12 9 = 12 3 ⇒ α = tan −1 4 10. On the trajectory there be two points P
and Q at height h from ground. P
At B :
u
Vertical velocity of particle 2 Horizontal velocity of particle 2
tan (90°− θ) =
or
cos θ =
gt u2
…(ii)
Multiplying Eq. (i) by Eq. (ii), gt gt 1= × u1 u2 t=
∴
u1u2 g
Distance between A and B = R1 + R2 = u1t + u2 t = (u1 + u2 ) t uu = (u1 + u2 ) 1 2 g
If particle takes t time to reach point A (i.e., vertical displacement of + h) 1 ( + h) = (u sin α) t + ( − g) t2 2 2 or …(i) gt − 2 (u sin α) t + 2h = 0 The above equation is quadratic in t. Two values of t will satisfy Eq. (i). One having lower value will be time ( = t1) to reach point P while the higher value will be the time ( = t2 ) to reach point Q. ∴ Time to reach point Q from point P = t2 − t1 = ( t2 + t1)2 − 4 t1t2 2
2u sin α 2h = − 4 g g [Using Eq. (i)] 4u2 sin2 α − 2 gh = g
x = tan α R (y, x)
or
u Y = 3m φ
α θ x =6m
a=–g
O
7 2 9.8 = 2.47 m
h
α
=
9. Y = x 1 −
Q
(R – x)
YR = tan α x( R − x)
( t2 − t1) =
16 gh sin2 α − 8 gh g (Qu = 2 gh)
Distance between P and Q : 2h = (u cos α) ( t2 − t1) ∴ 4 h2 = u2 cos2 α ( t2 − t1)2 16 gh sin2 α − 8 gh or 4 h2 = ( 4 gh) cos2 α g2
Projectile Motion 2
or or or or ⇒
2
1 = cos (16 sin α − 8) 1 = cos2 α [16 (1 − cos2 α) − 8 ] 1 = cos2 α [8 − 16 cos2 α ] 16 cos4 α − 8 cos2 α + 1 = 0 ( 4 cos2 α − 1)2 = 0 1 ⇒ 4 cos2 α = 4 1 cos α = 2 cos α = 60° 1 (cos α = − being not possible). 2 16 gh sin2 60° − 2 gh ∴ ( t2 − t1) = g2 =2
h g
u β
R+ b= i.e.,
aR + ab =
⇒
θ=
13. R1 = v1
2h g v1
or u sin α cos β − u cos α sin β = gt cos β gt cos β or u= sin (α − β)
h
R1 = 10 m v2
R2
i.e.,
u sin 2α g ∴ u
α R
…(i)
1 R1 g 2 v1
2
1 10 × 9.8 2 5
2
= 19.6 m 2h T= g =
a
bu2 sin 2α bR − ab = g
h= =
2
i.e.,
b sin 2α + a sin 2β 1 sin −1 2 a+b
a=–g
At time t vertical velocity = v sin β …(i) ∴ v sin β = u sin α = ( − g) t Now, as horizontal acceleration will be zero. v cos β = u cos α Thus, Eq. (i) becomes u cos α sin β = u sin α − gt cos β
12. R − a =
…(ii)
(Proved.) a=0
α
au2 sin 2β g
( b sin 2α + a sin 2β) u2 sin 2θ = u2 g ( a + b) g
β
t
u sin 2β g
Adding Eqs. (i) and (ii), we have b sin 2α + a sin 2β bR + aR = u2 g ( b sin 2 α + a sin 2 β ) or R = u2 ( a + b) g
v v
b
R 2
(Proved)
11. Initial vertical velocity = usin α
59
2 × 19.6 9.8
=2s R2 = v2 ⋅ T = 7.5 × 2 = 15 m
60 | Mechanics-1 = 1.69 s 2u2 sin (α − β) cos α R= g cos2 β
14. Vertical velocity of balloon (+ bag) α
O 12 km/h
2 × (20 2)2 × sin 15° × cos 45° 10 cos2 30° = 39 m 2 u sin (α + β) T= g cos β
A
=
50 m
vw = 20 km/h α
= 12 ×
5 m/s 18
10 m/s 3 Horizontal velocity of balloon (+ bag) 5 m/s = Wind velocity = 20 km / h = 20 × 18 50 m/s = 9 = 5.55 m/s 12 ∴ tan α = . i.e., sin α = 0.51 20 Bag is released at point A. Let t be time, the bag takes from A to reach ground. 1 Using, s = ut + at2 2 10 1 ( −50) = sin α t + ( − g) t2 3 2
16.
=
i.e., t=
∴
5t2 − 1.7 t − 50 = 0 1.7 + ( −1.7)2 − 4 × 5 × ( −50) 2×5
= 3.37 s Vertical velocity of bag when it strikes ground 10 vB = − + (10) (3.37) 3 = 37.03 m/s vω = 5.55 m/s ∴ Velocity of bag with which it strikes ground vnet = v2B + v2ω
=
= 6.31 s u2 R= [sin (2α + β) + sin β ] g cos2 β (20 2)2 [sin (90° + 30° ) + sin 30° ] 10 cos2 30° = 145.71 m 2 u sin (α + β) 17. T= g cos β 2u sin β (Qα = 0°) = g cos β 2u = tan β g 2 × 20 = tan 30° 10 = 2.31 s u2 R= [sin (2α + β) + sin β ] g cos2 β =
u2 ( 2 sin β) g cos2 β 20 × 2.31 uT = = cos 30° cos β
2 × 20 2 × sin ( 45° − 30° ) 10 cos 30°
[as α = 0°]
=
= 53.33 m u2 18. R = [sin (2α + β) + sin β ] g cos2 β
= 37.44 m/s 2 u sin (α − β) 15. T = g cos β =
2 × 20 2 × sin ( 45° + 30° ) 10 cos 30°
=
u2 [sin {2 (α + β) − β } + sin β ] g cos2 β
=
u2 [sin ( π − θ) + sin θ ] g cos2 θ [Q(α + β) =
or
R=
2 u2 tan θ sec θ g
π ] 2
Projectile Motion 19. (a) Acceleration of particle 1 w.r.t. that of
particle 2 = ( − g) − ( − g) =0
61
Vertical acceleration of stone = g Horizontal velocity of stone = v ∴ Path of the stone will be parabolic. 21. (a) g eff = g − ( − a)
^
(b) Initial velocity of 1st particle = 20 j m/s
a m/s2
w m/s
Initial velocity of 2nd particle = (20 2 cos 45° ^i + 20 2 sin 45° ^j) m/s
u
= (20 ^i + 20 ^j) m/s
θ
∴ Initial velocity of 1st particle w.r.t. that of 2nd particle =g+a
= [ (20 ^j) − (20 ^i + 20 ^j ) ] m/s
= 10 + 1
= − 20 ^i m/s
= 11 m/s 2 2 u sin θ T= g eff
= 20 m/s (downward) (c) Horizontal velocity of 1st particle = 0 m/s Horizontal velocity of 2nd particle
=
= 20 ^i m/s ∴ Horizontal velocity of 1st particle w.r.t. that of 2nd particle = 0 − (20 ^i)
= 0.18 s (b) Dotted path [(in lift) acceleration upwards] u
= − 20 ^i m/s Relative displacement of 1st particle w.r.t. 2nd particle at t = 2 s = − 20 ^i × 2 = − 40 ^i m/s ∴ Distance between the particles at t = 2 s = 40 m
θ
Full line path [In lift at rest or moving with constant velocity upwards or downwards]. (c) If lift is moving downward with acceleration g. Path of particle u
20. (a) As observed by passenger
Vertical acceleration of stone = g −0= g Horizontal velocity of stone =v−v=0 ∴ Path of the stone will be a straight line (downwards). (b) As observed by man standing on ground
2 × 2 × sin 30° 11
θ
g eff = g − g = 0 22. Horizontal motion :
∴
x1 = u1 cos θ1 and x2 = u2 cos θ u1 cos θ1 + u2 cos θ2 = 20
…(i)
62 | Mechanics-1 Vertical motion :
20 + (u1 sin θ1) t + u2 θ2
or
θ1
1
20
1 ( − g) t2 2 …(ii) (u1 sin θ1 − u2 sin θ2 ) t = 10 + (u2 sin θ2 ) t +
t
u1
x1
1 ( − g) t2 = 30 2
x2
d = 20 m
Objective Questions (Level 1) →
^
→
^
^
^
=
1. v = 3 i + 4 j and F = 4 i − 3 j → →
v ⋅ F = (3 ^i + 4 ^j) ⋅ ( 4 ^i − 3 ^j)
= 14.14 m/s = 14 m/s (approx) Option (b) is correct. u2 sin2 α 4. H (maximum height) = 2g
= 12 − 12 = 0 →
→
F⊥v
∴
Path of the particle is circular. Y
∴ H1 =
→ v
u2 sin2 (90° − θ) u2 sin2 θ and H2 = g 2g =
X
Option (c) is correct. 5. Equation to trajectory is
→ F
1 gx2 2 2 u cos θ sin θ x2 Y = x tan θ − R
Y = x tan θ −
Option (c) is correct. 2. Projectile motion is uniformly accelerated
everywhere even at the highest point. Option (a) correct. Option (b) incorrect. At the highest point acceleration is perpendicular to velocity. Option (c) incorrect. 3. For range to be maximum
i.e., or
θ = 45° ux 1 = cos 45° = u 2 u vx = 2
u2 cos2 θ g
H1 sin2 θ = H2 cos2 θ
Thus,
j i
20 2
(Qvx = ux )
Area = =
R
∫ 0 Y dx x2 x tan θ − dx R
R
∫0
R
x2 x3 A = tan θ − 3 R 0 2 R2 R3 = tan θ − 2 3 1 2 tan θ =R − 3 2
Projectile Motion =
4v40
2
2
sin θ cos θ tan θ 1 − 2 3 g2
4v40 sin 3 θ cos θ sin2 θ cos2 θ − 2 3 g2 2 v40 = [3 sin2 θ cos θ − 2 sin2 θ cos2 θ ] 3 g2
9. Rmax = 1.6 m
u2 = 1.6 g
=
v φ β
u
or
u cos 60° cos 30° u v= 3
v=
4 2 10 10 2 10 2 Number of jumps = = T 4 2 / 10
1 mv2 2
2
= =
1 u2 m⋅ 2 3 K = 3
1 2 or 2θ = 30° ∴ θ = 15° Option (a) is correct. 2 u sin θ 8. T1 = g 2 u sin (90° − θ) ∴ T2 = g 2u cos θ = g 2 2 u2 sin θ cos θ Thus, T1T2 = g g 2R or T1T2 = g 1 ⇒ R = g T1T2 2 Option (d) is correct. sin 2θ =
=
Q 1 mu2 = K 2
∴ Option (b) is correct. u2 sin 2θ 1 u2 7. = g 2 g or
∴
= 25 ∴ Grass hopper would go = 25 × 1.6 m i.e., 40 m. Option (d) is correct. |Displacement| 10. |Av. velocity|= time
θ
or KE at B =
u = 4 m/s 2u sin 45° T= g
⇒
=
6. v cos φ = u cos θ
A
63
1 R 2 +H T / 2 2 1 u sin θ g
2
u2 sin θ cos θ u2 sin2 θ + g 2g
= u cos2 θ +
sin2 θ 4
u 1 + 3 cos2 θ 2 Option (b) is correct. =
11. d2 = R2 + u2 T2 T u
2 /g
R
Ball
=u
45° 90°
α
uT T 2 2
u 2 u sin 45° = + u2 g g =
u4 2 u4 + 2 g2 g
=
3u4 g2
2
2
64 | Mechanics-1 d=
∴
u2 g
3=
302 10
3 = 90 3 m
Option (b) is correct. 12. At maximum height
dy =0 dt d (10t − t2 ) = 0 dt or 10 − 2t = 0 or t = 5s ∴ Maximum height attained = 10 (5) − 52 = 25 m Option (d) is correct. u2 13. R = [sin (2α + β) + sin β ] g cos2 β i.e.,
u
a = –g
1 ( −10) t2 2 or 5t2 − 25t − 70 = 0 or t2 − 5t − 14 = 0 ( t − 7) ( t + 2) = 0 t=7s (− 2 s not possible) Option (c) is correct.
β
As α = 0 (according to question) u2 R= [2 sin β ] g cos2 β
=
1000 m 3
sin 2φ = sin 3α 2φ = π − 2α π or φ= −α 2 π π π = − = = 30° 2 3 6 ∴ Maximum height attained by second particle u2 sin2 φ = 2g 102 × 2 g sin2 30° = × 2g sin2 60° 102 × 1 / 4 = = 34 m 3/4
∴
R
(50)2 × 2 sin 30° 10 × cos2 30°
u2 sin 2φ u2 sin 2α = g g
Option (d) is correct. 1 15. s = ut + at2 2
α
=
if,
( −70) = (50 sin 30° ) t +
16. Initial separation
T
53°
Option (b) is correct.
x R
14. First particle :
H max = 102 m ∴ ⇒
u2 sin2 α = 102 2g 102 × 2 g u2 = sin2 60°
[As α =
π = 60°] 3
Second particle : Range of the second particle will be equal to that of particle
5T
x = R + 5T = (u cos α) T + 5T = T [(u cos α) + 5] 2u sin α = [(u cos α) + 5] g 4 2 × 80 × 5 80 × 3 + 5 = 256 m = 10 5 Option (d) is correct.
Projectile Motion
65
JEE Corner Assertion and Reason 1. Assertion is wrong while the explanation
and
as given in reason is correct. ∴ Option (d) is correct.
Thus,
2. Assertion and
Reason both are correct. both carry the same meaning. Option (a) is correct. u2 sin2 α 3. H = [H = maximum height] 2g 2
=
g 2 u sin α g 2 = T 8 g 8
i.e., H ∝ T2 (Reason) ∴ H will become four times if T is made two times. Thus, assertion is correct and also reason is the correct explanation of the assertion. Option (a) is correct. → →
→
4. Reason is correct as A ⋅ B = 0 when A and →
B are perpendicular to each other.
y=
2y 1 2y + ax ay 2 ay
x = ux
or
x = k1 y + k2 y
or
( x − k2 y)2 = k12 y
or
x2 + k22 y2 − 2k2 y − k12 y = 0
or
x2 + k22 y2 − (2 k2 + k12 ) y = 0
→
6.
1 a y t2 2
→
v2 − v1 t2 − t1 →
=
→
→
→
( u + a t2 ) − ( u + a t1) t2 − t1 →
= a (which is a constant quantity in projectile motion). ∴ Assertion is correct and the reason for this also correct, as explained. Thus, option (a) is correct. 7. Let initial velocity of A = u A
Initial velocity of B = uB B is projected at an angle α with horizontal. For time (TA ) taken by A to return to the point of projection :
→ u 45° 45° → v →
In the case mentioned v is perpendicular → to u only at t = T and not at any time t. Option (b) is correct. →
→
[Note : v will never be perpendicular to u if angle of projection is less than 45°.] 1 5. x = ux t + ax t2 2 y
ay
⇒
( − u A ) = ( + u A ) + ( − g) TA 2u TA = A = 4 (s) g
Time taken by B to reach ground 2 uB sin α TB = = 4 (s) g (Q TB = TA given)
a
∴ uB sin α = 2 g ux
Height attained by B
ax x
H=
u2B sin2 α 2g
66 | Mechanics-1 =
(2 g)2 2g
Reason is correct as the given relation under heading reason leads to 2v Range = vH × V g
= 2g = 20 m ∴ Assertion is correct. Reason for this incomplete as then TA will not be equal to TB . Thus, option (c) is correct. u2 8. H= V 2g H1 : H2 = 4 H : H u12V u22 V : = 4 :1 2g 2g
∴ i.e., or
energy …(i) ∴
1 1 mv2 + mgh = mu2 2 2 2 v = u − 2 gh
Thus, assertion is correct.
For R1 = R2 u1V u1H = u2 V u2 H
or
which would decrease by 10 m/s in the downward direction. Acceleration is nothing but rate of change of velocity written against reason is correct. ∴ Option (d) is correct. 10. Using law of conservation of mechanical
u1V : v2 V = 2 : 1 2 uV uH R= g
u1H u2 V = u2 H u1H u1H 1 = u2 H 2
∴ Option (a) is correct. 9. Assertion is incorrect as it is the velocity
If the particle is projected with vertical component of velocity as u y the vertical component of velocity of the particle at height h would be u2y − 2 gh is correct as [Using relation (i)]
∴ Assertion is correct.
written against reason but it is not the correct explanation of the assertion as nothing is given regarding the change in the horizontal component of velocity.
Objective Questions (Level 2) Single Correct Option 1. s = ut +
1 2 at 2
Time for attaining maximum height u sin θ = g
R
u P
16.8 m
Q 4.8 m
θ O
D
1 = (u sin θ) t + ( − gt)2 2 = 24 sin θ − 7 × 2 For θ = 30° : s = 4.8 m
= 1s ∴ Above displacement is for point Q. For θ = 90° : s = 16.8 m For θ = 30° : OD = (u cos θ) t = (20 cos 30° ) 1 × 2 = 20.8 2 ∴ QR = (20.8) + (16.8 − 4.8)2 = 24 m Option (c) is correct.
Projectile Motion 2. t =
d 1 also h = gt2 u 2 u
d
h time = t
5.
1 d2 g⋅ 2 2 u 2 u2 h 2 d = g
h=
∴ ⇒
Option (b) is correct. u2 3. R = [sin (2α + β) + sin β ] g cos2 β Substituting v = 10 m/s g = 10 m/s2 α + β = 90° and β = 30° (10)2 R= [sin (90° + 60° ) + sin 30° ] 10 cos2 30° 10 1 1 40 m = + = 3 2 2 3 4
y = βx2 dy dx ∴ = β ⋅ 2x dt dt d2 x dx 2 d2 y and = 2β x ⋅ 2 + dt dt2 dt d2 x dx 2 or α = 2β x 2 + dt dt 2 d x Now, as 2 = 0 dt (acceleration being along y-axis only) dx α = dt 2β Option (d) is correct.
6. As the projectile hits the inclined plane
horizontally P
u
Option (c) is correct. 1 x2 4. y = x tan θ − g 2 2 u sin θ cos θ
R'
α β
Q
O
Slope
PQ = H max = x
dy 1 2x = tan θ − dx 2 u2 sin θ cos θ Slope
OQ =
u2 sin α cos α g
2
u2 sin2 α u2 sin α cos α = + g 2g
tan θ
O
and
u2 sin2 α 2g
∴ Range on inclined plane R ′ = ( PQ)2 + (OQ)2
T = Time of flight
T/2
67
x Slope of trajectory = tan θ − 2 u sin θ cos θ Substituting x = (u cos θ) t (u cos θ) t dy = tan θ − 2 dx u sin θ cos θ t = tan θ − u sin θ Option (a) is correct.
T t
=
u2 g
sin4 α + sin2 α cos2 α 4
=
u2 g
3 1 2 1 3 + 4 4 2 2
4
2
2
68 | Mechanics-1 =
u2 g
=
u2 21 8g
9 3 + 64 16
Acceleration of the particle d2 x = 2 dt = 50 m/s2
Option (d) is correct. 7. Let the projectile hits the inclined plane at
[as
dy = constant] dt
9. tan α = tan θ + tan φ
P at time t u = 10 √3 m/s i.e., v = u
v = 10 √3 m/s v
H
α θ R/2
φ R/2
u β
α R
∴ P (time = t) β (u cos α) t vt
Further, and i.e.,
or
or or
or or or
dy dy dx +2 = dt dt dt dy dx (2 y + 2) = dt dt dx = 10 ( y + 1) dt d2 x dy = 10 dt dt2
2y
= 50 m/s 2
10. At any time t, 20 m/s
60°
30° 20√3m
OQ tan β = PQ 1 t [v − cos α ] tan β = u sin α − gt t 2 10 [1 − cos 60° ] tan 30° 1 = 10 sin 60° − × 10 × 2 5 = 5 3 − 5t 3 1 t= 3 − 3 2 s = 3
R
Option (c) is correct. 20√3 m/s
1 PQ = (u sin α) − gt2 2 QR = vt − (u cos α) t PQ tan β = QR
8. y2 + 2 y + 2 = x
∴
Q
= tan θ + tan θ = 2 tan θ tan α θ = tan −1 2
t
Horizontal distance between particles x = 20 3 − (10 3 + 10 3 ) t = 20 3 (1 − t) Vertical distance between particles at time t y = (30 − 10) t = 20 t Distance between particles at time t D = x2 + y2 D2 = [20 3 (1 − t)]2 + [20 t ]2 = 400 [3 (1 − t)2 ] + 400 t2 For D to minimum dD =0 dt 3 × 2 (1 − t)( −1) + 2 t = 0 or 6t − 6 + 2t = 0 3 i.e., t= s 4 2 2 3 3 ∴ D2min = 400 3 1 − + 400 4 4 or
= 75 + 75 × 3 ⇒ Dmin = 10 3 m Option (b) is correct.
Projectile Motion
Substituting v = u and α = β = 30° in Eq. (ii) u cos 30° + u cos 30° T= g 2 u cos 30° = g
11. Let time of flight = T u α P
β α
θ
Q β α
69
u
=
β
As, horizontal component of the velocity of the particle will not change u cos (90° − α) = v cos (90° − β) or …(i) u sin α = v sin β Using v = u + at for the horizontal component of velocity ( − v cos β) = ( + u cos α) + ( − g) T ⇒ gT = u cos α + v cos β u cos α + v cos β or …(ii) T= g Now, as α = β ( = 30° ), u = v from Eq. (i)
u 3 g
Option (b) is correct. 2 u2 sin θ cos θ 12. PQ = g or2a cos α =
2 u2 sin (90° − α) cos (90° − α) g
or
a=
u2 sin α g
or (as α = 30°) u2 = 2ag or u2 = 2 × 4.9 × 9.8 or u = 9.8 m/s Option (a) is correct.
More than One Correct Options 1. Two particles projected at angles α and β
with same speed ( = µ ) will have same range if α + β = 90° Option (a) is correct. 2 u2 sin α cos α R= g h1 (maximum height attained by first) u2 sin2 α = 2g h2 (maximum height attained by second) u2 sin2 β = 2g
Option (b) is correct.
P
Wind a=g
a = constant
e
u2 sin α cos α R = 2g 4
flight)
l tic ar fp
=
Option (d) is correct. 2. Horizontal displacement in time t (Time of
o th
u2 sin α sin β 2g u2 sin α sin (90° − α) = 2g
h1h2 =
Option (c) is correct. h1 sin α Also, = = tan α h2 sin β
Pa
∴
2u sin α g 2u sin β t2 (time of flight of second) = g t1 sin α sin α ∴ = = = tan α t2 sin β sin (90° − α) t1 (time of flight of first) =
Q
R
x=
1 2 at 2
…(i)
70 | Mechanics-1 Vertical displacement in time t 1 y = gt2 2 y g ∴ = x a or y = constant x [as a is constant] Option (a) is correct. Substituting y = 49 m in Eqs. (ii) 1 49 = × 9.8 × t2 2 ⇒ t = 10 = 3.16 s PQ > PR ( = 40 m ) Option (d) is correct.
…(ii)
3. As there will be no change in the
horizontal component of the velocity of the particle
v P
θ v cos θ
θ + φ = 90°
Q
φ
w cos φ w
w cos φ = v cos θ or w cos (90° − θ) = v cos θ or w sin θ = v cos θ or w = v cos θ Option (b) is correct. Option (a) is incorrect. Vertical velocity of particle at Q = wsin φ = wsin (90° − θ) = wcos θ Using, v = u + at ( − w cos θ) = ( + v sin θ) + ( − g) T [where T = time from P to Q] ⇒ gT = v sin θ + w cos θ = v sin θ + v cot θ cos θ cos2 θ = v sin θ + v ⋅ sin θ = v cosec θ v cosec θ ∴ t= g
Option (c) is correct. Option (d) is incorrect. →
^
^
4. v = 10 i + 10 j v H O a = – 10 m/s2
– 15m
t
If θ be the angle of projection v y 10 tan θ = = =1 vx 10 i.e., θ = 45° Option (a) correct. 1 2 at 2 1 ( − 15) = 10 t + ( −10) t2 2 2 i.e., t − 2t − 3 = 0 or ( t − 3) ( t + 1) = 0 ∴ t=3s Option (b) is incorrect. Horizontal range of particle = vx ⋅ t = 10 × 3 = 30 m Option (c) is incorrect. Maximum height of projectile from ground = H + 15 u2 sin2 θ = + 15 2g Using relation, s = ut +
= =
v2y 2g
+ 15
(10)2 + 15 2 × 10
= 20 m Option (d) is correct. 5. Average velocity between any two points
can’t remain constant as in projectile motion velocity changes both in magnitude and direction. Option (a) is incorrect. Similarly option (b) is also incorrect.
Projectile Motion In projectile motion →
a = constant
⇒
→
dv = constant dt Option (c) is correct.
∴
→
d2 v =0 dt2 Option (d) is correct. 1 6. ( + h) = ( + u sin α) t + ( − g) t2 2 t = t1 A
α
2
h
gt2 − (2 u sin α) t + 2h = 0 2 u sin α 2h , t1t2 = t1 + t2 = g g
( t2 − t1)2 = ( t2 + t1)2 − 4 t1t2 4u2 sin2 α 2h = = 4⋅ g g ∴ t AB =
40 2 = 20 m/s
ucos α =
⇒
40 m
20 m
⇒
u sin α = 40 g = 20 m/s Option (c) is correct. (u cos α) tAB = 40
Bt=t
15 m
[where H = Maximum height] 8 ( H − 15) 2= 10
∴ H = 20 m Option (b) is correct. u2 sin2 α i.e., = 20 2g
also
v
71
8 ( H − h) g
…(i)
Option (d) is correct. Substituting values of g, usin α and h in Eq. (i) 10t2 − 40t + 30 = 0 i.e., t2 − 4 t + 3 = 0 ∴ t = 1 or 3 t = 1 s for A and t = 3 s for B. Option (a) is correct.
Match the Columns 1. (a) At t = 2 s S1
S2
2 (t = 1 s) 10 m 1 (at t = 1s) 10 m/s
30 m 1 (at t = 2 s)
S3
2 at t = 2 s
S1 : S2 : S3 =1:3:5
(b) Vertical distance between the two at t = 2 s = 30 m ∴ (b) → (s) (c) Relative velocity
horizontal
component
of
= 10 − 0 = 10 m/s ∴ (c) → (p) (d) Relative vertical component of velocity (at t = 2 s) = Velocity of 1 at t = 2 s
50 m
horizontal distance between 1 and 2 = horizontal displacement of 2 = 10 × 1 = 10 m ∴ (a) → (p)
− Vertical velocity of 2 at t = 2 s = Velocity of 1 at t = 2 s − Velocity of 1 at t = 1 s = (2 g) − (1 g) = g = 10 m/s (d) → (p).
72 | Mechanics-1 2. Given, H = 20 m
∴
u2 sin2 α = 20 2g usin α = 20
∴ (d) → (s) Time of flight =
2u sin α =4s g
(a) → (s) R (given) 2 u2 sin2 α 1 2 u2 sin α cos α ∴ = 2g 2 g u sin α ⇒ =2 u cos α ∴ (b) → (q) u sin α u cos α = = 10 m/s 2 ∴ (c) → (r) u2 sin (2 × 15° ) 3. = 10 2g H=
2
∴ ⇒ (a) → (q)
u = 20 m 2g Rmax = 20 m
(b) Maximum height =
u2 = 10 m 4g
∴ (b) → (p) (c) Range at ( 45° + θ) = Range at ( 45° − θ) ∴ Range at 75° = Range at 15° = 10 m (given) (c) → (p) u2 sin2 30° (d) Height at 30° = 2g =
1 u2 4 2g
= 5m ∴ (d) → (s) 2 u sin α 4. (a) T = g 2u⊥ = g
∴ On increasing u⊥ to two times the T (time) of flight will also become two times i.e., (a) → (q). (b) H =
u2 sin2 α u⊥ = 2g 2g
2
H max will become 4 times. ∴ (b) → (r) 2(u⊥ ) (u||) (c) R = g R will become two times (c) → (q) (d) Angle of projection with horizontal u tan α new = ⊥ (new) u|| 2u⊥ = u|| = 2 tan α α new = tan −1 (2 tan α) ≠ 2α ∴ (d) → (s) →
^
^
5. (a) u (at 0) = (u cos θ) i + (u sin θ) j →
v (at A) = (u cos θ) ^i
|Change in velocity between A & B| = usin θ (a) (q) ∴ → (b) |Average velocity between 0 and A| → |OA| = T /2
=
=
R H2 + 2
2
T /2 u2 sin2 θ 2 1 2 u sin θ cos θ 2 + 4 g 2 g u sin θ g
u 1 + 3 cos2 θ 2 ∴ (b) → (s) =
Projectile Motion (c) Velocity at B → vB →
73
∴ Vertical displacement of (2) w.r.t. (1), ^
^
= (90 ^i − 40 ^j) m
= (u cos θ) i − (u sin θ) j
→
= 50 ^j m
∴|v B − u| = 2 u sin θ Thus, (c) → (s). (d) |Average velocity between 0 and B| Range = Time of flight (u cos θ) T = = ucos θ T ∴ (d) → (p). 6. (a) Horizontal displacement of (1) at t = 2 s Y
∴ (b) → (r) (c) Horizontal velocity of (1) at t = 2 s →
v1H = 30 ^i
Horizontal velocity of (2) at t = 2 s →
v2 H = − 20 ^i
Relative horizontal component of the velocity of (2) w.r.t. 1 →
→
→
v RH = v2 H − v1H
20 j
= ( − 20 ^i) − ( + 30 ^i)
(2) – 20 i
= − 50 ^i m/s
30 j
→
∴|v RH|= 50 m/s X
(1)
30 i
Thus, (c) → (r) (d) Vertical velocity of (1) at t = 2 s
→
→
S1H = 30 ^i m/s × 2 s
v 1 V = 30 ^j + ( − 10 ^j) (2)
= + 60 ^i m
= 10 ^j
Horizontal displacement of (2) at t = 2 s →
^
Vertical velocity of (2) at t = 2 s →
^
v 1 V = 20 ^j + ( − 10 ^j) (1)
S2 H = 130 i m + ( −20 i m / s) (1 s) = 110 ^i m
= 10 ^j
Horizontal displacement of (2) w.r.t. (1) ^
^
^
= (110 i − 60 i) m = 50 i m ∴ (a) → (r) (b) Vertical displacement of (1) at t = 2 s → 1 S1V = 30 ^j m/s × 2 s + ( − 10 ^j m / s2 ) (2 s)2 2 = 60 ^j m − 20 ^j m = 40 ^j m Vertical displacement of (2) at t = 2 s →
S2 V = 75 ^j m + (20 ^j m / s) (1 s) 1 + ( −10 ^j m / s2 ) (16)2 2 = 90 ^j m
Relative vertical component of the velocity of (2) w.r.t. 1 →
v RV = (10 ^j) − (10 ^j) = 0
→
∴ |v RV|= 0 m/s Thus (d) → (s) 2 uy 7. T = ∴ T ∝ uy g H=
u2y 2g
∴ H ∝ u2y
H A = HB = HC ∴ TA = TB = TC Further, R = ux T T is same. But R A < RB < RC ∴ u XA < u XB < u XC
5
Laws of Motion Introductory Exercise 5.1
1. N = Normal force on cylinder by plank R
θ
w
N ′ sin θ + w A = N
Force acting on cylinder f
N
R = Normal force on cylinder by ground, f = Force of friction by ground by cylinder, w = Weight of cylinder. N cos θ = f w + N sin θ = R, N ( R) = Reaction to N,
N (R) B R'
θ
Force of sphere B
wB
wall,
O (Force acting on plank)
R' w
θ P
f'
Ground
i.e.,normal force on plank by cylinder R′ = Normal force on plank by ground, w = Weight of plank, f ′ = frictional force on plank by ground. Resultant of f ′ and R′, N ( R) and w pass through point O.
N ( R) = Reaction to N i.e. normal force on sphere B by sphere A, wB = Weight of sphere B, R′, N ( R) and wB pass through point O, the centre sphere B. 3. N = Normal force on sphere by wall, C θ
N
2.
B A
A θ
O
R Force on sphere A w
N'
R
O
R′ = Normal force on sphere B by right
N(R)
Q
R = Normal force on sphere A by left wall, N = Normal force on sphere A by ground, N ′ = Normal force on sphere A by sphere B, w A = Weight of sphere A. N ′ cos θ = R
w
N
Laws of Motion w = Weight of sphere, T = Tension in string. 4. Component
7. R cos 30° + 3 = f cos 60° R
f
→ of F1
60°
30° 3N
along x-axis : 4 cos 30° = 2 3 N along y-axis : 4 sin 30° = 2 N Component
i.e.,
along y-axis : 4 sin 120° = 2 3 N
or and
→ of F3
along x-axis : 6 cos 270° = 0 N
i.e.,
along y-axis : 6 sin 270° = − 6 N
or
Component
10 N
60°
→ of F2
along x-axis : 4 cos 120° = − 2 N
Component
→ of F4
along x-axis : 4 cos 0° = 4 N along y-axis : 4 sin 0° = 0 N 5. Taking moment about point A T
O A
T sin 30°
30° AB = l
B
w
R 3 f +3= 2 2 R 3 +6= f R sin 30° + f sin 60° = 10 1 3 R +f = 10 2 2 R + f 3 = 20
( T sin 30° ) l = w
…(i)
…(ii)
Substituting the value of f from Eq. (i) in Eq. (ii) R + ( R 3 + 6) 3 = 20 4 R + 6 3 = 20 20 − 6 3 ⇒ R= = 2.4 N 4 ∴ f = (2.4) 3 + 6 = 10.16 N 8. At
point B (instantaneous acceleration only)
vertical
N
AB = l
A
l 2
T T
T=w
⇒
75
B
45°
6. See figure (answer to question no. 3)
OA OB + BC a 1 = = a+a 2
sin θ =
or or
T cos 30° = w 3 T =w 2 2 T= w 3
mg
∴
mg − T sin 45° = ma
…(i)
At point A (instantaneous horizontal acceleration only) …(ii) ∴ T cos 45° = ma Combining Eqs. (i) and (ii) mg − ma = ma g ⇒ a= 2
76 | Mechanics-1 Introductory Exercise 5.2 1. Acceleration of system
a=
( + 120) + ( − 50) 1+ 4 +2
n co v(
sta
nt)
= 10 m/s2 Let normal force between 1 kg block and 4 kg block = F1 ∴ Net force on 1 kg block = 120 − N 120 − F1 ∴ a= 1 or 10 = 120 − F1 i.e., F1 = 110 N Net force on 2 kg block = 2 × a = 2 × 10 = 20 N 2. As, 4 g sin 30° > 2 g sin 30°
The normal force between the two blocks will be zero. mg 3. N ( R) = 4
T θ mg
θ =30°
Tension in spring = mg = 1 × 10 = 10 N 5. Pseudo force ( = ma) on plumb-bob will be
as shown in figure a T φ ma θ °– 90 mg
N a
θ = 30°
ma A mg N (R) = N
∴
N=
mg 4
As lift is moving downward with acceleration a, the pseudo force on A will be ma acting in the upward direction. For the block to be at rest w.r.t. lift. N + ma = mg mg or + ma = mg 4 3 ⇒ a= g 4 4. Angle made by the string with the normal
to the ceiling = θ = 30° As the train is moving with constant velocity no pseudo force will act on the plumb-bob.
T cos φ = mg + ma cos (90° − θ) i.e., …(i) T cos φ = mg + ma sin θ and T sin φ = ma cos θ Squaring and adding Eqs. (i) and (ii), T2 = m2 g2 + m2 a2 sin2 θ + 2m2 ag sin θ …(iii) + m2 a2 cos2 θ 2 2 2 (Qθ = 30°) T = m g + m2 a2 + m2 ag g g2 g = m2 g2 + m2 + m2 ⋅ g (Q a = ) 4 2 2 7 m2 g2 = 4 mg 7 or T= 2 =5 7 N Dividing Eq. (i) by Eq. (ii), ma cos θ tan φ = mg + ma sin θ
Laws of Motion a cos θ = g + a sin θ cos θ = 2 + sin θ cos 30° = 2 + sin 30°
6.
2kg
1kg
F=8N
8 = 2 m/s 2 2+1+1 B
T1 T1
1kg T2 T2
A
Net force on 1 kg mass = 8 − T2 ∴ 8 − T2 = 1 × 2 ⇒ T2 = 6 N Net force on 1 kg block = T1 ∴ T1 = 2a = 2 × 2 = 4 N
3 φ = tan −1 5
i.e.,
B
a=
3 5
=
A
2kg
Introductory Exercise 5.3 1. F = 2 g sin 30° = g T1
T2
T1 T1
T2 F
T1
T1
T1
mg
T1
T2 T2
4g
T2
2g
1g
For the system to remain at rest
or
3g
T2
T2 = 2 g
…(i)
T2 + F = T1
…(ii)
T2 + g = T1
…[ii (a)]
T1 = mg
…(iii)
Substituting the values of T1 and T2 from Eqs. (iii) and (i) in Eq. [ii(a)]
∴ and
T1 = 4 g T2 = 1 g
∴
T1 =4 T2
3. 2 g − T = 2a
2g + g = mg i.e.,
77
m = 3 kg
2. As net downward force on the system is
T
zero, the system will be in equilibrium
T
T
T
a
a 1g 2g
8N
78 | Mechanics-1 T − 1 g = 1a Adding above two equations 1g = 3a g ∴ a= 3
=
u2 ( g / 3)2 g = = 2a 2g 18
Downward displacement of 2 kg block 2 1 1 g g = at2 = g ⋅ = 3 2 2 18
Velocity of 1kg block 1 section after the system is set in motion v = 0 + at g = ⋅1 3 g (upward) = 3
As the two are just equal, the string will 1 again become taut after time s. 3 …(i)
F + 1 g − T = 1a
4.
On stopping 2 kg, the block of 1kg will go upwards with retardation g. Time ( t′ ) taken by the 1 kg block to attain zero velocity will be given by the equation. g 0 = + ( − g) t ′ 3 1 ⇒ t′ = s 3
T
T T
a
F
1g
T
a
2g
If the 2 kg block is stopped just for a moment (time being much-much less than 1 s), it will also start falling down when 3 the stopping time ends. 1 In t′ = s time upward displacement of 3 1 kg block
and
…(ii)
T − 2 g = 2a
Adding Eqs. (i) and (ii), F − 1g = 3a 20 − 10 10 ms −2 ∴ a= = 3 3
Introductory Exercise 5.4 …(i)
2T = 2 × a
1. a 2 kg
T
T
T
T
∴ Acceleration of 1 kg block 2 g 20 2a = = ms −2 3 3 Tension in the string g 10 N T= = 3 3
2T 2T T T 2a 1g
and
1 g − T = 2a
Solving Eqs. (i) and (ii), g a= 3
…(ii)
…(i)
Mg − T = Ma
2.
T T
T
T
M a
T T a mg
Laws of Motion …(ii)
T = Ma Solving Eqs. (i) and (ii) g a= 2 and T = Mg
T T M
or
Substituting value of above value of T in Eq. (i), 24 M= 5 = 4.8 kg T 4. = m1 ⋅ 2a 2
3. Block of mass M will be at rest if
T
T
T/2
Mg
a
T/2 T
2a
T
T/2 T/2
T/2
F
T/2
i.e.,
T/2 T/2
a
79
g T = 4 g + 5 24 g = 5
…(ii)
F − T = m2 a
a
or
3g
or
2g
T = Mg
…(i)
T = 4 m1a
…(i)
For the motion of block of mass 3 kg T …(ii) 3g − = 3a 2 For the motion of block of mass 2 kg T …(iii) − 2 g = 2a 2 Adding Eqs. (ii) and (iii), g = 5a g i.e., a= 5 Substituting above value of a in Eq. (iii), T = 2 ( g + a) 2
F − 4 m1a = m2 a F a= 4 m1 + m2 =
0.40 ( 4 × 0.3) + 0.2
0.40 1.4 2 = ms −2 7 =
∴
T = 4 m1a = 4 × 0.3 × 2.4 7 12 N = 35
=
2 7
80 | Mechanics-1 Introductory Exercise 5.5 1 2 at 2 Displacement of block at time t relative to car would be
1. Block on triangular block will not slip if m1a cos θ
2. (a) Using s = s0 + ut +
N
m1a N sin θ
m1 g
m1g sin θ cos
m1a sin θ
a = 5i ms–2
a = – 5i ms–2 u = 10i ms–1
θ
v=0
(at t = 0)
θ
O
T N
Car
x0
T
1 ( −5) t2 2 x = x0 + 10 t − 2.5 t2 x = x0 + 10 t +
a
M Mg
m1a cos θ = m1 g sin θ i.e., …(i) a = g tan θ N = m1 g cos θ + m1a sin θ …(ii) For the movement of triangular block …(iii) T − N sin θ = m2 a For the movement of the block of mass M …(iv) Mg − T = Ma Adding Eqs. (iii) and (iv), Mg − N sin θ = ( m2 + M ) a Substituting the value of N from Eq. (ii) in the above equation Mg − ( m1 g cos θ + m1a sin θ) sin θ = ( m2 a + Ma) i.e., M ( g − a) = m1 g cos θ sin θ + ( m2 + m1 sin2 θ) a Substituting value of a from Eq. (i) in the above equation, M (1 − tan θ) = m1 cos θ sin θ + ( m2 + m1 sin2 θ) tan θ m cos θ sin θ + ( m2 + m1 sin2 θ) tan θ ∴M = 1 (1 − tan θ) Substituting θ = 30° , m1 = 1 kg and m2 = 4 kg (cos 30° sin 30°+4 + sin2 30° ) tan 30° M= (1 − tan 30° ) 0.443 + ( 4.25) (0.577) = 0.423 = 6.82 kg
x
or
Velocity of block at time t (relative to car) will be dx v= = 10 − 5t dt (b) Time ( t) for the block to arrive at the original position (i.e., x = x0 ) relative to car x0 = x0 + 10 t − 2.5 t2 ⇒ t=4s 3. (a) In car’s frame position of object at time
t would be given by In car’s frame (at t = 0 s) O x0 z0 a = –5 ms–2
x
a = 5i ms–2 v=0
u = 10k ms–1
z
1 ( − 5) t2 2 i.e., …(i) x = x0 − 2.5 t2 and …(ii) z = z + 10 t Velocity of the object at time t would be dx …(iii) vx = = − 5t dt dz and …(iv) vz = = 10 dt (b) In ground frame the position of the object at time t would be given by x = x0 + 0 × t +
Laws of Motion
∴ Net deceleration = (3 + 5) m/s = 8 m/s 2 ∴ Displacement of object at any time t (relative to car) 1 x = x0 + 10t + ( −8) t2 2 or x = x0 + 10t − 4 t2
In ground frame O x0
At t = 0 a = 5i ms–2 v=0
z0 u = 10k ms–1
Thus, velocity of object at any time t (relative to car) dx vx = = 10 − 8 t dt
x = x0 z = z0 + 10 t
and
Velocity of the object at time t would be dx vx = =0 dt dz and vz = = 10 ms −1 dt
The object will stop moving relative to car when 10 − 8 t = 0 i.e., t = 1.25 s ∴ vx = 10 − 8 t for 0 < t < 1.25 s
4. m = 2 kg O
For block not to slide the frictional force ( f ) would be given by
u = 10i ms–1
x0
x
a = 5i ms–2 v=0
f
a = 3 ms–2
ma mg sin θ θ
θ = 37°
z
Normal force on object = mg Maximum sliding friction = µ s mg = 0.3 × 2 × 10 = 6 N 6 Deceleration due to friction = = 3 m/s2 2 Deceleration due to pseudo force = 5 m/s2
f + ma cos θ = mg sin θ f = mg sin θ − ma cos θ 3 4 = m × 10 × − m × 3 × 5 5 18 m 9 = = mg 5 25
or
AIEEE Corner Subjective Questions (Level 1)
T1 and T2 = 2 T At point Q, F2 = 1 and W = 2
1. FBD is given in the answer.
At point P, F1 =
2. FBD is given in the answer. 3. FBD is given in the answer. 4.
5. T2 F1
P
81
2
NB 45° T1
30° T1 45°
NA F2
Q
W
W
N A sin 30° = N B and N A cos 30° = W
T1 2 T1 2
82 | Mechanics-1 3T =W 2 T H= 2 Net moment about O = zero l 3T ∴. W × = ×l 2 2 100 − 40 10. (a) a = = 3 m/s2 6 + 4 + 10
6. T2
V +
45° T1
W
T2 = W and 2 T2 = T1 2
(c) N − 40 = F10 = 30 ∴ N = 70 N. F 60 11. a = = = 1 m/s2 m1 + m2 + m3 60 (a)
40 N x N 10 cm
N = 40 f = 20 f × 10 = N × x
…(i) …(ii) …(iii)
T1 = m1a = 10 N T2 − T1 = m2 a ∴ T2 − 10 = 20 ∴ T2 = 30 N (b) T1 = 0 . New acceleration 60 a′ = = 1.2 m /s2 50 T2 = m2 a ′ = 24 N 12. (a)
N
8.
T
T1
a
30°
0.1 kg
1.9 kg
2g
f
T1 − 2 g = 2a 30°
(b)
T2
W
9.
T + f cos 30° = N sin 30°
…(i)
N cos 30°+ f sin 30° = W T × R = fR
…(ii) …(iii)
V
O
3T 2 H
T/2 W
…(iii)
F6 = 18 N, F4 = 12 N and F10 = 30 N
∴ f
and
…(ii)
(b) Net force = ma
7.
W = 20 N
…(i)
2.9 kg a
0.2 kg 1.9 kg
T2 − 5 g = 5a
5g
Laws of Motion 200 − 16 g 13. (a) a = 16 (b) T1 − 11 g = 11a (c) T2 − 9 g = 9 a
20.
83
a T2 = T1 1 T1
14. If the monkey exerts a force F on the rope upwards, then same force F transfers to bananas also. If monkey releases her hold on rope both monkey and bananas fall freely under gravity.
T1
a 2
2T1
15. Tension on B = T Tension on A = 3 T 1 Now in these situations a ∝ T
T1 ar
T1 2
16. x A + xC + 2xB = constant.
3
Differentiating twice w.r.t. time we get the acceleration relation.
θ
aA
T
2T a 2
M
aA = sin θ aB
g M
a A = aB sin θ
∴
a
21.
aB θ
…(i) …(ii) …(iii)
T1 = 1a T1 − 20 = 2 ( ar − a / 2) 30 − T1 = 3( ar + a / 2)
17.
18. x + y = 6
30
2M
°
2 Mg
…(i)
T − Mg sin 30° = Ma a 2Mg − 2T = 2M ⋅ 2
…(i)
y−x=4 Solving, we get x = 1 m/s2
n si
…(ii)
…(ii)
22. T = 1a 10 − T = 1a
x
23.
ar
…(i) …(ii)
2T
2T
B
A
50
40
1 x 2a
a y 2 y 3
19.
7g − 3g = 4 m/s2 10 40 − T1 = 4 a 30 + T1 − T2 = 3 a T3 − 10 = 1a
a=
50 − 2T = 5a T − 40 = 4(2a) 24. (a) N = 40 N, µ s N = 24 N F < µsN ∴ f = 20 N and a = 0 (b) N = 20 N, µ s N = 12 N and µ k N = 8 N
…(i) …(ii)
84 | Mechanics-1 F > µsN f = µ kN = 8 N 20 − 8 a= = 6 m/s 2 2
∴ and
27. f = (06 . )(2)(10) = 12 N a2 2 kg
1 kg
µ s N = 8 N and Since, F cos 45° > µ s N ∴ and
f = µ kN = 4 N 20 − 4 a= 6 8 = m/s2 3
25. a = µ g = 3 m/s2 (a) v = at ∴ 6 = 3 t or t = 2 s 1 (b) s = at2 2 1 ∴ s = × 3 × 4 = 6 m. 2 26. f = 0.4 × 1 × 10 = 4 N a1 1 kg f a2 2 kg f
a1 =
f = 4 m/s2 1
a2 =
f = 2 m/s2 2
(a) Relative motion will stop when v1 = v2 or 2 + 4 t = 8 − 2t ∴ t = 1s (b) v1 = v2 = 2 + 4 × 1 = 6 m/s 1 (c) s1 = u1t + a1t2 2 1 s2 = u2 t − a2 t2 2
+ve
a1
(c) N = 60 − 20 = 40 N µ kN = 4 N
–ve
f f
12 = 6 m/s2 2 12 and a1 = = 12 m/s2 1 (a) Relative motion will stop when v1 = v2 or u1 + a1t = u2 + a2 t or 3 − 6t = −18 + 12t 7 ∴ t= s 6 (b) Common velocity at this instant is v1 or v2 . 1 (c) s1 = u1t + a1t2 and 2 1 s2 = u2 t + a2 t2 2 a2 =
28. N = 20 N µ s N = 16 N and µ k N = 12 N Since, W = 20 N > µ s N, friction µ k N will act. 20 − 12 ∴ a= = 4 m/s2 2 29. N = 20 N µN = 16 N Block will start moving when F = µN or 2t = 16 or t = 8 s. After 8 s 2t − 16 a= = t−8 2 i.e., a-t graph is a straight line with positive slope and negative intercept. 30. N = 60 N, µ s N = 36 N, µ k N = 24 N Block will start moving when F = µsN
Laws of Motion or ∴
(b)
4 t = 36 t=9s
85
F
v
12 N
After 9 s a=
4 t − 24 2 = t−4 6 3
52 N
F + 12 = 52 F = 40 N
31. N = mg cos θ = 30 N ∴
mg sin 30° = 30 3 N ≈ 52 N. µ s N = 18 N and µ k N = 12 N F (a)
(c)
F
a
52 N
18 N
12 N
F − 52 − 12 = 6 × 4 ∴ F = 88 N
52 N
F = 52 − 18 = 34 N. Objective Questions (Level-1) Single Correct Option
mg − F F =g− m m m A > mB ∴ a A > aB or ball A reaches earlier. 4 g − 2g g 2. a = = 6 3 g Now, 2 g − T = 2 3 4g ∴ T= = 13 N 3
mg − T = ma
1. a =
∴
Tmax m
2 mg g =g−3 = m 3 10 g − 5 g g 5. a = = 15 3 g 10 g − T = 10 × 3 20 g ∴ T= = Reading of spring 3 balance. 2mg sin 30°− mg 6. a = =0 3m ∴ T = mg
T1
3.
amin = g −
30° T2
7. a1 = g sin θ − µ g cos θ = g sin 45°−µ g cos 45° 100 N
3 T1 = 100 2 T1 = T2 2 4.
T a
mg
…(i) …(ii)
a2 = g sin θ = g sin 45° 2s 1 Now t = or t ∝ a a t1 a2 ∴ = t2 a1 g sin 45° g sin 45°− µ g cos 45° 3 Solving, we get µ = 4 or
2=
86 | Mechanics-1 8. F1 = mg sin θ + µmg cos θ
14. f = mg sin θ
F2 = mg sin θ − µmg cos θ Given that F1 = 2F2
(if block is at rest)
15. 2T cos 30° = F 60°
9. For equilibrium of block, net force from plane should be equal and opposite of weight.
T 30°
a
F
10. No solution is required. 11. Angle of repose θ = tan −1(µ ) = 30°
∴
θ h
F 3 T cos 60° a= m F = 2 3m
T=
F
16. ∴
3 h = R − R cos θ = 1 − R 2
θ
12. Net pulling force = 15 g − 5 g = 10 g = F Net retarding force = (0.2)(5g) = g = f F−f 9 ∴ a= = g 25 25 9 T1 − 5 g = 5a = g 5 34 ∴ T1 = g 5 27 15 g − T2 = 15a = g 5 48 g ∴ T2 = 5 T1 17 ∴ = T2 24 13.
N = mg − F sin θ µN = (tan φ) N = (tan φ)( mg − F sin θ) F cos θ = µN 17. µmg = 0.2 × 4 × 10 = 8 N At t = 2 s, F = 4 N Since F < µmg ∴ Force of friction f = F = 4 N 18. a1 = g sin θ a2 = g sin θ − µg cos θ t=
t1 = t2
a
1 = 2 θ
v
Relative to lift, ar = ( g + a) sin θ along the plane. 2s Now, t = ar 2L = ( g + a) sin θ
25 1 or t ∝ a a a2 a1
g sin θ − µg cos θ g sin θ
19. Net accelaration of man relative to ground = a + a = 2a T − mg = m (2a) ∴ T = m( g + 2a) 20.
a=0 3 F = (50 + 25) g = 75 g ∴ F = 25 g = 250 N
Laws of Motion 21.
A
N1
1 + µ2 θ
W
B
3m
O
1
µN2
or
22.
tan θ = µ
=
1 + µ2 µ
mg
∴ f = 30 N 30 N > µmg ∴ Normal reaction N = mg = 40 N
N
N2 = W = 250 fmax = µN2 = 75 N
1 Mg g 2 = M 2
Mg −
(always)
∴ Net contact force = (30)2 + ( 40)2 = 50 N 26.
2 kg
a
0.2
8 kg
25 N
F f
f
∴
1 + µ2
Since
θ f
T−
µ2
+
25. µmg = 32 N
N2
∴
1
1 + µ2 1 = mg 2
N1
W
µmg
Fmin =
N1 = µ N2 N2 = W Net moment about B should be zero. 3 ∴ W × = N1 × 4 2
23. a =
µ
N2
4m
87
1 M M Mg × ×g= ×a= 2 2 2 4 Mg T= 2 F
24. θ
N = mg − F sin θ F cos θ = µ N = µ ( mg − F sin θ) µ mg F= cos θ + µ sin θ dF For F to be minimum, =0 dθ
0.5
fmax = (0. 5)(8 + 2)(10) = 50 N > 25 N ∴ Blocks will not move and therefore force of friction between two blocks = 0. 3 27. mg sin θ = 10 × 10 × = 60 N 5 This 60 N > 30 N ∴ Force of friction is upwards. Net contact force is resultant of friction and normal reaction. 28. In first figure T = F and in second figure, T = 2F . mg 29. F = 4 → upwards 2 W = 2mg → downwards F =W ∴ a=0
88 | Mechanics-1 30.
t=
∴
m m
2 µm g a
31. a1 = g sin 30° = g/2
F = at
a2 = g sin 60° =
Maximum acceleration of upper block due to friction µmg amax = =µ g m F at = =µ g 2m 2m
3 g 2 →
→
Angle between a 1 and a 2 is 30°. ∴ →
→
ar = |a 1 − a 2|=
( g/2)2 + ( 3 g/2)2 − 2 ( g/2) ( 3 g/2) cos 30°
= g/2 JEE Corner Assertion and Reason
1. Even if net force = 0, rotational motion can take place. 2. No solution is required. 3. a1 = g sin θ + µ g cos θ a2 = g sin θ − µ g cos θ a1 1 + µ as θ = 45° and sin θ = cos θ = a2 1 − µ 1 we have, 3 a1 1 + 1 / 3 2 = = a2 1 − 1 / 3 1
Substituting µ =
∴
T = m1( g − a) 2m1m2 g = ( m1 + m2 )
Mathematically, we can prove that m2 g < T < m1 g Similarly if m2 > m1, then we can prove that, m1 g < T < m2 g
4. There is no force for providing ( − 2 i$) m/s2 to the block. 5. If we increase F1, maximum value of friction will increase. But if we increase F2 friction acting on the block will increase. 6. No solution is required. ( m1 − m2 ) g 7. a = m1 + m2
m1 g − T = m1a
(if m1 > m2 )
8. If accelerations of both the frames are same then one frame as observed from other frame will be inertial. Further, a frame moving with constant velocity is inertial. 9. No solution is required. 10. No solution is required. 11. Force of friction is in the direction of motion.
Laws of Motion
89
Objective Questions (Level 2) Single Correct Option 3. N sin θ will accelerate B towards left.
1. Tmax = Fmax + fmax + 2T ′ max A T
T
B
T'
T'
f
T T
N
2T
N sin θ
F
T
90° – θ ma (pseudo force)
A
T' T' θ
mg co
N
T mg
= µ ( m A + mB ) g + µ m A g + 2µm A g = µ ( 4m A + mB ) g But, Tmax = M max ⋅ g ∴ M max ⋅ g = µ ( 4 m A + mB ) g i.e., M max = µ ( 4 m A + mB ) = 0.3 × [( 4 × 100) + 70] = 81 kg Option (c) is correct.
Option (b) is correct. 4. If a ≤ µg
v = 8 t ^i − 2 t2 ^j
m2
m1
a
→
→
a=
∴
θ
Let a be the acceleration of B. Due to acceleration of B towards left pseudo force equal to ma will act on block, toward right. Thus, N + ma sin θ = mg cos θ ⇒ N = mg cos θ − ma sin θ
→
2.
sθ
B
dv dt
m2a
^
m1a
^
= 8 i − 4t j y
m1
2i 4j → g
x → a
∴ Force (pseudo) on sphere →
Fs = 1 ⋅ ( − 8 ^i + 4 ^j ) = − 8 ^i + 4 ^j Gravitational force on sphere = − mg ^j = − 10 ^j ∴ Net force on sphere →
F = − 8 ^i + 4 ^j − 10 ^j = − 8 ^i − 6 ^j →
|F |= 10 N Option (b) is correct.
N=0 f2
f1
N=0
→ a
m2
When a is f1
N
f2
Zero µg 2
Zero µg m1 2
Zero Zero
Zero µg m2 2
µg
m1 µg
Zero
m2 µg
(Not yet attained limiting value which is 2 µm1 g)
(Attained limiting value)
Now, if a > µg, the block of mass m2 would be just greater like to move towards left than µg as a then (max) m2 a > f2 While the block of mass m1 will remain at rest as then f1 has margin to increase. Thus, m1 will apply normal force ( N ) or m2 and so will do m1 on m2 to stop its motion (towards left). Option (d) is correct.
90 | Mechanics-1 Fig. 3
5. N + ma sin θ = mg cos θ N
T a3
ma sin θ ma (pseudo force)
a
in gs
m1
mg sin θ mg
The block will fall freely if N = 0 i.e., ma sin θ = mg cos θ ⇒ a = g cos θ
and ∴
Option (c) is correct. 6. Speed of A w.r.t. C = v1
A
Speed of C w.r.t. ground = v0 ∴ Speed of A w.r.t. ground = v1 − v0 Now, speed of B w.r.t. ground = speed of A is w.r.t. ground ∴ v2 = v1 − v0 ⇒ v1 − v2 = v0 Option (a) is correct.
C
B
7. Fig. 1 m1
T
T T T
T
T
T
θ θ
a3
m2g
m2 g − T = m2 a3 T − m1 g sin θ = m1a3 m − m1 sin θ a3 = 2 g m1 + m2
Substituting m1 = 4 kg, m2 = 3 kg and θ = 30° 3 a1 = g 4+3 3 = g 7 4−3 a2 = g 4+3 1 = g 7 1 3 − 42 ⋅ 2g a3 = 4+3 1 = g 7 ∴ a1 > a2 = 3 Option (b) is correct. 8. F − fmax − T = ma (For lower block)
m2g
and
T
T
⇒
T
T T
T mg 2 mg 1
F
T
T − fmax = ma (For upper block) ⇒ F − 2 fmax = 2ma or F − 2 µmg = 2ma F or a= −µ g 2m Option (c) is correct.
Fig. 2 and
f
f
m2 g − T = m2 a1 m2 a1 = g m1 + m2 m1 g − T = m2 a2 T − m2 g = m1a2 m − m2 a2 = 1 g m1 + m2
T
a2
a1
9. a2 cos(90° − θ) = a1
i.e., a2 sin θ = a1 [θ < 0°] Option (b) is correct.
θ
a2 θ
Laws of Motion 10. Had θ been 90°.
⇒ ∴
v1 v1 90° – θ θ m
v2
B w
T = m ( g + a) M−m = m g + g M + m M − m = mg 1 + M + m
ma cos θ
12. At the position of maximum deflection the
net acceleration of bob (towards its mean position will be) g sin θ − a cos θ as explained in figure.
θ 10° 20°
F (force of friction) mg sin 10° = 0.174 mg mg sin 20° = 0.342 mg
30° 40°
µmg cos 30° = 0.500 mg µmg cos 40° = 0.442 mg
60° 90°
µmg cos 60° = 0.287 mg Zero
F
θ a cos θ O a 90°– θ
T − mg = ma
T m
T
φ = 30°
O
ma cos θ > mg sin θ i.e., a > g tan θ ∴ amin = g tan θ [cylinder will be at the point of rising up the inclined plane]
13. Mg − T = Ma
T
T
For θ ≥ φ frictional force ( F ) = µ mg cos θ [φ = angle of repose] For example 1 1 Let µ = ∴ tan φ = 3 3
inclined plane if
g
2T 2T
14. For 0° < θ ≤ φ frictional force ( F ) = mg sin θ
11. The cylinder will start rising up the
ma (Pseudo furce)
91
mg = mg [1 + 1] {as m << M , M − m ≈ M and Mg M + m ≈ M} = 2 mg ∴ Tension in the string suspended from ceiling = 2T = 4 mg Option (a) is correct.
For every x displacement of wedge ( w) the vertical fall in mass would have been 2x as the string passes through pulley B. i.e., v1 = 2 v2 For the situation given the speed of mass would thus be v1 = 2 v2 cos (90° − θ) i.e., v1 = 2 v2 sin θ Option (c) is correct.
M−m a= g M+m
g cos (90° – θ) = g sin θ
φ
90° θ
Option (b) is correct. µ mg 15. aB = 2m µg = 2
Condition Increase but not linearly only Decrease but not linearly only
92 | Mechanics-1 µmg
F F
Am
cos ma F
B µmg
2m
a
F F
mg
For no slipping a A = aB F − µ mg µ g = m 2 3 i.e., F = µmg 2 2 F µ= 3 mg
16. Fnet (downward) = mg sin θ + ma cos θ a ma
θ
2×1 3 1 10 3 × _ 10 × 2 2 1 s = 5
t=
Option (b) is correct. w w 18. N + sin 30° = w + 2 2
ma cos θ mg sin θ θ
= m ( g sin θ + a cos θ) g eff = g sin θ + a cos θ
Time ( T) required to cover 2L distance along inclined would be 2L T= g eff 2L ( g sin θ + a cos θ)
Option (c) is correct. 17. Fnet on block along incline in the upward
direction
∴
θ
Substituting s = 1 m, θ = 30°, a = 10 3 m/s 2 and g = 10 m/s 2
For no slipping. Option (c) is correct.
=
sin
∴ Time ( t) to move s distance would be given by 1 s = g eff t2 2 2s 2s i.e., t = = g eff ( a cos θ − g sin θ)
Slipping will obviously be m2 there if µ is greater than above mentioned value 2 F ∴ µ min = 3 mg
∴
θ
= ma cos θ − mg sin θ = m ( a cos θ − g sin θ) g eff = a cos θ − g sin θ
N
w/2
w sin 30° w/2 2 30°
w cos 30° 2
f w
f = frictional force 5w 4 5w fmax = µ N = µ ⋅ 4 The block will remain stationary if µ 5w w cos 30° ≤ 2 4 w 3 5w or ≤µ 2 2 4 or 3 ≤ 5µ 3 or ≤µ 5 3 ∴ Block will move if µ < 5 Option (d) is correct. N=
Laws of Motion | T
19. F1 = 2t f1
T
F = 15N
20. 2 mg sin α − T = 2ma y
f2 µ2 = 0.5
µ1 = 0.6
60°
At t = 2 s F1 = 4 N 4N
B
j
P T
2m
T
A
T T
15N
T
α = 30°
f1 and f2 are the frictional forces
and ∴
( f1) max = 0.6 × 1 × 10 = 6N = 10 N
∴
At t = 2 s
Force ( R) applied by clamp on pulley would be
Net external force ( Fnet ) on system = 15 N − 4 N
→
= 11 N As Fnet > ( f1) max + ( f2 ) max , the system will remain at rest and the values frictional forces on the blocks will be given T = 4 + f1 and T = 15 − f2 4 + f1 = 15 − f2
…(i)
f1 + f2 = 11 N
→
|T1|= |T2|= T → mg mg T1 = (cos 30° ) ^i + (sin 30° ) ^j 3 3 mg 3 ^ mg ^ = i+ j 6 6 → mg ^ |T2|= j 3 Force by clamp on pulley P y → T2
Let direction being + ive for Eq. (i)
→ T1
Option (a) f1 = − 4 N, f2 = = 5 N ⇒ f1 + f2 = 1 N
wrong
60° x
Option (b) f1 = − 2 N, f2 = + 5 N ⇒ f1 + f2 = 3 N
wrong
Option (c) f1 = 0 N,f2 = + 10 N ⇒ f1 + f2 = 10 N
wrong
Option (d) f1 = + 1 N, f2 = + 10 N ⇒ f1 + f2 = 11 N correct. OR As the likely movement would be towards right f2 will be at its maximum. ∴
f2 = 10 N
⇒
f1 = 1 N
Option (d) is correct.
m
T
T = ma 2 mg sin α = 3 ma 2 g sin 30° g a= = 3 3 mg T= 3
⇒
( f2 ) max = 0.5 × 2 × 10
x
i
T T
93
=
→ T1
+
→ T2
mg 3 ^ mg ^ mg ^ i+ j+ j 6 6 3 mg 3 $ 3 mg $ mg = i+ j = ( 3 ^i + 3 ^j) 6 6 6 =
Option (b) is correct. 21. f1(max) = 0.3 × 4 × 10 = 12 N 4 kg
2 kg f2
µ1 = 0.6
f1
µ2 = 0.3
f1 and f2 are frictional forces. f2 (max) = 0.6 × 2 × 10 = 12 N As, f1(max) + f2 (max) < 16 N ( Fext )
16N
94 | Mechanics-1 T
2 kg
T
f2
4 kg
16N
x=
∴
f1
The system will remain at rest. For the equilibrium of 4 kg mass : …(i) ∴ 16 = T + f1 As f1 will be at its maximum value f1 = 12 N ∴ T = 16 − 12 = 4 N [from Eq. (i)] Further, for the equilibrium of 2 kg mass. T = f1 ∴ f1 = 4 N Option (c) is correct. 22. For the rotational equilibrium of rod
Option (a) is correct.
A ma mg
…(i)
25. N = ma sin θ + mg cos θ N ma sin θ ma a mg sin θ
mg cos θ
θ
Now, as the block does not slide ma cos θ = mg sin θ i.e., a = g tan θ Substituting the found value of a in Eq. (i) N = m ( g tan θ) sin θ + mg cos θ sin2 θ = mg θ + cos θ = mg sec θ cos
Taking moment about O. l l R × cos θ = s sin θ 2 2 s (= ma) R (= mg) O
1 m 2
When, the block stops a = 0, the value of normal force will be N ′ = mg cos θ N ′ mg cos θ ∴ = N mg sec θ
a
θ
or mg cos θ = ma sin θ ⇒ a = g cotθ Option (d) is correct.
Option (c) is correct. 26. For the rotational equilibrium of the block N
2
23. v = 2 t
dv d = (2 t2 ) = 4 t dt dt At t = 1 s, a = 4 ms −2 As a = µsg a 4 µs = = = 0.4 g 10 a=
∴
φ
A mg
be zero. P
R2 8m
B
1m 10 g
x 1m
∴ Taking moment about point Q (10 g) ( 4) = (80 g) ( x)
mg cos θ
θ = 45°
24. Just at the position of tipping off, R1 will
A
x
mg sin θ
Option (c) is correct.
R1
f
O
B
Taking moment about O. a Nx = f 2 a or ( mg cos θ) x = ( mg sin θ) 2 a or x = tan θ 2 x or = tan θ a/2 or or
tan φ = tan θ φ=θ
Laws of Motion | Thus, the normal force ( N ) will pass through point A.
9 ms–2
N
Option (a) is correct.
A
[Note : The cube will be just at the point of tilting (about point A). The cube will tilt if θ is made greater than 45°].
N sin θ
Moment of couple ( N , mg) = Moment of couple ( F , f )
mg F (= 3 )
f=F
a
mg
Option (b) is correct.
→
→ aA
aA = | a A |
28. Taking moment about point O. N1
→
l/4
l/6
mB a A cos θ sin θ tan θ mB a A = tan2 θ 10 × 9 = = 160 N 2 3 4
l 6
5 37° 4
31. Net downward force on ring = mg − µ ma
= m ( g − µa) T
T
N = ma
f = µN = µma
a = 4 ms–2 L = 1m
mg
ma
∴
mg
( T / 2) = ma → Box ma mg T+ = → Pendulum with 2 2
∴ ∴
respect to box ma mg 2ma + = 2 2 a= g/3
3
Option (c) is correct.
a T
→ aB
N cos θ =
mg
l l l N1 − = N2 − 2 4 2 l l N1 = N2 4 3 ∴ N1 : N2 = 4 : 3 Option (c) is correct.
aB = | a B |
θ
N2
29.
…(ii)
a A = aB tan θ aA N sin θ = mB tan θ mB a A N= sin θ tan θ ∴ Force on rod by wedge ⇒
N (= mg)
x
N
θ = 37° B
27. For the rotational equilibrium of the cube
∴ mgx = Fa mg or mgx = ⋅a 3 a i.e., x = 3
…(i)
N sin θ = mB aB
30.
95
g eff = g − µ a 2L t= g eff =
2L g −µa
=
2×1 10 − (0.5 × 4)
=
1 = 0.5 s 2
96 | Mechanics-1 32. The direction of the normal reactions
33. T − Mg = Ma
between any one hemisphere and the sphere will be along the centres of the two. The three centres of the hemisphere and that of sphere will form a tetrehadron of edge equal to 2R. In figure, C1, C2 and C3 are the centres of the hemispheres and C is the centre of the sphere C1C2 = C2C3 = C3C1 = C1C = C2C = C3C = 2R N
φ
θ
C
O
C3
C2
∠COC2 = 90° 2R C2O = 3 C2O cos θ = C2C 2R / 3 = 2R 1 = 3 φ is the angle which any N makes with vertical φ = 90° − θ √3 1 sin φ = cos θ = 1 3 φ 2 ∴ cos φ = 2 3 For the vertical equilibrium of the sphere. 3 N cos φ = mg 2 or 3N × = mg 3 mg or N= 2 3 Option (b) is correct.
( 500 + 80n )g
i.e.,
N N
C1
T
T = M ( g + a)
∴
2 × 104 ≥ M ( g + a)
or
2 × 104 ≥ (500 + 80n)(10 + 2)
or
14.58 ≥ n
or
n = 14
Option (b) is correct. [Note : Tension in lift cable will increase when the lift is accelerated upwards]. 34. Normal reaction between the surface and
the particle will be zero throughout the motion if the path of the particle is that of a projectile motion (particle is free from surface). v sin φ
v φ
u sin θ u
v cos φ
h
+
a=–g
φ – + u cos θ
v2 = u2 + 2as (v sin φ)2 = (u sin θ)2 + 2 ( − g) h v sin φ = u2 sin2 θ − 2 gh = (20)2 (sin2 60° )2 − 2 ⋅ 10 ⋅ 5 3 = 400 × − 100 4 = 10 2 v cos φ = u cos θ = 20 cos 60° = 10 v sin φ 10 2 = v cos φ 10
Laws of Motion | tan φ = 2 φ = tan −1 2 Option (c) is correct. 35. Acceleration
of block B will be g throughout its motion while that of block A will increase from 0 to g and as such t A < tB Option (b) is correct.
vertical
Option (c) is correct. 38. String is winding on the motor shaft the
36. f1 (max) = 0.5 × 10 × 10 = 50 N 100 N
A
µ = 0.5f1
Velocity along PQ = 20 m/s. ∴ Velocity along PO = 20sin θ 3 = 20 × = 12 m/s 5 Velocity of sphere (along direction) 12 12 = = = 15 m/s cos θ 4 / 5
97
B
µ = 0.25f2
C
block B will move up. Further, as shaft is also moving down, B will further.
f1
2 m/s f2
Here, f1 and f2 are friction forces. As, f1 (max.) < Fext . (100 N), block A will move. f2 (max.) = 0.25 (10 + 20) 10 = 75 N As, f1 (max.), [driving force for block B] < f2 (max.), the block will not slop over block C. As, there is no friction between block C and surface below it, both the blocks B and block C will move together with acceleration f1 (max.) a= (mass of B + C) 50 = (20 + 30) ∴
= 1 ms −2 a A = 1 ms −2
B
2 m/s
Thus, Velocity of block B = Velocity of lift + Velocity of winding of string on shaft + Velocity of moving down of shaft q= 2 m/s + 2 m/s + 2 m/s = 6 m/s. Option (d) is correct. 39. F ′ ′ = F ′ cos θ
F is resultant of two equal forces F ′ and F ′ 2 φ F = 2F ′ cos 2
Option (c) is correct. R R− 5 =4 37. cos θ = R 5
F φ
F' 20 m/s N O
θ
Q
P
R/5
F''
F''
= 2F ′ cos φ = 2F ′ cos (90° − θ) = 2F ′ sin θ F′ ′ = 2 sin θ cos θ
5 4 θ 3
98 | Mechanics-1 = 2F ′ ′ tan θ = 2( ma ′ ′ ) tan θ 5 4 = 2 × 03. × × = 2 N 2 3
For A to remain in contact with B, B must accelerate in the − ive x-direction. → aA (new)
15 j
Option (b) is correct. φ
40. Velocity of block A = 2 v cos 37 °
bi
15 i
Let acceleration of B = − b ^i
2v
Due to this pseudo force with act on A in the + ive direction. 15 tan φ = 15 + b 3 15 or = 4 15 + b
37° A v
B
= 2 × 10 ×
4 5
= 16 m/s Option (d) is correct. 41. As
the mass is applying maximum possible force without moving, the blocks would at the point of slipping, fA T' T'
T T
A
fC
fB
θ = 45°
As,φ < θ the block will leave contact with B. aA θ
B 2F
A N sin θ
N sin θ θ
N
Fnet on block A = Fnet on block B F − N sin θ = N sin θ − 2F 2 N sin θ = 3 F 3F N= 2 sin θ (as θ = 30°)
= 3F Option (d) is correct.
→
a A = |5 ^i + 15 ^j|
15 j
Option (d) is correct.
F
Option (d) is correct. [Note : The value fB will be 300 N and the values of T and f A will be zero] ∴
∴ Acceleration of B = − 5 ^i
N
T ′ = fC (max) = 0.5 × 60 × 10 = 300 N fB (max) = 0.3 × (60 + 60) × 10 = 360 N As, T ′ < fB (max) the value of T will be zero.
42.
45 + 3 b = 60 3 b = − 15 b= −5
43. For block A not to slide on block B
fAT T
B
or or or
44. Equation to circle is Y (x, y) B
y
φ
15 i x
O θ
B φ = 37°
A 3 tan φ = tan 37°= 4
x
Laws of Motion | 2
2
2
x + y = r (where r = OB) dx dy 2 +2 =0 dt dt dx dy =− dt dt
∴ ⇒
…(i)
l1
99
l2 2T 2T
M1
= − ( −u) = + u =u
T
T
T
T
Speed of bead B 2
dx dy = + dt dt 2
M3g
2
M2 g
M1 gl1 = 2Tl2 2M1M 3 M1 gl1 = 2 g l2 M2 + M 3
2
= (u) + ( − u)
or
=u 2
M2 M2 ⋅ 3 3l M1l1 = 4 1 M2 + M2 3 M1 =3 M2
Option (a) is correct. 45. At maximum acceleration value of a, the
or
block would be in a position to move upwards. ma cos θ
⇒
N
47. fmax = µ k N 2
ma ma sin θ
N√2 mg sin θ fm
mg cos θ
ax
a
Trongh
θ = 45°
f = frictional force N = ma sin θ + mg cos θ ma cos θ = fmax + mg sin θ or
ma cos θ = µ N + mg sin θ
or
ma cos θ = µ ( ma sin θ + mg cos θ) + mg sin θ (Qθ = 45° )
or
a = µ ( a + g) + g
or
a (1 − µ ) = (1 + µ ) g 1+µ a= g 1−µ
i.e.,
Block N = mg cos θ
NN
Option (b) is correct. 46. For the beam to have no tendency to
= µ k 2 mg cos θ ∴ ma = mg sin θ − 2 µ k mg cos θ i.e., a = g (sin θ − 2 µ k cos θ) Option (c) is correct. 48. f1 (max) = 0.5 × 3 × 10
= 15 N f2 (max) = 0.3 × (3 + 2) × 10 = 15 N F 3kg f1
f1 2kg
f2
f2 1 kg
B f3
rotate f3 (max) = 0.1 × (3 + 2 + 1) × 10 =6N
100 | Mechanics-1 Value of maximum frictional force is between block 1 kg and the ground. Increasing from zero when F attains 6 N, the block of mass 1 kg will be at the point of slipping over ground below it. Option (c) is correct. 49. f2 (max) = µ ( m1 + m2 ) g m1 µ1
f1 m2 f2
B
A
f1 F (= 30N)
Option (a) is correct. 50. F = µ s mg cos θ − mg sin θ
4 3 1 = mg − 2 3 3 2 mg = 6 Option (b) is correct.
51. µ s = 2 µ k F
a
µ2 s mg
= 0.5 (1 + 2) 10 = 15 N aS = Acceleration of both as one a A = Acceleration of A f1 (max) = µ 1m1 g = 0.2 × 1 × 10 = 2N F − f2 (max) 30 − 15 aS = = = 5 m/s2 m1 + m2 3 µ m g a A = 1 1 = µ 1 g = 0.2 × 10 = 2 m/s2 m1 As, F > f2 (max.) both will move. Further, as aS > a A both will accelerate as one unit. F − µ 2 ( m1 + m2 ) g − µ 1m1 g aB = m2
2s 2.1 2 = = s a AB 9/2 3
∴ Required time t =
in
µ SN
θ
θ = 30°
Force required downwards a=
to 0
in gs m 2
just
slide F
θ
µ KN
θ = 30°
not
Acceleration of A w.r.t. B a AB = a A − AB f _ µ 2 ( m1 + m2 ) g − µ 1m1 g = µ1 g − m2 µ 1m2 g − F + µ 2 ( m1 + m2 ) g + µ 1m1 g =− m2 F − (µ 1 + µ 2 ) ( m1 + m2 ) g =− m2 30 − (0.2 + 0.5) (1 + 2) 10 =− 2 9 2 = − m/s 2 Negative sign indicates that the direction of a AB will be opposite to that of a A .
F = µ s N − mg sin θ = µ s mg cos θ − mg sin θ F = mg sin θ − µ k N = mg sin θ − µ kmg cos θ Thus, µ s mg cos θ − mg sin θ = mg sin θ − µ kmg cos θ (µ s + µ k) mg cos θ = 2mg sin θ µ + µ s = 2 tan θ s 2 3 2 4 ⇒µs = µs = 2 3 3 3 Option (a) is correct. F + mg sin θ − µ k mg cos θ 52. a = m mg mg 2 3 + − ⋅ mg 2 3 3 2 = 6 m g = 3 Option (d) is correct.
block
Laws of Motion | 53. Minimum force required to start the
For retarted motion
motion upward = mg sin θ + µ k mg cos θ 1 4 3 = mg + ⋅ 2 3 3 2 7 = mg 6
02 = v2max − 2a ′ s2 v2 s2 = max 2a ′ v2max 1 1 s1 + s2 = + 2 a a ′ v2 8 6 25 = max + 2 g g
⇒
54. Minimum force required to move the block
up the incline with constant speed = mg sin θ + µ k mg cos θ 1 2 3 = mg + ⋅ 2 3 3 2 5 = mg 6 (5.22)2 55. S1 = = 14.3 m 9.8 2 ⋅ 8
vmax =
50 × 9.8 = 5.92 ms −1 14
Option (c) is correct. 57. mg sin θ = 170 × 10 ×
8 = 906.67 N 15 µ = 0.4 T
µ = 0.2 mg
Option (c) is correct. 1200 g − 1000 g 56. a′ = 1200 g = 6
101
sin
θ
si mg
T
A
B
f2
nθ
f1
θ
15 = 300 N 17 15 f2 (max) = 0.4 × 170 × 10 × 17 = 600 N f1 (max) = 0.2 × 170 × 10 ×
17 T' = 1000g
θ
– a'
a
Speed = vmax T = 1350g
15 25 m
1200 g
1200 g
1350 g − 1200 g 1200 g = 8
a=
For accelerated motion
⇒
8
Stops
v2max = 02 + 2as1 v2 s1 = max 2a
The whole system will accelerate as mg sin θ is greater than both f1 (max) and f2 (max). Total force of friction = f1 (max) + f2 (max) = 900 N Option (a) is correct. 58. mg sin θ − 300 − T = ma
and mg sin θ + T = ma
…(i) …(ii)
Substituting Eq. (i) by Eq. (ii), 2T + 300 = 0 T = − 150 N = 150 N, compressive. Option (a) is correct.
102 | Mechanics-1 More than One Correct Options 1. (a) Normal force between A and B = m2 g F 2N
A
T
T 2N
B
i.e., T =3N Option (d) is correct. 2. At point A
T T
α
= 1 × 10 = 10 N ∴ Force of limiting friction by B on A (or by A on B) = µ × 10 = 2 N
T1 T1 A T2
β
Total force opposing applied external force F = 2N + T = 2N + 2N = 4N Thus, if F ≤ 4 N The block A will remain stationary and so block B also. The system will be in equilibrium. ∴ Option (a) is correct. (b) If F > 4 N F − T −2=1a and T − 2 = 1a Adding above equation F − 4 = 2a i.e., F = 4 + 2a
…(i)
T2 B F
mg
…(i)
T1 cos α = T2 cos β + mg and T1 sin α = T2 sin β
…(ii)
At point B T2 cos β = mg
…(iii)
T2 sin β = F − mg
…(iv)
Using Eq. (iii) in Eq. (i), T1 cos α = 2T2 cos β
…(ii)
= mg
…(v)
Dividing Eq. (ii) by Eq. (v), …(vi)
2 tan α = tan β Option (a) is correct.
For F > 4 N
Squaring and adding Eqs. (iii) and (v), 2a + 4 > 4
or
a>0
⇒
T − 2> 0
i.e., T > 2N ∴ Option (b) is incorrect. (c) Block A will move over B only when F > 4 N and then the frictional force between the blocks will be 2 N if a is just 0 [as explained in (b)]. Option (c) is correct. (d) If F = 6 N using Eq. (ii) 2a = 6 − 4 ⇒
a = 1 m/s 2
∴ Using Eq. (i), T −2=1
T12 = 4 T22 cos2 β + T22 sin2 β Dividing Eq. (iii) by Eq. (iv)
…(vii)
√2 1 β 1
∴
tan β = 1 1 cos β = = sin β 2
Substituting the values of sin β and cos β in Eq. (vii) 1 1 T12 = 4 T22 + T22 2 2 5 = T22 2
Laws of Motion | ⇒ 2T1 = 5T2 Option (c) is correct. 3. Displacement of block in 4 s v (ms–1)
φ
θ O
4
t (s)
S = Area under curve = 16 m. ∆K =Workdone by frictional force 1 × 1 × 42 = µ × 1 × 10 × 16 2 ⇒ µ = 0.1 Option (a) is correct. Option (b) is incorrect. Acceleration, a = tan φ = tan ( π − θ) = − tan θ = − 1 m/s 2 If half rough retardation = 0.5 m/s2 1 ∴ 16 = 4 t + ( − 0.5) t2 2 2 i.e., t − 16t + 64 = 0 or t=8s Option (d) is correct. Option (c) is incorrect. 4. Let acceleration of wedge ( A) = a ma sin θ N F = mg cos θ
a
ma (Pseudo force)
N sin θ F θ A
M
103
2
mg θ
N + ma sin θ = mg cos θ N = mg cos θ − ma sin θ Acceleration of N sin θ a= M or Ma = ( mg cos θ − ma sin θ ) sin θ
or a ( M + m sin θ) = mg cos θ sin θ mg cos θ sin θ i.e., a = M + m sin2 θ 0.6 × g × cos 45° sin 45° = 1.7 + (0.6 × sin2 45° ) 3g = 17 + 3 3g = 20 Let aB = Acceleration of block B Net force on B (along inclined plane) maB = ma cos θ + mg sin θ ⇒ aB = a cos θ + g sin θ Thus, ( aB ) V = ( a cos θ + g sin θ) cos θ = a cos2 θ + g sin θ cos θ 1 = ( a + g) 2 3g 1 = + g 20 2 23 g = 40 ( aB ) H = ( a cos θ + g sin θ) sin θ − a 23 g 3 g = − 40 20 17 g = 40 5. f1 (max.) = µ 1m A g T(pull) = 125 N f1
A
f1 B
f2
= 0.3 × 60 × 10 = 180 N Fnet on B = f1 (max.) + T = 180 + 125 = 305 N A will remain stationary as T < f1 (max.) ∴ f1 = 125 N Force of friction acting between A and B = 125 N ∴ Options (c) and (d) are incorrect.
104 | Mechanics-1 (a)
f2 (max) = µ 2 ( m A + mB ) g
7.5 N
= 0.2 (60 + 40) 10
17.5 N F
= 200 N
C
B
T 17.5 N
f1 + T = 125 + 125 = 250 N As, f1 + T > f2 (max. block B /along the A as A is stationary) will move towards right with acceleration. Option (a) is correct. ( f + T) − f2 (max.) aB = 1 mB + m A =
250 − 200 40 + 60
Option (b) is correct.
9. N sin θ = ma = 1 × 5 = 5 N cos θ = mg = 1 × 10 = 10
6. (See solution to Question no. 4). A
T = 17.5 + 7.5 = 25 N F = T + 37.5 + 17.5 = 80 N (c) T − 7.5 − 17.5 = 4 a F − T − 37.5 − 17.5 = 8 a F = 200 N Solving these equations we get, a = 10 m/s 2
m a
N = mg cos θ − ma sin θ Option (c) is correct and option (d) is incorrect. As angle between the directions of a and g sin θ will be less than 90°, acceleration of block A will be more than g sin θ. A
θ anet > g sin θ
Option (a) is correct and option (b) is incorrect.
…(i) …(ii)
Solving these two equations we get, 1 tan θ = and N = 5 5 N. 2 10. Let f1 = friction between 2 kg and 4 kg f2 = friction between 4 kg and ground ( fs1) max = 0.4 × 2 × 10 = 8 N ( f k1) = 0.2 × 2 × 10 = 4 N ( fs2 ) max = 06 . × 6 × 10 = 36 N Fk2 = 0.4 × 6 × 10 = 24 N (b) At t = 1 s, F = 2 N < ( fs2 ) max ∴ Both the blocks are at rest. ∴ f1 = 0
a
7. Maximum value of friction. f1 = between A and B = 0.25 × 3 × 10 = 7.5 N f2 = between B and C = 0.25 × 7 × 10 = 17.5 N f3 = between C and ground = 0.25 × 15 × 10 = 37.5 N
…(i) …(ii) …(iii)
8. Maximum value of friction available to block is less than the maximum value of friction available to man.
= 0.5 m/s 2
ma
T
37.5 N
(c) At t = 4 s, F = 8 N < ( fs2 ) max ∴ Both the blocks are at rest. f2 = F = 8 N, 11. a = 0, T1 = 10 N, T2 = 20 + T1 = 30 N, T3 = 20 N. 12. fmax = 0.3 × 2 × 10 = 6 N (a) At t = 2 s, F = 2 N ∴ f = 2N (b) At t = 8 s, F = 8 N > 6 N ∴ f =6N
Laws of Motion | (c) At t = 10 s, F = 10 N and f = 6 N 10 − 6 ∴ a= = 2 m/s2 2 (d) Block will start at 6 s. After that, net impulse 1 = × 4 × (6 + 10) + 2 × 10 − 6 × 6 2 = 16 N-s = mv 16 ∴ v= = 8 m/s. 2 13. fmax = 0.4 × 2 × 10 = 8 N (b) At t = 3 s, F = 6 N ∴ Common acceleration
6 a = = 1 m/s 2 6 ∴ Pseudo force on 2 kg =2×1=2 N
105
(backward)
14. N = Mg − F sin θ F cos θ = µN = µ ( Mg − F sin θ) µMg ∴ F= cos θ + µ sin θ For F to be minimum, dF =0 dθ
Match the Columns 1. Acceleration after t = 4 s
At t = 4 s, F = 8 N ∴ Fmax = 8 i.e., µ s mg = 8 8 8 ⇒ µs = = = 0.4 mg 2 × 10 ∴ (a) → (r) At t = 4 s,
a = 1 ms −2
t = 4 s, F =8N F − µ k N = ma F − ma F − ma i.e., µ k = = N mg 8 − (2 × 1) = = 0.3 2 × 10 (b) = (q) At t = 01 . s, F = 0 . 2 N ∴ Force of friction (at t = 01 . s) = 0.2 N (c) → (p) ∴ At t = 8 s, F = 16 N F − µ kmg ∴ a= m 16 − (0.3 × 2 × 10) = =5 2 a i.e., = 0.5 10 (d) → (s). ∴ 2. At θ = 0°, dragging force = 0
∴
Force of force = 0
∴ (a) → (s) At θ = 90° Normal force on block by plane will be zero. ∴ Force of friction = 0 ∴ (b) → (s) At θ = 30° Angle of repose = tan −1 µ = tan −1(1) = 45° As θ < angle of repose, the block will not slip and thus, force of friction = mg sin θ = 2 × 10 × sin 30° = 10 N ∴ (c) → (p) At θ = 60° As θ > angle of repose Block will accelerate and thus force of friction = µ N = 1 × 2 × 10 × cos 60° = 10 N ∴ (d) → (p). 3. All contact forces (e.g., force of friction and
normal reaction) are electromagnetic in nature. ∴ (a) → (q), (r) (b) → (q), (r).
106 | Mechanics-1 Nuclear force is the force between nucleons (neutrons and protons). Between two protons field force also acts.
5. (a) As block A is stationary, f = 10 N T
T
T
T
B
4. (a) N R − 10 = 2 ( + 5)
T T=F
C A
5 m/s2
F = 10N
f
+ ive
∴ (a) → (p) 10 N
(b) As block C is stationary force of friction between C and ground will be zero. ∴ (b) → (s) (c) Normal force ( N C ) on C from ground N C = N B + mC g = mB g + mC g = ( mB + mc ) g = (1 + 1) 10 = 20 N
NR
N R = 20 N
⇒ ∴ (a) → (f)
(b) mg ( = 20 N) > F ( = 15 N)
∴ (c) → (q). (d) As block A is stationary T = F (as shown in figure) = 10 N
f NR 20 N
∴ (a) → (p). 6. If friction force ( f ) is less than the applied
F (= 15 N)
force ( F ).
Block would be slipping in the downward direction. Force of friction will be in the upward direction. Frictional force = 20 − 15 =5N ∴ (b) → (p) (c) If F = 0, the block will slip downwards due to mg ( = 20N) Limiting friction = µ s N R = 0.4 × 20 =8N Minimum value of F for stopping the block moving down = 20 − 8 = 12 N ∴ (c) → (s). (d) F = mg + 8 = (2 × 10) + 8 = 28 N ∴ (d) → (s)
mg F
fmax = 8 N
F
Net force on body = F − f The body will be in motion and thus the friction will be kinetic. ∴ (a) → (q) (b) If friction force ( f ) is equal to the force applied, the body will be at rest. If the body is at the point of slipping the force of force will be limiting too. Emphasis is being given to the word “may be” as when a body is moving and the external force is made just equal to the frictional force, the body would still be moving with friction force at its limiting (kinetic) value. ∴ (b) → (p), (r) (c) If object is moving the friction would be kinetic as explained in (a). (c) → (q). ∴ (d) If the object is at rest, then friction may be static and limiting as explained in (b). ∴ (d) → (p), (r)
Laws of Motion | 7. (a) Normal force between A and B = m A g
f1 (max.) = µ 1m A g (towards left) f1
A
µ1
f1( R) B
3 kg
f2 C
(d) Friction force on 3 kg block due to 5 kg block = f2 (max) = 5 N, towards left. ∴ (d) → (q), (s). 8. (a) and (b)
F µ2
5 kg
T
f2(R)
= 0.2 × 2 × 10 =4N Normal force between B and C = ( m A + mB ) g f2 (max) = µ 2 ( m A + mB ) g = 0.1 (2 + 3) × 10 =5N Total friction force on 3 kg block = f1 (max) + f2 (max) = 4 + 5 = 9 N towards left ∴ (a) → (q), (s) (b) Friction force on 5 kg block = f2 ( R) = f2 (max) = 5 N, towards right ∴ (b) → (p), (s) (c) Friction force on 2 kg block due to 3 kg block = f1( R) = f1 (max) = 4 N, towards right ∴ (c) → (p), (s)
107
2 kg s mg
in
° 30
A
T µ=
√3 2
30°
µ=
T T T B 3kg
3 2
3 = 40.89 ° 2 Now as, angle of incline (30° ) < angle repose (= 40.89 °) The block A and so also B will remain stationary. ∴ (a) → (r), (b) → (r) (c) Tension ( T) in the string connecting 2 kg mass = mg sin 30° 1 = 2 × 10 × 2 = 10 N ∴ (c) → ( s) (d) Friction force on 2 kg mass = zero. ∴ (d) → (r). ∴ Angle of repose = tan −1
6
Work, Energy and Power
Introductory Exercise 6.1 →
→ →
→
1. Work done by F = F⋅ x
→ →
2. Work done by F = F ⋅ l
→ N → T
→ T
→ F → W
→ –x →
→
→
|F|= F ,|N|= N, |W |= W →
→ T β
→ N
→ F α
→ → → |F| = F, |N| = N → → |W| = W, |T| = T → |l| = l
→ W
→
→
|T|= T and |x|= x →
→
=−Fl →
=−Fx → → →
= |N||x| cos
→
→
π 2
→
→
=0 →
→ →
Work done by T = T ⋅ x →
→
= |T||x|cos 0 =Tx
→ →
→ → π = |W || l | cos + α 2
Work done by W = W ⋅ x →
π =0 2
Work done by |W |= W ⋅ l
→
= |W || ⋅ x| cos
→
= |N|| ⋅ l | cos
=0 →
→ →
Work done by N = N ⋅ l
Work done by |N|= N ⋅ x →
→
= |F|| ⋅ l | cos π
= |F ||x| cos π →
→ l
π 2
= − W l sin α →
→ →
Work done by |T|= T ⋅ l →
→
= |T|| l |cos β = T l cos β
Work, Energy and Power | →
→
→
Substituting value of N from Eq. (ii) in Eq. (i). µ (W − F sin 45° ) = F cos 45° 1 1 1 or W − F =F 4 2 2 F F or W − =4 2 2 5F or =W 2 2 2 or F= W = mg 5 5
g 4
3. W − T = m
g a=4
→ T
→ W = mg
→ →
Work done by force F = F ⋅ s
→ l
→
2 1 mgs = mg s = 2 5 5 1.8 × 10 × 2 = 5 = 7.2 J
|g|= g and | l |= l →
→
→
|T|= m g −
∴
=
mg 4
→ →
Work done by friction = µ N ⋅ s → →
3 → mg 4
= µ |N||s|cos π = − µ Ns = − F cos 45° s = − 7.2 J
→ →
Work done by string = T ⋅ l →
→
= |T|| l |cos π 3 → → = m g l cos π 4 3 = − mgl 4 F sin 45° N
→ →
Work done by gravity = g ⋅ s → → π = |g|| ⋅ s | cos 2 =0 …(i)
µ N = F cos 45°
4.
→
= |F||s |cos 45° = F s
→
→
F
5.
F = mg sin 45°
F
2ms–1
45°
F
F cos 45°
µN
mg sin 45°
s
W →
mg cos 45° 45°
→
N = |N|, W = |W | →
→
= 1 × 10 ×
→
s = |s |, F = |F|, g = |g| N + F sin 45° = W
109
…(ii)
1 2
=5 2N Displacement of lift in 1s = 2 m
1 2
110 | Mechanics-1 → →
Work done by force of friction ( F ) = F ⋅ s →
7. Work done = Area under the curve F(N)
→
= |F||s |cos 45° = F s cos 45° = 5 2 ⋅ 2 ⋅
6. Total work-done by spring on both masses m
10
1 = 10 Nm 2 –4
–2
A2
A3
A4 2
A1
x (m)
4
m – 10
x0 m
m
= A1 + A2 + A3 + A4
= PE of the spring when stretched by 2 x0 1 = k (2x0 )2 2 = 2 k x20
− 2 × − 10 − 2 × 10 = + + [10 × 2] 2 2 2 × 10 + 2
∴ Work done by spring on each mass 2 k x20 = = k x20 2
= 30 Nm
Introductory Exercise 6.2 1. a =
−20 ms 2s
−1
= − 10 ms −2
B
fP
P
F
−2
F = ma = 2 kg × − 10 ms = − 20 N s = Area under curve 1 = × 25 × 20 ms −1 2 = 20 m ∴ Work done = F s = ( −20 N) (20 m ) = − 400 Nm ∴
Q
According to B (non-inertial frame) work done = Work done by F + Work done by f p (pseudo force) = ( mas) + ( − mas) = 0 As P is at rest , ∆K = 0 ∴ ∆K = W (Work -Energy theorem)
2. According to A (inertial frame) P
A Q
m a
Acceleration of P = a Force on P = ma ∴ Work done over s displacement = mas Now, v2 = u2 + 2as (Qu = 0) = 2as 1 1 2 ∴ Gain in KE = mv = m 2 as = mas 2 2 ∴ ∆K = W (Work -Energy theorem)
a
Note In inertial frames one has to also to consider work-done due to pseudo forces, while applying Work-energy theorem. 3.
∴
∴
v=α x dv 1 dx a= =α dt 2 x dt 1 α2 =α α x= 2 x 2 2 α F = ma = m 2 mα2 W = b 2
Work, Energy and Power |
111
2
4. ∆K = Work done by F
or
+ Work done by gravity F = 80 N
⇒
−
1 x mv20 = − A 2 2 m x = v0 A
7. (a) If T = mg, the block will not get
accelerated to gain KE. The value of T must be greater that Mg.
g
4m
F
5g
= 80 ⋅ 4 ⋅ cos 0 + 5 g ⋅ 4 ⋅ cos π = 320 + ( − 200) K f − K i = 120 J (as K i = 0) K f = 120 J
or or
T
F
T
T
5. Change in KE = Work done R(1 – cosθ)
ma = mg mg R sin θ mg θ R mg O
Hand Mg
∴ Ans. False (b) As some negative work will be done by Mg, the work done by T will be more that 40 J. ∴ Ans. False
1 mv2 = mgR (1 − cos θ) + mgR sin θ 2 ⇒ v = 2 gR (1 − cos θ + sin θ) 6. ∆K = W
or
0−
1 mv20 = 2
x
∫ 0 − Ax dx
(c) Pulling force F will always be equal to T, as T is there only because of pulling. ∴ Ans. True (d) Work done by gravity will be negative Ans. False
Introductory Exercise 6.3 Spring is having its natural length.
1. In Fig. 1
T
T
T
A
T A
x
mAg
B m Ground
T T B m Ground mg
Fig. 1
Fig. 2
In Fig. 2 A is released. A goes down byx . Spring get extended by x. Decrease in PE of A is stored in spring as its PE. 1 ∴ mAg x = k x2 2 Now, for the block B to just leave contact with ground kx = mg i.e., 2m A g = mg m ⇒ mA = 2
112 | Mechanics-1 1 1 mv2C = × 500 × 00924 . 2 2 500 or vc = × 0.0924 = 2.15 m s −1 100 m 4. Work done by man = gh + Mgh 2 m = + M gh 2
l 2 1 l mv2 = mg 2 2 v= gl
or
2. Decrease in PE = mg
∴ i.e.,
3. OA = 50 cm C 20 cm B 40 cm A
5. When block of man M goes down by x, the
30 cm
spring gets extended by x. Decrease in PE of man M is stored in spring as its PE.
O
∴ Extension in spring (when collar is at A) = 50 cm − 10 cm = 0.4 m Extension in spring (collar is at B) = 30 cm − 10 cm = 20 cm = 0.2 m KE of collar at B = PE of spring − PE of spring (collar at A)
m
37° T
KE of collar at C = PE of spring − PE of spring (Collar at A)
(Collar at C)
1 = × 500 × [(0.4)2 − (026 . )2 ] 2
T
mg sin 37°
=
Extension in spring (collar arrives at C) = [ (30)2 + (20)2 − 10] cm = 0.26 m
T T
T
(collar at B)
1 × K × [(0.4)2 − (0.2)2 ] 2 1 1 or mv2B = × 500 × 12 m 2 500 × 012 . or vB = = 2.45 s −1 10
m
x
T
µmg cos 37° 37° mg
1 2 kx 2 or kx = 2Mg For the block of man m to just slide k x = mg sin 37 ° + µ mg cos 37 ° 3 3 4 or 2Mg = mg + mg 5 4 5 3 or M= m 5 Mgx =
∴
Introductory Exercise 6.4 1. Velocity at time t = 2 s −1
v = g t = 10 × 2 = 20 ms Power = Force × velocity = mgv = 1 × 10 × 20 = 200 W
2. Velocity at time = a t =
Ft (acceleration being constant) 2m F2 t = F × vav = 2m
∴ vav = Pav
F ⋅t m
Work, Energy and Power | Instantaneous power = Force × instantaneous velocity Ft F 2 t =F⋅ = m m Energy 3. Power = Time KE ⇒ KE = P ⋅ t P= t 1 ∴ mv2 = Pt 2 2 Pt or v= m ds 2 P 1/ 2 or = ⋅t dt m 2P 1/ 2 or ∫ ds = m ∫ t dt 2 P t 3/ 2 or s= +c m 3/2 At t = 0, s = 0, ∴ c = 0 8 P 3/ 2 Thus, s= t 9m 4. P = 2 t
KE = ∫ P dt = ∫ 2t dt = t2 + c
113
As at t = 0, KE = 0, c = 0 ∴ KE = t2 1 2 i.e., mv2 = t2 or v = t m 2 2.0 + 2 t Pav = =t 2 5. U = − 20 + ( x − 2)2
PE( = U) is minimum at x = 2. ∴ Equilibrium position is at x = 2 m dU = 2 ( x − 2) dx d2U =2 dx2 ∴ Equilibrium is stable. 6. F = x − 4
For equilibrium, F = 0 i.e., x− 4=0 i.e., x=4m dU As, F = − dx dU = − ( x − 4) dx d2U ∴ = −1 dx2 Thus, equilibrium is unstable.
AIEEE Corner Subjective Questions (Level I) (a) Work done by a constant force
3. m1 g − T = m1a
…(i)
1. (a) Work done by a constant force
T − m2 g = m2 a
…(ii)
Work done by applied force = F s cos 0 = 40 × 2 = 80 Nm Work done by force of gravity = mgs cos π = 2 × 10 × 2 × − 1 = − 40 Nm →
→
→
2. r21 = r2 − r1 ^
^
= (2 ^i + 3 ^j − 4 k) − (1 ^i + 4 ^j + 6 k) ^
^
^
= i − j − 10 k →
→
Work done = F ⋅ r21 ^
^
= (6 ^i − 2 ^j + k) ⋅ ( ^i − ^j − 10 k) = 6 + 2 − 10 = − 2 Nm
Solving Eq. (i) and Eq. (ii), m − m2 a= 1 g m1 + m2 4−1 = g 4+1 Q m1 = 4 kg and m2 = 1 kg T a 3 = g T 5 1 s = ut + at2 2
T a T m2g m1g
114 | Mechanics-1 =
1 3 2 g 2 2 5
= 2 × 10 × 2 sin 60° = 20 3 Nm = 34.6 Nm Work done by force of friction = f s cos π = µ Ns cos π = − µ ( mg cos θ) s 1 = − (2 × 10 × cos 60° ) × 2 2 = − 10 Nm
= 12 m Work done by gravity on 4 kg block = 4 g × 12 cos 0 = 480 Nm Solving Eq. (i) and Eq. (ii), 2 m1m2 T= g m1 + m2 4 =2× g 5 = 16 N Work done by string on 1 kg block = 16 × 12 cos 0 = 192 Nm
(b) Work done by a variable force. W =
∫ F dx x =− 4 =∫ − 2x dx x=2
6.
x= −4
x2 = − 2 2 x = 2
4. Work done by applied force = F s cos 45° N
F = 16 N
= − [( −4)2 − (2)2 ] = − 12 Nm
F
45°
W =
∫ F dx x=2 4 =∫ dx x = 4 x2
7. s = 2.2 m
=
mg
x=2
∫x =4
4x −2 dx x=2
x− 2 + 1 = 4 − 2 + 1 x = 4
1 = 16 × 2 ⋅ 2 × 2 = 24.9 Nm Work done by normal force = N s cos 90° = 0 Work done by force of gravity = mgs cos 90° = 0 Total work done on the block = 24.9 + 0 + 0 = 24.9 Nm
x=2
1 = − 4 x x = 4 1 1 = − 4 − = − 1 Nm 2 4 (c) Work done by area under F- x graph 8. (a) W = 3 × (5 − 10) = − 15 Nm Fx(N)
5. Work done by gravity = mgs sin θ N
µ f=
3 N
x(m)
s mg θ = 60 °
2
4
6
8
10
12
(b) W = 3 × (1 0 − 5) = + 15 Nm
14
16
Work, Energy and Power | 3 × (12 − 10) (c) W = = 3 Nm 2 (10 − 4) + (12 − 0) (d) W = ×3 2 = + 27 Nm (3 − 2) + (3 − 0) 9. (a) W = ×2 2
∴ minimum PE = ( 4 − 4) − 16 = − 16J Now, (KE) max + (PE) min = mechanical energy or (KE) max + ( − 16) = − 4 or (KE) max = 12 J (c) PE max = KE max = 12 J
Fx(N)
PEmax 2
O
x(m) O
1
2
3
4
5
6
115
2
7
–1
= 4 Nm (b) W = 0 Nm 1 (c) W = (6 − 4) ( − 1 − 0) = − 1 Nm 2 (d) W = 4 Nm + 0 Nm + ( − 1) Nm = 3 Nm. Conservative force field and Potential Energy. d 10. F=− U dr d =− Ar−1 dr A = ( −) ( − A ) r − 1 − 1 = 2 r 11. U = ( x − 4)2 − 16 ∴ PE (at x = 6.0 m) = (6 − 4)2 − 16 = − 12 J KE (at x = 6.0 m) = 8 J (a) ∴ Total mechanical energy = ( − 12) + (8) = − 4 J (b) KE will be maximum where, PE, is minimum. For U to be minimum, dU =0 dx d i.e., [( x − 4)2 − 16] = 0 dx or 2 ( x − 4) = 0 or x=4m
PEmax
Total
x
x=4 KEmax
U = ( x − 4)2 − 16 or 12 = ( x − 4)2 − 16 or x2 − 8 x − 12 = 0 ⇒ x= 4+2 3 and = 4−2 3 (d) U = ( x − 4)2 − 16 dU = 2 ( x − 4) dx dU ∴ Fx = − = − 2 ( x − 4) dx or Fx = 8 − 2x (e) Fx = 0 i.e., 8 − 2x = 0 or x=4m Kinetic energy and Work-energy theorem p2 12. K = 2m 2 p + p 2 ∴ K′ = 2m (K ′ is the KE when momentum p is increased by 50%) 9 p2 or K′ = 4 2m 9 or K′ = K 4 9 5 or K′ − K = K − K = K 4 4 K′ − K 5 ∴ = K 4 5 = × 100% = 125 % 4
116 | Mechanics-1 13. p = (2 mK )1/ 2
∴ p′ = [2m ( K + 1% of K ) ]1/ 2 ( p′ is the momentum when KE i.e., K is increased by 1%) 1
= [2 mK (1 + 1%) ]2 1 = (2 mK )1/ 2 1 + 100
1/ 2
1/ 2
1 p′ = p 1 + 100 1 1 = p 1 + × 2 100 1 = p 1 + % 2 1 p′ = p + % of p 2 ∴ Increase in momentum = 0.5 %. i.e.,
102 = 2.5 ms −2 2 × 20
Now, mg − F = ma ∴ F = m ( g − a) = 5 (10 − 2.5) = 37.5 N Work done by push of air = F s cos π = − (37.5 × 20) = − 750 Nm
F (Push of air) a
mg
17. (a) W =
∫ F dx 2 = ∫ (2.5 − x2 ) dx 0 2
x3 = 2.5x − 2 0 23 = (2.5 × 2) − 3 = 2.33 Nm (1 Nm = 1 J) ∴ ∆K = 2.33 J i.e., KE (at x = 2) − KE (at x = 0) = 2.33 J ∴ KE (at x = 2) = 2.33 J (b) Position of maximum KE F = 2.5 − x2 ∴ F decreases as x increases and F is zero when x = 2.5 m Thus, work will be +ive from x = 0 to x = 2.5 m an so KE will be maximum at x = 2.5 m. KE (at x = 2.5 m)
14. s = (2t2 − 2t + 10) m
ds = 4t − 2 dt d2 s = 4 ms −2 dt2 d2 s F=m 2 dt = 2 × 4= 8 N s (at t = 0 s) = 10 m s (at t = 2 s) = 2.22 − 2.2 + 10 = 14 m ∴ ∆s = 14 m − 10 m = 4 m Work = F ∆s =8N× 4m = 32 Nm v2 (0.4)2 0.16 15. a = = = = 0.04 m s −2 2s 2 × 2 4 T mg − T =a m i.e., T = m( g − a) = 30 (10 − 0.04) = 298.8 N Work done by chain = T s cos π = − (298.8 × 2) Nm = − 597.6 Nm
16. a =
=
∫0
2.5
(2.5 − x2 ) dx
x3 = 2.5x − 3 0 a
2.5
2.5 2.5 = 2.5 2.5 − 3 2 = × 2.5 2.5 3 = 2.635 J
Work, Energy and Power | 18. a =
F − mg m g F = −g 10 m 11 F= mg 10
or or
T F F
mg
(a) ∴ Work done on astronaut by F 11mg = × 15 10 11 = × 72 × 10 × 15 10 = 11642.4 Nm (b) Work done on astronaut by gravitational force = mghcos π = − 72 × 9.8 × 15 = − 10584 Nm (c) Net work done on astronaut = (11642.4) + ( − 10584) = 1058.4 Nm ∴ KE = 1058.4 J 1 (d) mv2 = 1058.4 2 1058.4 × 2 ∴ v= = 5.42 ms −1 72 19.
and
…(i) …(ii)
T = m2 a m1 g − T = m1a a m2
T
T
Adding Eq. (i) and Eq. (ii), m1 g = ( m1 + m2 ) a m1 ∴ a= g m1 + m2 1 = g (Q m1 = 1 kg and m2 = 4 kg) 1+ 4 g = = 2 ms −2 5 Now, v2 = u2 + 2as i.e., (Qu = 0 ms −1) v2 = 2as or v = 2as = 2×2×1 = 2 ms −1 50 20. T − 50 = a 10 T
T
a
a 50 (N)
T' A 300 (N)
or and i.e.,
T − 50 = 5a T ′ = 2T T′ T= 2
…(iii)
300 − T ′ = 30a
…(iv)
a = 5 ms −2 1m
…(ii)
T ′ − 100 = 10a 300 300 − T ′ = a 10
Solving Eq. (iii) and Eq. (iv),
m1g
…(i)
From Eq. (i) and Eq. (ii), T′ − 50 = 5a 2
i.e., a
B
T'
Also,
T
T
T T
T
or T
117
T ′ = 100 + 10a = 150 N
118 | Mechanics-1 1 mv2s 2 1 ( mg cos α) R ( − cos α) + mgR = mv2s 2 1 2 2 or mvs = mgR (1 − cos α) 2
21. NR cos (180° − α ) + mgR =
1 × 10 × 1 = µ k × 4 × 10 × 2 + or
1 × 1 × (0.3)2 2
10 = 80 µ k + 0.045 10 − 0.045 µk = = 0.124 80
or
24. f = force of friction while disc in slipping
over inclined surface = µ mg cos 30° α α
N
N
f
s( say )
R
s/2
mg
30°
2R
Stops
f' 0.50 m
22. For 45 kg mass to drop 12 mm, the
increase in length of the spring will be 24 mm. Now, decrease in PE of 45 kg mass = Increase in KE of 45 kg mass + Increase in PE of spring 1 1 45 × 9.8 × 12 × 10−3 = × 45 × v2 + × 1050 2 2 × [(75 + 24)2 − 752 ] × 10−6 i.e., 5.292 = 22.5 v2 + 2.192 or 22.5 v2 = 3.0996 or v2 = 0.13776 or v = 0.371 ms −1 (b) With friction (when mechanical energy does not remain conserved) 23. Decrease in PE of 1 kg mass
= Work done against friction due to 4 kg mass + Increase in KE of 1 kg mass
25. m A g sin θ − µ m A g cos θ − 2 T = m A a T T
T T T
2T
a
2T
T
θ
or = 2 gR sin α or vs = 2 gR sin α (vs is the speed with which sphere hits ground) 1 1 mv2w = mgR − mv2s 2 2 1 = mgR − m 2 gR sin2 α 2 = mgR (1 − sin2 α) = mgR cos2 α ∴ vw = 2 gR cos α (vw is the speed of wedge when the sphere hits ground)
f ′ = force of friction while disc is slipping over plane surface = µ mg Now, decreases in PE of disc = Work done against frictional force s mg = fs + f ′ (0.5) 2 s or mg = (µ mg cos 30° ) s + µ mg (0.5) 2 or mgs (0.5 − µ cos 30° ) = µ mg (0.5) µ × 0.5 ⇒ s= 0.5 − µ cos 30° 0.15 × 0.5 = = 0.2027 m 3 0.5 − 0.15 × 2 Work performed by frictional forces over the whole distance s 50 0.2027 = − mg = − × 10 × 2 1000 2 = − 0.051 J
co s
2
µmAg cos θ
A
Ag
v2s
m
α
S
3
θ
B mBg
4
or 300 ×
3 4 − 0.2 × 300 × − 2 T = 30a 5 5
a
Work, Energy and Power | or 180 − 48 − 2T = 30a or …(i) 132 − 2T = 30a Also, T − mB g = mB a or T − 50 = 5a or …(ii) 2T − 100 = 10a Adding Eq. (i) and Eq. (ii), 32 = 40a or a = 0.8 ms −2 Speed (v) of block A after it moves 1 m down the plane v2 = 2as 2 or v = 2 × 0.8 × 1 or v = 1.12 ms −1
Three types of Equilibrium dU 28. (a) F = − dr
26. Work done by frictional force acting on
block = − µ mgs = − 0.25 × 3.5 × 9.8 × 7.8 = − 66.88 J ∴ Increase in thermal energy of block-floor system = 66.88 J As the block stopped after traversing 7.8 m on rough floor the maximum kinetic energy of the block would be 66.88 J (just before entering the rough surface). Maximum PE of spring = Maximum KE of block 1 2 kxmax = 66.88 2 2 × 66 . 88 m ∴ xmax = 640 Maximum compression in the spring = 0.457 m 27. Decrease in PE of mass m2 = Work done
against friction by mass m1 + Increase KE of mass m1 + Increase in KE of mass m2 10 kg
m1
m2 = 5 kg
119
1 1 m2 g × 4 = µ m1 g × 4 + m1v2 + m2v22 2 2 1 5 × 10 × 4 = 0.2 × 10 × 10 × 4 + (10 + 5) v2 2 Solving, v = 4 ms −1
Point
dU dr
F
A
+ ive
− ive
B
+ ive
− ive
C
− ive
+ ive
D
− ive
+ ive
E
zero
zero
(b) x = 2 m point is of unstable equilibrium (U being + ive) x = 6 m point is of stable equilibrium (U being lowest − ive) x3 29. U = − 4x + 6 3 For U to be maximum (for unstable equilibrium) and minimum (for stable equilibrium) dU =0 dx d x3 i.e., − 4 x + 6 = 0 dx 3 x2 − 4 = 0 x= ±2 d2U d 4 = ( x − 4) = 2x dx dx2 At x = + 2 m, d2U = 2 × ( + 2) = + 4 dx2 ∴ U is minium. At x = − 2 m, d2U = 2 × ( − 2) = − 4 dx2 ∴ U is maximum. ∴ x = + 2 m point of stable equilibrium. x = − 2 m point of unstable equilibrium. or or
120 | Mechanics-1 30. F = −
dU dx
Thus, the equilibrium of the charge − q is unstable if it is slightly displaced along x-axis. (b) If charge − q is displaced slightly along Y axis, the net force on it will be along origin O and the particle will return to its original position. And as such the equilibrium of the − q is stable.
i.e.,
U = − ∫ F dx = − (Area under F - x graph) The corresponding U Vs x graph will be as shown in figure
F
B
–q
F'
F'
+q E
C
A
+q Fnet
B (– a, 0, 0)
O
A
x
32. (a) Velocity at t = 0 s is 0 ms −1 U
Velocity at t = 2 s is 8 ms −1 (using v = 0 + at) ∴vav = 4 m s −1 (as acceleration is constant) Pav = F × vav = ma vav = 1 × 4 × 4 = 16 W (b) Velocity at t = 4 s is 16 ms −1 (using v = u + at) ∴ Instantaneous power of the net force at t = 4 s will be P = mav = 1 × 4 × 16 = 64 W
D
+
x
–
Thus, point C corresponds to stable equilibrium and points A and E correspond to unstable equilibrium. charge placed at origin is in equilibrium as is two equal and opposite forces act on it.
31. − q
y –q
+q B (– a, 0, 0)
F
+q F
A (+ a, 0, 0)
(a) But, if we displace it slightly say towards +ive x side, the force on it due to charge to B will decreases while that due to A will increase. –q O F2
x
33. Power = Fmin vmax
Fmin = (r)
Fmax r
Initially
+q
r
Finally
P = r vmax
F1 > F2
Due to net force on − q towards right the change − q will never come back to original O, its origin position.
vmax
∴
vmax =
P r
Objective Questions (Level 1) Single Correct Option
1 mv2 2 m is always +ive and v2 is +ive. (even if v is − ive). ∴ KE is always + ive.
1. In KE =
Below reference level (PE = 0) PE is − ive. ∴ Mechanical energy which is sum of KE and PE may be − ive. ∴ Correct option is (a).
Work, Energy and Power | 2. Yes, this is work-energy theorem.
∴ Correct option is (a). 3. On a body placed on a rough surface if an
external force is applied the static body does not move then the work done by frictional force will be zero. →
→
4. W = F ⋅ r21 →
→
→
= F ⋅ ( r2 − r1) ^
^
^
= ( ^i + 2 ^j + 3 k) ⋅ [( ^i − ^j + 2 k) − ( ^i + ^j + k)] ^
^
= ( ^i + 2 ^j + 3 k) ⋅ ( − 2 ^j + k) = − 4+3 = −1J ∴ Correct option is (b). 5. ∴ W =
∫ F dx =
=
t =2
∫t = 0 4 ⋅ t
3
dt
2
5
t4 = 4⋅ 4 0
2
∫ 0 7 − 2x + 3 x dx
= [7 x − x2 + x 3 ] 50 = 135 J ∴ Correct option is (d). → →
= 16 J ∴ Correct option is (c). 10. Range = 4 × height
6. P = F ⋅ v
= (10 i + 10 j + 20 k) ⋅ (5 i − 3 j + 6 k)
or
u2 sin 2θ u2 sin2 θ = 4⋅ g 2g
= 50 − 30 + 120
or
sin 2θ = 2 sin2 θ
= 140 W
or
2 sin θ cos θ − 2 sin2 θ = 0
or
sin θ (cos θ − sin θ) = 0
^
^
^
^
^
^
∴ Correct option is (c). 7. Work done in displacing the body
= Area under the curve 1 × 10 = (1 × 10) + (1 × 5) + (1 × − 5) + 2 = 15 J ∴ Correct option is (b). 1 mgh + mv2 W 2 8. P = = t t 800 kg × 10 ms −2 × 10 m + 1 × 800 kg × (20 ms −1)2 2 = 1 min =
121
80000 + 160000 = 60 = 4000 W Option (c) is correct. t3 9. x = 3 dx ∴ v= = t2 dt dv and a= = 2t dt F = ma = 2 × 2 t = 4 t W = F v = 4 t × t2 = 4 t 3 ∴ Work done by force in first two seconds
(800 × 10 × 10) + [ 400 × (20)2 ] 60
or
cos θ − sin θ = 0 (as sin θ ≠ 0)
or
tan θ = 1
i.e.,
θ = 45°
Now, K = KE at highest point 1 = m (u cos θ)2 2 1 = mu2 cos2 θ 2 1 = mu2 cos2 45° 2 1 1 = mu2 ⋅ 2 2 1 2 ∴ mu = 2 K 2 i.e., Initial KE = 2K ∴ Correct option is (b).
122 | Mechanics-1 11. P = 3 t2 − 2t + 1
15. x = 2.0 m to 3.5 m
dW = 3 t2 − 2t + 1 dt dW = (3 t2 − 2 t + 1) dt
i.e.,
W =
4
2
∫ 2 (3 t
− 2t + 1) dt 4
t 3 2t2 = 3 ⋅ − + t 2 3 2 = [ t 3 − t2 + t ] 42 = ( 4 3 − 42 + 4) − (23 − 22 + 2) = 52 − 6 = 46 ∆K = 46 J Correct option is (b). 1 12. K i = × 10 × 102 = 500 J 2 Work done by retarding force, or
W =
30
∫ 20 − 0.1 x dx 30
x2 = − 0.1 2 20
= − 0.05 × [(30)2 − (20)2 ] = − 25 J Final kinetic energy = K i + W = 500 + ( − 25) = 475 J ∴ Correct option is (a). 13. KE of 12 kg mass : KE of 6 kg mass
1 1 m12v2 : m6v2 2 2 [Acceleration being same (equal to g) both will have same velocities] = m12 : m6 = 12 : 6 = 2 :1 1 2 14. W = k [ x2 − x12 ] 2 10 2 5 2 1 = × 5 × 103 − 100 2 100 5 × 103 = (100 − 25) = 18.75 Nm 2 × 104 =
∴ Correct option is (c).
dU 6 =− =−4 dx 1.5 ∴ F =+4N x = 3.5 m to 4.5 m dU 2 = =2 dx 1 ∴ F = −2N x = 4.5 m to 5.0 m dU =0 dx ∴ F =0N Work done = ( 4 × 1.5) + ( − 2 × 1) + 0 =4J 1 2 ∴ mv = 4 2 1 or × 1 × v2 = 4 2 ∴ v = 2 2 ms −1 16. KE at highest point
1 m (u cos 45° )2 2 2 1 1 = mu2 2 2 1 1 = mu2 2 2 E = 2 ∴ Correct option is (a). =
17. Work done by person
= − [Work done by gravitational pull on rope + gravitational pull on bucket] h = − mg + ( − Mgh) 2 m = M + gh 2 ∴ Correct option is (a). 1 18. mv2 = Fx 2 1 mv ′2 = Fx ′ 2 x ′ v ′2 (2 v)2 ∴ = 2 = 2 = 4 ⇒ x′ = 4 x x v v ∴ Correct option is (b).
…(i) …(ii)
Work, Energy and Power | 19. Vertical velocity (initial) = v0 sin θ
K
At an altitude h (vertical velocity)2 = (v0 sin θ)2 + 2 ( − g) h (horizontal velocity)2 = (v0 cos θ)2 (Net velocity)2 = v20 − 2 gh Net velocity = v20 − 2 gh
φt02
21.
v = 0 + at F = at = ⋅t m Instantaneous power, P = Fv F =F ⋅t m F3 or P= ⋅t m or P = constant t. ∴ Correct option is (b).
22. While ball comes down
t0
O
t
Fig. 1
20. Maximum power will be at 2 s
velocity (at t = 2 second) = a ⋅ 2 F = ⋅2 m 2F 2F 2 ∴ Power (at t = 2 s) = F × = m m ∴ Correct option is (d).
123
While ball goes up after elastic collision with the surface it strikes a = + g H u = gt –0
Using,
∴ i.e.,
v = u + at v = ( − gt0 ) + ( + g) ( t − t0 ) v = − g ( t − 2t0 ) 1 KE = mg2 ( t − 2t0 )2 2 K = φ ( t − 2t0 )2 K φt02
u=0 +
a=+g
H
O
t0
2t0
t
Fig. 2 v = gt0
at t = t 0 when ball strikes the surface Using i.e., ∴
where,
v = u + at v = 0 + gt v = gt. 1 1 KE = mv2 = mg2 t2 2 2 K = φ t2 1 φ = mg2 = constant. 2
From the expression for KE at t = 2t0 , K = 0 as shown in Fig. 2. ∴ Correct option is (b). 23. ρ (block) = 3000 kgm −3 and σ (water)
= 1000 kgm −3
vσg
vρg
F
(ρ − )gσ
124 | Mechanics-1 σ 1 = ρ 3
i.e.,
3 y + αx = 5 y x + =1 5/3 5/α
(External force applied to move the block upward with constant velocity).
From above figure 15 3 = 20 4 Work done will be zero, if φ=θ i.e., tan φ = tan θ 5/α 3 or = 5/3 4 3 3 or = α 4 ⇒ α=4 ∴ Correct option is (d).
Work done = F s
tan θ =
= v (ρ − σ) gs = v (ρ − σ) × 10 × 3 5 = (ρ − σ) × 10 × 3 ρ
(Qvρ = 5)
σ = 5 1 − × 10 × 3 ρ = 5 1 −
1 × 10 × 3 = 100 J 3
∴ Correct option is (a). Net work done 24. pav = Time of flight ( T) mgH cos π + mgH cos 0 = T H = Maximum height attained by the projectile (0) = zero = T 1 25. W ( a) = 2 k′ ( x)2 2
27. Let x be the elongation in the spring.
Increase in PE of spring = Decrease in PE of block 1 2 ∴ k x = mgx sin θ 2 2mg sin θ ⇒ x= k ∴ Correct option is (a). s = 2 t2 ^i − 5 ^j
→
l
∴ ds = 4 t ^i dt
2k'
k'
l/3 (a)
2l/3 (b)
Force Constant
→
28.
→ →
W = ∫ F ⋅ ds
1 k′ ( x)2 2 W ( a) > W ( b)
W ( b) = ∴
Given Straight line along which work is zero. → s → F = 20i + 15j
5/3
29.
X
∴
^
=
∫ 0 12 t
2
2
dt
= [ 4 t 3 ] 20 = 32 J ∴ Correct option is (b).
or θ 5 α
^
2
Y
φ
^
∫ 0 (3 t i + 5 j) ⋅ 4 t i dt
t3 = 12 3 0
∴ Correct option is (c). 26.
2
=
W = mgh + mgd = mg ( h + d) Fav d = mg ( h + d) h Fav = mg 1 + d
Work, Energy and Power | 1 30. Energy stored in A = E = kA x2A 2 kA
125
34. Decrease in PE of mass m
kB d
A
B
2kA = 2kB F F
F A
=
F 1 kA 2 kA
k (spring) = 100 N/m θ
F
B 2
i.e.,
2
=
F 2kA
or
Energy stored in B =
F2 2kB
=
F2 4 kA
= (Q kB = 2 kA )
36. Work done by normal force is zero, being
…(i) …(ii)
perpendicular to the displacement. Work done by string is + mgh while that due to gravity it is − mgh. Net work done = ( + mgh) + ( − mgh) =0 ∴ Correct option is (c). 37. Work done on floor = 0 (as displacement is
⇒ n=8 ∴ Correct option is (d). 1 1 33. mgh = mv2 + kx2 2 2 k 2 2 i.e., v = 2 gh − h m
or v = 0.866 m s −1 ∴ Correct option is (b).
∴ Correct option is (c). then the KE gained by the block will change into the PE of the spring. Due to inertia the spring will not start compressing the moment the block just touches the spring and as such the block will still be in the process as increasing its KE. Thus v of the block will be maximum when it compresses the spring by some amount. ∴ Correct option is (b).
∴ Correct option is (a). 1 2 1 31. kx = mv2 2 2 m x=v k 0.5 = 1.5 = 0.15 m 50
= 2 × 10 × 0.15 −
100 × 22 = 4 m. 2 × 10 × 10 × sin 30°
35. PE of block will change into its KE and
E 1 F2 = = 2 2kA 2
∴ Correct option is (a). 1 32. mv2 = F (2x) 2 1 m (2 v)2 = F ( nx) 2 Dividing Eq. (i) by Eq. (ii), (2 v)2 n = 2 v2
= Increase in PE of spring 1 mgd sin θ = kx2 2 kx2 d= 2mg sin θ
zero). 38. Acceleration =
10 × (0.15)2 0.1
(0 − 20) ms −1 (10 − 0) s
= − 2 ms −2 Therefore, net force on particle = 2 kg × − 2 m s −2 =−4N i.e., the net force on the particle is opposite to the direction of motion. ∴ Correct option is (a).
126 | Mechanics-1 Net work done = F × s = F × (Area under v-t graph) 1 = − 4 × × 20 × 10 = − 400 J 2 Thus net work done may not be wholly due to frictional force only. Further net force − 4 N may not be wholly due to friction only. 39. Height of bounce = (100 − 20) % of 10 m
=8
∴ Correct option is (c).
1 mv2 = mgh 2 KE just after bounce = 80% of mgh When the ball attains maxmum height after bounce Gain in PE = mgh′ = Loss of KE or mgh′ = 80% of mgh or h′ = 80% of h = 80 % of 10 m =8m KE just before bounce =
JEE Corner Assertion and Reason 4. In circular motion only the work done by
1. P = Fv
For power to be constant, the velocity must also be constant. Thus, assertion is false. According to 2nd low of motion, net constant force will always produce a constant acceleration. Reason is true. ∴ Correct option is (d). 2. As displacement is opposite to force
(reason) the work done by force will be negative. Thus, Assertion is true. Further, as reason is the correct explanation of the Assertion. ∴ Correct option is (a). 3. Conservative force has nothing to do with
kinetic energy. (If a non-conservative force acts on a particles, there would be loss of KE). Thus, assertion is false. Work done by conservative force decreases PE (reason is true). A
The centripetal force is zero. Assertion is false, centripetal force acts towards centre while the velocity acts tangentially. Reason is true. ∴ Correct option is (d). 5. As the speed is increasing (slope of graph
being increasing) there must be net force in the +ive direction of displacement. Thus, work done to all forces will be positive, Assertion is true and also as explained above reason besides being true is the correct explanation for the Assertion. ∴ Correct option is (a). →
6. Work done by constant force F when the
body shifts from A to B. → F θ
→ F
→ S
A
→ S'
mg (a conservative force) h
B
→ F
→ S'
→ F
→ S D
B
C → →
mg
W = mgh If PE at A is zero. The PE at B would be − mgh. ∴ Correct option is (d).
W AB = F ⋅ s = F s cos θ → →
Similarly, W AC = F ⋅ s
= F s′ cos (90° + θ) = − F s′ sin θ → →
WCD = F ⋅ s = F s cos θ
Work, Energy and Power | → →
WDB = F ⋅ s cos(90° − θ) = F s′ sin θ W AC + WCD + WDB = F s cos θ = W AB ∴ Work done by a constant force is path independent. thus, Assertion is true. The Reason is false. Kinetic frictional force remains constant but is a non-conservative force. ∴ Correct option is (c).
127
greater than 90° and that between N (normal force on wedge) and its displacement will be less than 90° as given in reason. N A
N B
7. It is true that work-energy theorem can be
applied to non-inertial frame also as explained in the answer to question no. 2 of introductory exercise 6.2. Earth is non-inertial which is also true is a separate issue and has nothing to do with the assertion. Thus, option (b) would be the answer. 8. When block is depressed the excess of
upthrust force will act as restoring force and will bring the block up. The velocity gained by block will take the block above its equilibrium and the block will oscillate about its equilibrium position (as given in reason). Thus, the block will be in equilibrium in the vertical direction. Thus, assertion is also true and the reason being correct explanation of the assertion. ∴ Correct option is (a). 9. As displacement ( = s2 − s1) is not equal to
zero, the work done by all forces may not be zero. Therefore, assertion is false. Work done by all the forces is equal to change in KE is as per work-energy theorem. Thus, reason is true. ∴ Correct option is (d). 10. When the block comes down the wedge,
the wedge will move towards left and the actually displacement of normal force will not be along the wedge (see chapter on CM). It will along AB as shown in figure. Thus, the angle between N (normal force on block) and its displacement AB will be
For this the work done byN (or block) will be negative and that by N (on wedge) will be positive as given in assertion. Reason also being the correct explanation of the assertion. 11. There will be increase in length of the
elastic cord. elastic cord
θ
Block Plank f2
f1 s
f1 f
Work-done by static force f1 (on block) = − f1s Work done by static force f1 (on plank) = + f1s ∴ Work done by static force f1 on the system (block + plank + cord) = 0 Thus, reason is true. The work done by the force F will be used up in doing work against friction f2 and also increasing the elastic potential energy of the cord. Thus, assertion is true. Further, as the reason is the correct explanation of the assertion.
128 | Mechanics-1 12. Decrease in KE =
1 mv2 − 0 2
v Slope
f Rough
Case-I gh s E=m p o P St
h
v' 30° f'
1 mv2 2 Change in mechanical energy =
1 mv2 2 Decrease in ME is used up in doing work against friction. In this case the mechanical energy is being used up in doing work against friction and in increasing the PE of the block. 1 ∴ Change ME = mv2 − mgh 2 Thus, assertion is true. As explained above the reason is false. (µ does not change with the increase in angle of inclination) ∴ Correct option is (c). PE + KE =
Objective Questions (Level 2) Single Correct Option 1. Increase in KE of bead = Work done by
gravity + work done by force F 1 ∴ mv2 = mgh + FR 2 (Displacement of force F is R) 1 = × 10 × 5 + 5 × 5 = 50 2 100 ∴ u= m = 200 = 14.14 ms −1 ∴ Correct option is (a). 2. P = Fv
= mav dv dv ds dv dv = m v v a = = ⋅ =v dx dt ds dt ds m 2 ∴ ds = v dv p m 2v 2 i.e., ∫ ds = v dv p ∫v 2v
or
m v 3 s= p 3 v =
m 7 mv 3 [(2v) 3 − (v) 3 ] = 3p 3p
∴ Correct option is (a).
3. Loss of PE of block = Gain in PE of spring d k A
1 k × 22 2 1 1 or 10 × 10 × ( d + 2) × = × 100 × 22 2 2 d = 2m Let, v A = velocity of block when it just touches spring mg ( d + 2) sin 30° =
d 2m m
A
30°
1 mv2A = mgd sin 30° 2 1 v2A = 10 × 2 × 2 × 2 v A = 20 ms −1 Correct option is (a).
∴
Work, Energy and Power | m 4. Mass per unit length of chain = πR/2 m dm = R dθ πR/2 yCM =
129
∴ a2 = 0 ∴ Correct option is (b). 6. Let x be the expansion in the spring. kx A
y dm
∫ ∫ dm
kx B
T T T
T
θ C
M1g
CM
(
R 1–2 π
y
R (1 – cos θ) π /2
=
∫0
R (1 − cos θ)
)
m R dθ πR/2
m 2R π / 2 = (1 − cos θ) dθ π ∫0 2R π = − 1 π 2 2 = R 1 − π Now, or
1 mv2 + mg yCM 2 v2 = 2 g yCOM 2 v = 2 gR 1 − π
∴ Correct option is (c). 5. The moment string is cut T T = mg
A mg F (= mg)
mg F
F = kx = mg F = mg B mg (weight)
Net force on A = 2mg ∴ a1 = 2 g Net force on B = 0
mg
(downward)
Increase in PE of spring = Decrease in PE of block C 1 2 kx = M1 gx 2 i.e., kx = 2M1 g For block A to remain at rest kx = µ min Mg or 2M1 g = µ min Mg 2M1 ∴ µ min = M ∴ Correct option is (c). 7. Ti = mg 2mg KX i = = mg 2 When one spring is cut. It means KX i becomes zero. Downward acceleration, KX i mg g a= = = 2m 2m 2 Now drawing FBD of lower mass : mg mg − Tf = m. a = 2 mg ∴ Tf = 2 mg or ∆T = Tf − Ti = 2 mg sin θ − µ mg cos θ 8. a = m = g sin θ − µ g cos θ For v to be maximum x a dv θ =0 os mg sin θ dx gc m µ dv or v =0 dx θ
130 | Mechanics-1 4 µ mg x ∴ Correct option is (c).
dx dv ⋅ =0 dt dx dv or =0 dt or a=0 i.e., g sin θ = µ g cos θ or sin θ = µ cos θ 3 3 4 or = x⋅ 5 10 5 10 ⇒ x= = 2.5 m 4 ∴ Correct option is (d). or
9. (a) Between points E and F,
x=
⇒
→ →
12. Power delivered by man = T ⋅ v
= T v cos θ
→ T → v
→
|T|= T m
→
and |v|= T 13. φ = 3 x + 4 y y 3N
dU is − ive. dr
dU , the force between E and F dr will be + ive i.e., repulsive. (b) At point C the potential energy is minimum. Thus, C is point of stable equilibrium. ∴ Correct option is (c).
P (6m, 8m)
4N Fnet = 5N
Now, F = −
→ →
10. Power = F ⋅ v → v → u
→ mg
θ → mg →
→
→
= m g ⋅ ( u + g t) → →
= mg u cos (90° + θ) + m g ⋅ g t P = − mg u sin θ + mg2 t
i.e., → → [|u|= u,|g|= g]
Therefore, the graph between P and t will be as shown in option (c). 11. x 2x µmg
PE of spring due to its compression by x = | Work done by frictional force when displaced by 2x| 1 2 i.e., kx = µ mg 2x 2
R
6m
Q
x
∂φ = −3N ∂x ∂φ and Fy = − =−4N ∂y PR 5 = PQ 4 5 ⇒ PR = × PQ = 10 m 4 ∴ Work done by the conservative force on the particle = Fnet × PR = 5 N × 10 m = 50 Nm = 50 J ∴ Correct option is (c). ∴
Fx = −
14. Both at x = x1 and x = x2 the force acting
on the body is zero i.e., it is in equilibrium. Now, if the body (when at x = x1) is moved towards right (i.e., x > x1) the force acting on it is + ive i.e., the body will not come back and if the body (when at x = x2 ) is moved toward rght (i.e., x > x2 ) the force acting on it is − ive i.e., the body will return back. Then,x = x2 is the position of stable equilibrium. ∴ Correct option is (b).
Work, Energy and Power | i.e.,
15. The man will stop when
6b x
−7
= 12a x
131
−12
1
x mg sin θ
i.e.,
µmg cos θ
θ
or or
2a 6 x = b
∴ Correct option is (a). 19. For the rotational equilibrium of the rod
µ mg cos θ = mg sin θ (µ 0 x) mg cos θ = mg sin θ tan θ x= µ0
kx
l l/2 mg
16. Taking moment about A
l = kx l 2 mg x= 2k
mg⋅ i.e.,
k1x
k2x
C
∴ PE stored in the spring =
1 2 kx 2
2
A
B
=
F = (k1 + k2)x
∴
AC ⋅ F = ( k2 x) l ( k x) l AC = 2 F ( k2 x) l = ( k1 + k2 ) x k2 = l k1 + k2
∴ Correct option is (d). 17. mgh + work done by the force of friction
1 = mv2 2 ∴ Work done by the force of friction 1 = mv2 − mgh 2 1 = × 1 × 22 − (1 × 10 × 1) 2
= − 8J Correct option is (c). ∴ a b 18. U = 12 − 6 x x U = ax −12 − bx −6 dU =0 dx − b ( −6) x −7 = 0
For stable equilibrium, i.e.,
a ( − 12) x −13
1 mg ( mg)2 k = 2 2k 8k
∴ Correct option is (c). m − m2 20. a = 1 g m1 + m2 Speed (v) with which mass m1 strikes the floor m − m2 = 02 + 2 gh = 2 1 gh m1 + m2 ∴ Correct option is (a). 21. F = − ax + bx2
dU = − ax + bx2 dx or dU = ( ax − bx2 ) dx or U = ∫ ( ax − bx2 ) dx ax2 bx 3 or U= − +c 2 3 At x = 0, F = 0 ∴U = 0 and so, c = 0 ax2 bx 3 Thus, U= − 2 3 ax2 bx 3 U = 0, when = 2 3 3a i.e., at x = 2b ∴
−
132 | Mechanics-1 dU = 0, when F = 0 dx i.e., − ax + bx2 = 0 a i.e., at x= b Graph between U and x will be
O
3a 2b
a b
1 2 …(i) k xmax = mg ( h + xmax ) 2 ∴ From above Eq. (i), xmax depends upon h and also xmax depends upon k. KE of the block will be maximum when it is just at the point of touching the plank and at this moment there would no compression in the spring. Maximum KE of block = mgh ∴ Correct option is (c). ∴
X
25. Gain in KE of chain = Decrease in PE of
chain
∴ Correct option is (c). W A Fs 1 22. = = WB Fs 1
CM (i)
∴ Correct option is (c). 1 mv2A WA 2 = =1 WB 1 4 mv2 B 2 v2A 4 ⇒ = v2B 1 vA 2 ⇒ = vB 1 WA 1 = WB 1 KA 1 ∴ = KB 1
2r π
πr/2 πr
CM (f)
23. Ui at (1, 1) = k (1 + 1) = 2k
U f at (2, 3) = k (2 + 3) = 5k W = U f − Ui = 5k − 2k = 3 k ∴ Correct option is (b). 24. Gain in PE of spring = Loss of PE of block
h
When the whole chain has justcome out of the tube. 1 2π πr or mv2 = mg + π 2 2 2 π ∴ v + 2 gr + π 2 ∴ Correct option is (b).
xmax PE = 0 level for block
26. Acceleration of the block will decrease as
the block moves to the right and spring expands the velocity (v) of block will be maximum, when 1 1 mv2 = kx2 2 2
Work, Energy and Power | At this moment, F = k x F i.e., x= k 2 k F F2 or v2 = = m k km F ∴ v= mk
29. Work done on block A in ground frame
= 0.2 × 45 × 10 ×
30. N = mg cos θ N φ = 53° 20 m → s
27. Force on each block = k x A
kx
kx
10 ms–1
B
1 = 20 N 10 Power of A = 20 × 4 = 80 W Power of B = 20 × 6 = 120 W ∴ Total power = 200 W i.e., rate of energy transfer = 200 Js −1 ∴ Correct option is (c). 28. From O to x compression in the spring A m kx
kx
B 3m
+ ive
Average acceleration of A kx − F aA = 2m Average acceleration of B kx aB = 2 (3 m) As at maximum compression of the spring both the blocks would be having same velocity. 2 a A x = 2 aB x [using v2 = u2 + 2as] i.e., a A = aB kx − F kx = 2m 6m kx or kx − F = 3 2 kx or F= 3 3F i.e., x= 2k ∴ Correct option is (c).
→ s=|s|
6ms–1
= 200 ×
F
(50 − 30) 100
= 18 J ∴ Correct option is (b).
∴ Correct option is (b). 4ms–1
133
θ = 37°
= 10 × 10 × cos 37 ° 4 = 10 × 10 × 5 = 80 N ∴ Work done by N in 2 s = N cos φ = 80 × 20 × cos 53 ° 3 = 80 × 20 × 5 = 960 Nm = 960 J Correct option is (b). ∴ T ′ + mg 31. (given) =5 m Upper spring cut
Initially
Lower spring cut
T = T' + mg + ive T' mg
T
T' mg
T ′ + m × 10 = 5 m i.e., T ′ = − 5 m mg − T a= m mg − ( T ′ + mg) = m T′ =− m
mg
134 | Mechanics-1 ( − 5 m) m = 5 ms −2 ∴ Correct option is (b).
33. Total ME = − 40 J
=−
∴ PE (max) = − 40 J U (x) 50 J
32. Total ME = 25 J
∴ PE (U) max = 25 J [as KE can’t be −ive.]
25
U (x)
– 10
50 J PE (max)
6
–5
10
15
PE (max) PE (max) 6
–5
PE (max)
x (m)
25 – 10
x (m)
– 35
10
15
Particle can’t be found in the region above PE (max) line. ∴ −10 < x < − 5 and 0 < x < 15 ∴ Correct option is (a).
Particle can't be found in the regions above PE (max) line. ∴ “It is not possible”. Option (d).
More than One Correct Options → →
1. (i) Acceleration
U = 7 x + 24 y ∂U Fx = − = −7 ∂x ∂U Fy = − = − 24 ∂y 7 ax = − 5 24 ay = − 5 → 7 ^ 24 ^ a=− i− j m/s2 5 5
∴
i.e., i.e., ∴
→
→ v = 3i + 4j – 7i
→
∴ Correct option is (b). (ii) Velocity at t = 4 s →
→
→
24 ^ 7 = ( 8.6 ^i + 23.2 ^j ) + − ^i − j 4 5 5 = 8.6 ^i + 23.2 ^j − 5.6 ^i − 10.2 ^j = 3 ^i + 4 ^j |v|= 5 ms −1
∴ Correct option is (a).
→ a
– 24j
2. U = 100 − 5x + 100x2
v=u+at
→
→
i.e., a is not perpendicular to v. Options (c) and (d) are incorrect.
|a |= 5 ms −2
i.e.,
∴
(iii) As a ⋅ v ≠ 0, the path of the particle can’t be a circle.
∴
Fx = −
∂U ∂ =− (100 − 5x + 100x2 ) ∂x ∂x
= − [ − 5 + 200 x ] = 5 − 200 x Fx 5 − 200 x ∴ ax = = 0.1 0.1 = 50 − 2000 x (i) At 0.05 m from origin x = + 0.05 m ax = 50 − 2000 (0.05) = 50 − 100
(first point)
Work, Energy and Power | 2
= − 50 m/s = 50 m/s 2 towards − ive x ∴ Correct option is (a). (ii) At 0.05 m from origin (second point) x = − 0.05 m ax = 50 − 2000( − 0.05) = 50 + 100 = 150 m/s 2 ∴ Correct option is (c). Mean Position 50 m = 0.025 m a = 0 at x = 2000 (iii) For point 0.05 m from mean position x = (0.05 + 0.025) m = 0.075 m ∴ ax ( at x = 0.075 m ) = 50 − 2000 (0.075) = 50 − 150 = − 100 m/s 2 = 100 m/s 2 towards − ve x-axis. ∴ Correct option is (b). (iv) For second point 0.05 m from mean position x = 0.025 m − 0.05 m = − 0.025 m ∴ ax = 50 − 2000 ( − 0.025) = 100 ms −2 ∴ Option (d) is incorrect. 3. (i) If the spring is compressed by x, elastic
1 2 kx gets stored 2 in the spring. Now, if the compressed spring is released the energy stored in the spring will be lost. When the spring attains to natural length. Work done by spring = Energy stored in the spring 1 = kx2 2 ∴ Correct option is (a). (ii) If the spring is extended by x, energy 1 stored in the spring would be kx2 . If the 2 extended spring is released the energy stored in the spring will be lost when the spring attains its natural length. potential energy equal to
135
Work done by spring = Energy stored in the spring 1 = kx2 2 ∴ Correct option is (b). (iii) If spring is initially its natural length and finally compressed. 1 Work done on (not by) the spring = kx2 2 will be stored in the spring. ∴ Option (c) is incorrect. (iv) If spring is initially at its natural length and finally extended. 1 Work done on (not by) the spring = kx2 2 will be stored in the spring. ∴ Option is (d) is incorrect. 4. (i) Work-Energy theorem states that Wnet
(Work done by all forces conservative or non-conservative, external or internal) = ∆ (KE) Correct option is (d) is incorrect. ∴ Correct option is (c). (ii) Work done by non-conservative forces (i.e., all forces except conservative forces) lead to decrease in KE and thus change in mechanical energy takes place. ∴ Correct option is (b). (iii) Work done by a conservative force may be + ve, − ve or zero. → s 90° mg h
h
mg
mg (a conservative force) → →
W = m g⋅h = mghcos π = − mgh (PE increases)
136 | Mechanics-1 → →
W = m g⋅h = mghcos 0 = + mgh (PE increases) → →
W = m g ⋅ h = mghcos
π =0 2
(PE remains same) Option (a) is incorrect. 5. F is the force by hand or upper disc (δ – l) k (l + δ) k (δ + l) F k (δ + l) N
6. At maximum extension x :
k(l + δ) 3mg
When F is removed, the upper disc accelerates upwards and when it attains the position as in figure 2, its acceleration reduces to zero and the velocity gained by it takes it further upwards. Restoring force on the upper plate now acts downwards and that on the lower plate acts in the upward direction and would lift it (lower plate) if k (δ − l) > 3 mg i.e., kδ > 3 mg + kl or kδ > 3 mg + mg 4mg or δ> k ∴ Correct option is (d). Correct option is (a) obviously being incorrect.
3mg kl = mg l
decrease in potential energy increase in spring energy 1 ∴ (2m)( g)( x) = kx2 2 4 mg or k= k
of
B=
7. Total work done by internal forces of a
mg
system, which constitute action and reaction pairs, is always zero and if it is not so the total work done will not zero. ∴ Correct options are (b) and (c).
Artist to make it 3/2 times of l
8. (i) Work done by conservative forces may
kl 3mg
Uncomprssed/ Unstretched spring
N = 3 mg + k ( l + δ) 2mg = 3 mg + k l + k = 3 mg + kl + 2mg = 3 mg + mg + 2mg 2mg = 6mg for δ = k ∴ Correct option is (b). N = kl + 3 mg = mg + 3 mg = 4mg ∴ Correct option is (c).
be + ive, − ive or zero as explained in the answer to question no. 4 ∴ Option (a) is incorrect. Correct option is (b). Correct option is (c). (ii) In pure rolling work done by frictional force (a non-conservative force) is always zero. ∴ Correct option is (d). 9. In moving from 1 to 2 work done by
conservative force = U1 − U2 = ( − 20) − ( − 10) = − 10 J ∴ Option (a) is incorrect.
Work, Energy and Power | Option (b) is correct. Work done by all forces = ( K 2 + U2 ) − ( K 1 + U1) = [20 + ( − 10)] − [10 + ( − 20)] = 20 J ∴ Correct option is (c). Option (d) is incorrect. N
10.
137
Work done by Normal force ( N ) on block A W = Ns cos θ In motion 2 → + ve
as θ < 90°
In motion 1 → + ve
as θ < 90°
∴ Correct option is (b). Work done by force of friction (f ) In motion 1 → may be − ive if f is directed upwards along the plane as shown in figure 1. (Motion 1 being retarded)
N 2
∴ Correct option is (c).
1
mg
mg
In motion 1 → may be + ive if f is directed downwards along the plane if motion 1 is accelerated.
Work done by gravity In motion 2 → − ive
[as θ = π]
∴ Correct option is (d).
π In motion 1 → zero[as θ = ] 2 ∴ Correct option is (a). Match the Columns 1. As body is displaced from x = 4 m to
x =2m
(d) → (p, s). 2. W = Fs cos θ
→
^
(a) Work done by N will be + ive as θ < 90° (a) → (p) (b) Work done by mg will be − ive as θ = π
s = −2im
→
(a) F = 4 ^i → →
W = F ⋅ s = 4 ^i − 2 ^i
∴
N
= − 8 unit (a) → (q) and |W |= 8 unit (a) → (s)
mg
→
(b) F = 4 ^i − 4 ^j → →
^
^
(b) → (q) (c) Work done by force of friction (f ) will be zero as f will be zero. (c) → (r) (d) Work done by tension (T) will be + ive as θ < 90° (d) → (p)
^
∴ W = F ⋅ s = ( 4 i − 4 j) ⋅ ( − 2 i) = − 8 unit (b) → (q, s) →
(c) F = − 4 ^i → →
^
W = F ⋅ s = 8 i unit
∴ and (c) → (p, s)
|W|= 8 unit
→
(d) F = − 4 ^i − 4 ^j → →
∴ W = F⋅ s
= ( − 4 ^i − 4 ^j) ⋅ ( − 2 ^i) = 8 unit and
T
|W |= 8 unit
Y
3. +q A (– a, 0)
FB
FA = FB +Q Fnet = 0
+q B (+ a, 0)
F A = force on + Q by + q placed at A FB = force on + Q by + q placed at B
138 | Mechanics-1 As FA = FB equilibrium. (a)
charge
A
+Q
will
be
FA'
FB'
+q
in
+q B
+Q
Due to increase in distance between + q (at A) and+ Q FA′ < FA Due to decrease in distance between + q (at B ) and + Q FB′ > FB Using FA = FB , we have FB′ > FA . As there will net force on + Q which will being + Q to origin. Thus, equilibrium will be a stable one. Y (b) Fnet FB
FA +Q
A
B
+q
+q
As Fnet will be along the increasing direction of Y , the charge + Q will not return to origin. Thus, equilibrium will be an unstable one. (b) → (q)
∴
(c) As explained in (b) the equilibrium will be an unstable on. (c) → (q)
∴ (d)
x = y line
Fnet
4. (a) From A to B :
FA
A
kx
Mean position
B
x
mg
C
1 2 kx 2 Decrease in gravitational. PE of block = mgx = ( kx) x = kx2 (a) → (q) ∴ (b) From A to B Increase in KE of block = Decrease in gravitational PE of block − Increase in spring PE 1 2 2 = kx − kx 2 1 2 = kx < Decrease in gravitational PE of 2 block. (b) → (p) ∴ (c) From B to C 1 1 Increase in spring PE = k (2x)2 − kx2 2 2 3 2 = kx 2 1 Decrease in KE of block = kx2 2 (KE of block at C will be zero) ∴ (c) → (p). (d) From B to C decrease in gravitational PE = mgx Increase in spring PE =
= ( kx) x
A +q
x
= kx2
+q
Increase in spring PE = ∴ (d) → (p)
3 2 kx 2
Work, Energy and Power | m − m2 2−1 5. a = 1 g= g m1 + m2 2+1
T
T a
T a
T m2
m1 1g
2g
=
1 g 3
10 = m s −2 3 2 m1m2 T= g m1 + m2 1×2 =2× × 10 1+2 40 N = 3 Displacement of blocks, 1 s = ut + at2 2 1 2 = at 2 1 10 = × × (0.3)2 2 3 = 0.15 m (a) Work done by gravity on 2 kg block = mgs cos 0 = 2 × 10 × 0.15 × (1) =3J ∴ (a) → (r) (b) Work done by gravity on 1 kg block = mgs cos π = 1 × 10 × 0.15 × ( − 1) = − 1.5 J ∴ (b) → (p).
139
(c) Work done by string on 2 kg block = Ts cos π 40 = × 0.15 × ( −1) 3 = − 20 J ∴ (c) → (s). (d) Work done by string on 1 kg block = Ts cos 0 40 = × 0.15 × 1 3 =2J ∴ (d) → (q). 6. (a) Work done by friction force ( f )
(w.r.t. ground) f f' (= f )
= f s cos π =−fs ∴ (a) → (q). (b) Work done by friction force on incline (w.r.t. ground) = f × 0 × cos 0 =0 [There being no displacement of incline w.r.t. ground] ∴ (b) → (r). (c) Work done by a man in lifting a bucket (T = Tension in rope) = Ts cos 0 = a + ive quantity θ = 0 as T and s both would be in upward direction. ∴ (c) → (p). (d) Total work done by friction force in (a) w.r.t. ground = − f s+0= − f s ∴ (d) → (q).
7
Circular Motion Introductory Exercise 7.1
1. In uniform circular motion the magnitude
v2 of acceleration = does not change r
(iii) Magnitude of net acceleration = ( ac )2 + ( aT )2 = ( 4 cms −2 )2 + (2 cms −2 )2
while its direction (being always towards the centre of the circular path) changes. 2. If ω 0 and ω are in rad s −1 the value of α −2
must be in rad s . But, if ω 0 and ω are in degree s −1 the value of α must also be in degree s −2 . Thus, it is not necessary to express all angles in radian. One way change rad into degree using π rad = 180°. 3. During motion of an object along a curved
path the speed and magnitude of its radial acceleration may remain constant. Due to change in direction of motion the velocity of the object will change even if its speed is constant. Further, the acceleration will also change even if the speed is constant. 4. (i) Radial acceleration ( ac )
=
v2 (2 cms −1)2 = r 1 cm
= 4 cms −2 (ii) Tangential acceleration (aT ) dv d = = (2t) dt dt = 2 cms −2
= 2 5 cms −2 →
→
5. |v1|= |v2|= v (say)
2πr T= v → → |PQ| |v av|= tPQ =
r 2 r 2 = T / 4 πr / 2 v
=
2 2 v π
→ v2
Q → v1
r O
r
P
→
∴
|v av| 2 2 = v π
6. ω t = 0 + 4 t
Centripetal acceleration = tangential acceleration r ω2t = r α ⇒ ω2t = α ( 4 t)2 = 4 1 t= s 2
Circular Motion |
141
Introductory Exercise 7.2 1. In uniform circular motion of a body the
(a) At rest :
body is never in equilibrium as only one force (centripetal) acts on the body which forces the perform circular motion. 9.8 × 250 × (1.5 / 2) gra 2. vmax = = h 1.5
Required CPF = w − N =
= 35 ms
−1
l l
T sin θ T θ T cos θ mg
r
3. (a) T sin θ = mg
T cos θ = mrω2 g ∴ tan θ = r ω2 g g = r tan θ ( l cos θ) tan θ
⇒
ω=
or
2π = T
∴
l sin θ g ⇒ T = 2π l sin θ g
1 2 2 = 2π 9.8 1 9.8 rve s −1 f = = 1 2π =
9.8 × 60 rev min −1 2π
= 29.9 rev min −1 mg (b) T = = 2 mg sin θ = 2 × 5 × 9.8 = 69.3N 4.
N=w 2
w' w
w
⇒
w 2 mv2 w = r 2
…(i)
At dip : Required CPF = N ′ − w mv2 …(ii) = N′ − w r Comparing Eqs. (i) and (ii), w N′ − w = 2 3w 3 ∴ N′ = = × 16 kN 2 2 = 24 kN (b) At crest on increasing the speed (v), the value of N will decrease and for maximum value of v the of N will be just zero. mv2max Thus, = w−0 r wr ⇒ vmax = m (as w = mg) = gr = 10 × 250 = 50 ms −1 (c) At dip : mv2 r = w + mg = 2w = 32 kN
N′ = w +
5. Case I. If the driver turns the vehicle r µ mg
142 | Mechanics-1 m v12 ≤ µ mg r [where v1 = maximum speed of vehicle] ⇒ v1 ≤ µ gr Case II. If the driver tries to stop the vehicle by applying breaks. Maximum retardation = µ g ∴ v22 = 0 2 + 2 (µ g) r ⇒ v2 = 2 µ gr = 2 v1 As v2 > v1, driver should apply breaks to stop the vehicle rather than taking turn.
ω=
g l sin φ
If θ is the angle made by the string with the vertical θ + φ = 90° i.e., φ = 90° − θ ⇒ sin φ = cos θ g ∴ ω= l cos θ g ⇒ cos θ = l ω2
6. In the answer to question 3(a) if we
replace θ by φ
Introductory Exercise 7.3 →
1. |v1|= u →
→ and |v2|=
→
2. Ball motion from A to B :
v (say)
B
→
∆v = v2 + ( − v1) → – v1
→ v2
→ ∆i
2R C v
h
L g
A
→ v1 →
→
= v2 − v1 →
→
→
→
→
→
∆v ⋅ ∆u = ( u 2 − v 1 ) ⋅ ( v 2 − u 1 ) →
→
→
→
→
→
= u2 ⋅ v2 + u1 ⋅ v1 − 2 v2 − v1 → |v2 |2
→
2
2
0 2 = u2min + 2 ( − g) (2R) ⇒ = 4 gR Ball motion from C to A : v2 = u2min + 2 ( − g) h = 4 gR − 2 gh ⇒ v = 2 g (2R − h) u2min
+ |v1| = v + u 2
3. Decrease in KE of bob
= Increase in PE of bob
2 2
= u − 2 gL + u → 2
|∆v| = 2 (u2 − gL) →
|∆v|= 2 (u2 − gL)
v0
KE of bob = 1/2 mv02 PE of bob = K (say)
Circular Motion |
143
v20
= 2 gh v0 = 2 gl (1 − cos θ) = 2 × 9.8 × 5 × (1 − cos 60° ) = 7 ms −1
PE of bob = K + mgh
h
1 mv20 = mgh 2
AIEEE Corner Subjective Questions (Level 1) 1. v = 4 t2
dv = 8t dt i.e., aT = 8 × 3 = 24 ms −2 v = 4 × 32 = 36 ms −1 (at t = 3 s) v2 (36)2 ∴ = 24 ms −2 ac = = 4 54 → → Angle between a net and a t ∴
θ=
→ −1 |a c| tan = → |a T|
tan −1 1 = 45°
2. v = 16 ms −1 and r = 50 m
v2 (16)2 = = 5.12 ms −2 r 50 = a2c + a2T = (5.12)2 + 82
ac =
∴
anet
−2
(given aT = 8 ms ) = 9.5 ms −2 3. Speed (v) at the highest point ( P) v = u cos θ
u θ
u cos θ
R
v2 = 21.65 R i.e., v = 21.65 × 2.5 (Q R = 2.5 m) = 7.36 ms −1 1 (b) aT = a sin 30° = 25 × = 12.5 ms −2 2
∴
5. (a) v = 2.0t = 2.0 × 1
= 2.0 cms −1 v2 (2.0)2 ∴ ac = = = 4 cms −2 R 1 (b) v = 2.0t dv ∴ = 2.0 dt i.e., aT = 2.0 cms −2 (c) anet = a2C + a2T = 4.47 cms −2 2h 6. R = uT = u g
ac = g
r P
Centre of curvature
Now, ∴ i.e.,
v = u cos θ ac = g v2 =g R v2 u2 cos2 θ R= = g g 3 2 = 21.65 ms −2
4. (a) ac = a cos 30° = 25 ×
u h = 2.9 m h Range (R) = 10 m
∴ i.e.,
u=R u2 =
g 2h
R2 g 2h
144 | Mechanics-1 Thus, centripetal acceleration of the stone (at point P) while in circular motion u2 R2 g 1 = = ⋅ r 2h r 2 10 × 9.8 = 2 × 2.9 × 1.5 = 112.6 ms −2 18.5 7. v = 18 km / h = ms −1 = 5 ms −1 18 v2 Angle of banking (θ) = tan −1 rg
mv2 = µ mg r i.e.,
9.
µ=
2 T1 cos θ − mg cos θ mr
mg sin θ 4 × 100 = 200 − 4 5
T2 = T1 −
2
v (5) 1 = = = 0.25 gr 10 × 10 4
v = r g tan θ
= 150 N
= 50 × 10 × tan 30° = 17 ms −1 10.
or ω =
= 8.37 rad s −1 1 8.37 × rev 2 π = 1 min 60 = 39.94 rev min −1. (b) From Eq. (ii),
(5)2 10 × 10 1 = tan −1 40
2
or T1 cos θ − mg cos θ = mrω2 − T1 cos θ
3 3 2 × 200 × − 4 × 10 × 5 4 = 4×3
= tan −1
8.
mr ω2min = N …(i) and mg = µ N …(ii) Solving Eqs. (i) and (ii), g ω min = µr =
12. (a) For the block not to slip N L
µN N
mg
∴
= 4.7 rads −1 11. (a) ( T1 + T2 ) cos θ = mrω2
…(i) T1
T1 sin θ
4m
4m
µN
mg
10 0.15 × 3
T1 cos θ
mrω2 − T1 cos θ T1 sin θ − mg
cos θ =
⇒
T2 cos θ T2 sin θ T2 mg
mL ω2max = µ N = µ mg µg ω max = L
(b) If the angular speed is gradually increased, the block will also have translational acceleration ( = α L) besides centripetal acceleration (= L ω2 ). αL anet 2
Lω
At the t) time of just slipping At t
and ( T1 − T2 ) sin θ = mg i.e., T2 cos θ = mrω2 − T1 cos θ and T2 sin θ = T1 sin θ − mg
…(ii)
=
0
(
Circular Motion | The block will be at he point of slipping, when manet = µ mg 2 2
L2ω4 + α2 L2 = µ 2 g2 µ 2 g2 ω 4 + α2 = 2 L
or
∴ if
2
i.e., ( L ω ) + (α L) = µ g or
1
µ 2 g 2 4 ω = 2 − α2 L
or
i.e.,
2
T
T
Ff
= 4 ( 4)2 = 64 N
1
2πr T a/ 3 2πr a3 T= = 2π = 2π 1 v 3 Gm GM 2 a
Now, v =
14. N sin θ = mrω2
T − Ff = 1 ⋅ 1 ⋅ ω2 64 − Ff = ( 4)2 ⇒ Ff = 48 N (b) T = 2 ⋅ 2 ⋅ ω2max T − Ff max = 1 ⋅ 1 ⋅ ω2max ∴ 4ω2max − Ff max = 1 ⋅ ω2max i.e., 3ω2max = µ mg µ mg ∴ ω max = 3 0.8 × 1 × 10 = 3
O'P = r
N cos θ
O N
θ
O'
P N sin θ mg
and N cos θ = mg 2 sin θ ( R sin θ) ω2 sin θ rω or ∴ = = cos θ g cos θ g g i.e., cos θ = Rω 2 For θ > 0° : cos θ < 1
and
= 1.63 rad s −1 2 ⋅ 2 ⋅ ω2max
R
1
2m
Gm 2 GM 2 v= = a r 3
or
θ = 60°
15. (a) T = 2 ⋅ 2 ⋅ ω2
60° =F 3 2 PQ = QR = RP = a = 2 PO cos 30° ∴ a=r 3 For the circular motion of P, Q or R mv2 = F′ = F 3 r mv2 G mm or = − 3 r ( r 3 )2 1
g R
i.e., bead will remain at the lowermost position. 2g g 1 For : cos θ = ω= = 2 g R R 2 R
13. F ′ = 2F cos
∴
g < 1 or ω > Rω2 g : θ = 0° ω≤ R
i.e.,
145
(c) 100 = ∴ ω max = 5 rad s −1 Therefore, if the string can sustain a tension of 1100 N, the angular speed of the block system will be 5 rad s −1. In this case the frictional force ( Ff ) on the block of mass 1 kg will be given by the relation 100 − Ff = 1 ⋅ 1 ⋅ 52 ⇒ Ff = 75 N gr l 16. vmax = h 10 × 200 × 0.75 (2 r = 1.5 m) = 1.5 = 31.6 ms −1
146 | Mechanics-1 17. If the block just leaves the surface of
19. At point P
sphere at point C mv2 …(i) = mg cos θ r [v = velocity of block at point C] 2 ⇒ v = gr cos θ
l θ T
v0 P mg sin θ mg
v0
Tmax = 2W = 2 mg
mv2 l m 2 = [v0 + 2 gh] l m 2 or 2mg − mg sin θ = [v0 + 2 gl sin θ ] l 52 + 2 g × 2 sin θ or 2 g − g sin θ = 2 1 Solving, sin θ = 4 1 i.e,. θ = sin −1 4 T − mg sin θ =
h
vel = v r
θ
Further, v2 = v20 + 2 gh …(ii) = v20 + 2 gr (1 − cos θ) Comparing Eqs. (i) and (ii), v20 + 2 gr (1 − cos θ) = gr cos θ or (0.5 gr )2 + 2 gr (1 − cos θ) = gr cos θ 1 + 2 (1 − cos θ) = cos θ 4 3 i.e., θ = cos −1 4 2 mv 18. T − mg cos θ = r θ
h = l sin θ
20. 8 = 5 + 5 cos θ v
H θ
T=0
5 cos θ 8m
θ = 30°
5 4
5m
θ 3
T θ
∴ mg cos θ
mg
∴
2.5 mg − mg cos θ =
∴
ac =
Thus,
mv2 r
v2 = 25 − 5 3 ms −2 r at = g sin θ = 5 ms −2
anet = ( ac )2 + ( at )2 = (25 − 5 3 )2 + (5)2 = 625 + 75 + 25 − 10 3 = 26.60 m/s2
3 5 mv2 T + mg cos θ = r cos θ =
∆t θ = cos −1
3 5
T is just zero. mv2 r 2 i.e,. v = 5 g cos θ = 30 ∴ v = 5.42 ms −1 The height (H) to which the particle will rise further. ∴
mg cos θ =
Circular Motion | 2
2
v sin θ 30 16 = × 2g 2 × 10 25 24 = = 0.96 m 25 mv2 8 gR 21. (a) N = =m = 8 mg R R v2 = 02 + 2. g.4 R = 8 gR
F=
∴
H=
2
147
2
N + ( mg)
= 64 m2 g2 + m2 g2 = 65 mg mv ′2 (b) N ′ + mg = R m mg + mg = 2 g ( h′ − 2R) R or 2R = 2 ( h′ − 2R) i.e., h′ = 3 R
4R h = 5R N mg F
Objective Questions (Level 1) Single Correct Option 1. (a) As the speed (v) is increasing uniformly
the magnitude of centripetal acceleration v2 = will also keep on increasing besides r its direction as usual. (b) v = kt (k = constant) dv ∴ =k dt i.e., at = k Although the magnitude of the tangential acceleration will remain constant but its direction will keep on changing as the direction of velocity would be changing. (c) v = kt Rω = kt ω = k′ t k (k′ = = constant) R dω = k′ dt α = constant. As the direction of α (angular acceleration) is perpendicular to the plane of rotation of the body, it will remain constant both in magnitude and direction. Option (c) is correct.
→
→
2. |p1|= |p2 |= mv → –p1
→ p2 → → –p1 ∆p → ∆p
→ p2
→ p1
θ = 45° → ∆p
→ p2 → –p1 → p2 → –p1
→ ∆p →
θ = 45°, |∆p|= 2 mv sin 22
1° 2
= 0.765 mv →
θ = 90°, |∆p|= 2 mv sin 45° = 1.414 mv →
θ = 180° , |∆p|= 2 mv sin 90° = 2 mv →
θ = 270°, |∆p|= 2 mv sin 135° = 1.414 mv
(max)
148 | Mechanics-1 →
θ = 360°, |∆p|= 2 mv sin 180° = 0 (min) Option (c) is correct. 3. θ
PEmax = mgh = mgl (1 – cosθ)
h
KE = 0
6. Normal acceleration
= Tangential acceleration v2 = 5 (cms −2 ) R (Q R = 20 cm) ∴ v = 10 cms −1 Using, v = 0 + 5t 10 = 5t ⇒ t =2s Option (b) is correct. 7. Mass = 2π kg
∴ Mass per unit length = ∴KE max = mgl (1 − cos θ) Option (c) is correct.
[as here PE = 0]
2π kg 2π (0.25) m T ∆θ
4.
B OB
=
OA l=
v2 = u2 + 2(– g) (2l) = 4gl – 4gl =0
∆θ
T Light O rigid rod
A
1 r ∆θ = ∆θ 0.25 1 = kgm −1 0.25 rev ω = 300 min 2π rad = 300 × 60 s
∆m =
u = √4gl
= 4 gl − 4 gl =0 If the mass m is given velocity u ( = 4 gl ) at point A it will complete circle to reach point B (the highest point) with zero velocity. [If in place of light rod these is light string the minimum value of u at A for the mass to reach point B will be 5gl and the minimum velocity at B will be g l.] 5. As explained in question 4 speed at lowest
point u = 5 gl At the lowest point mu2 T − mg = l m 5 gl = l = 5 mg ⇒ T = 6 mg Option (d) is correct.
(Q r = 0.25)
= 10π rad s −1 From figure 180° − ∆θ 2 2T cos = ( ∆m) rω 2 ∆θ or 2T sin = ( ∆m) rω2 2 ∆θ or 2T = ( ∆m) rω2 2 or T ∆θ = ( ∆θ) rω2 or T = r ω2 = 0.25 × (10π)2 (Q π2 ≈ 10) ≈ 250 N Option (d) is correct. 8. Maximum speed of car = µgr
= 0.3 × 10 × 300 = 30 m/s
Circular Motion | 18 km/h = 30 × 5 = 108 km/h Option (c) is correct.
149
Frictional force = manet = 0.36 × 10−3 kg ×
9. T sin θ = mrω2 = m ( l sin θ) ω2
725 × 10−2 ms −2 3
= 32.4 µN Option (a) is correct. 12. At the lowermost point
θ θ l
T
T cos θ
l
θ r
T
T sin θ
u=0 h v2 = 2gh
and ∴
∴ i.e.,
T cos θ = mg g cos θ = l ω2 2 ω l = revs −1 π 2 = (2π rad) s −1 π = 4 rads −1 10 5 cos θ = = 1 ( 4)2 8 5 θ = cos −1 8
mg
mv2 l m ⋅ 2 gh T= + mg l 2h = mg 1 + l
T − mg = or
Option (d) is correct. 2π 13. ω A = rad min −1 3
Option (d) is correct. A
10. From above question no. 9.
B
2
t = mlω 100 8 = × 1 × 42 = N 1000 5 Option (b) is correct. 1 11. α = rads −1 and R = 25 cm 3 25 ∴ at = Rα = cms −2 3 Thus, ωt = 0 + α t 1 [Q t = 2 s] = ×2 3 2 2 ∴ aN = R ω2t = 25 × cms −2 3 725 Thus, anet = a2N + a2t = cms −2 3
ω B = 2π rad min −1 ω BA = ω B − ω A 2π −1 = 2π − rad min 3 4π rad min −1 = 3 Time required for B to complete one revolution w.r.t. A 2π 3 = = min 4π / 3 2
and ∴
= 1.5 min Option (c) is correct.
150 | Mechanics-1 14. Let all the particles meet at time t
(seconds) B 2.5 ms–1
A 1 ms–1
∴ Distance travelled by B in t second : Distance travelled by C in t second = 2.5t : 2t = 5:4 Option (c) is correct.
C 2 ms–1
JEE Corner Assertion and Reason 1. For stopping car :
Maximum retardation Maximum frictional force = m µ N µ mg = = =µ g m m v2 = u2 + 2as 02 = v2 + (2)( − µ g) d [Q Initial velocity = v] v2 ∴ d= 2µ g For circular turn of car : Centripetal force = Maximum frictional force = µ mg mv2 ∴ = µ mg d′ v2 i.e., safe radius = d′ = ⇒ d′ = 2d µg Thus, Assertion and Reason are both correct and Reason is the correct explanation of the Assertion. Option (a) is correct. →
2. a av =
→
→
vB− v A t AB
→
→ |BD| v 2 = t AB t AB
→
→ |AB| R 2 = t AB t AB
i.e.,
|a av |=
Now,
|v av|=
→
∴
|a av| →
|v av|
=
v 2 = ω (angular velocity) R 2
Thus, assertion is correct. In circular motion, when speed is constant, the angular velocity will obviously be constant; but this reason does not lead to the result as explained. Option (b) is correct. 3. A frame moving in a circle with constant
speed can never be an inertial frame as the frame is not moving with constant velocity (due to change in direction). Reason that the frame is having constant acceleration is false. Option (c) is correct. →
will always be 90° →
ω
→ vB →
vA
D
θ → →
B
→ BD = t AB
→
4. If speed is constant angle between v and a
a = ac
→
vA
A
∴
→ →
v⋅a = 0
Circular Motion | →
If speed is increasing angle θ between v → and a will be less than 90°. → v
=
(2)2 =2m 2
→
a T = − ^i ms −2
θ → ac
→
Speed is decreasing (as a T is −ive) at a rate of 1 ms −1 per second i.e., 1 ms −2 . Both Assertion and the Reason are correct but reason has nothing to do with the assertion. Option (b) is correct.
→ →
v ⋅ a will be positive.
∴
v ac
R=
⇒
→ aT
→ a
→
→
aT
v
→
→
→
6. a = a T + a c (Reason)
θ →
→
a
ac
ac →
Assertion is correct. → →
Reason as both are ω⋅v =0 perpendicular to each other. Reason is also true but not the correct explanation of the assertion. Option (b) is correct.
→
v2 + g2 r
∴
|a |=
⇒
|a |> g (Assertion)
→
We see that both Assertion and the Reason are correct and the reason is the correct explanation of assertion. Option (a) is correct. 7. At
points A and C : momentarily at rest.
^
5. v = 2 i ms −1
→
aT = g
a
→ →
∴ v ⋅ a will be negative.
A →
→
If speed is decreasing angle between v and → a will be greater than 90°.
→
151
2
The
bob
is
R →
anet
x 2j v = 2z
–i →
a = − ^i + 2 ^j ms −2
∴
→
^
→ ac
→ = |a c |=
a c = 2 j ms
→
−2
2 ms
v2 R
→
g
i.e.,
−2
v = |v|= 2 ms −1 ac =
C
A B
→
g
v = 0 (Reason) v2 ∴ =0 R but net acceleration is not zero (see figure) i.e., Assertion is false. ∴ Option (d) is correct. → |a c |=
152 | Mechanics-1 8. v (speed) = 4 t − 12
For t < 3 (time unit) speed is negative, which can’t describe a motion. Thus, assertion is correct. As speed can be changed linearly with time, the reason is false. Option (c) is correct. 9. In
circular motion the acceleration changes regularly where as in projectile motion it is constant. Thus, in circular → → → motion we can't apply v = u + a t directly, whereas in projectile motion we can say reason that in circular motion gravity has no role is wrong. Option (c) is correct.
10. N = mg cos θ N O N sinθ
θ
mg cos θ
mg
Therefore, assertion is wrong. Particle performs circular motion due to N sin θ mv2 ∴ = N sin θ r
N cos θ is balanced by mg (weight of particle). Acceleration is not along the surface of the funnel. It is along the centre O of the circle. Thus, reason is true. Option (d) is correct. mv2 11. Centripetal force = N + mg r N mg
i.e., Centripetal force (reason ) ≥ wt ( mg) of water for N = 0, v = gr If at the top of the circular path v ≥ gr i.e., if bucket moved fast, the water will not fall (Assertion). As assertion and reason both are true and reason is the correct explanation of the assertion of the option would be (a).
Objective Questions (Level 2) Single Correct Option
1 2
1 2
1. KE mv2 = Change in PE k( ∆x)2
∆x = Length of spring (Collar at B) − Length of spring (Collar at A) = (7 + 5)2 + 52 m − 7m = 6 m 1 1 Thus, × 2 × v2 = × 200 × 62 2 2 2 i.e., v = 3600 mv2 ∴ Normal reaction = r 2 × 3600 = = 1440 N 5 Option (a) is correct.
2. The particle will remain in equilibrium till
ω is constant. Any change in the value of ω will displace the particle up (if ω increase) and down (if ω decreases). Thus, the equilibrium is unstable. Option (b) is correct. 3. Centripetal force = µ mg
or or or
mrω2 = µ mg 5a 2 ω =µ g 4 4g ω2 = 15a
Option (d) is correct.
1 as µ = 3
Circular Motion | 4. Acceleration at B = Acceleration at A
v2 ∴ = g sin θ r 2 gr (1 − cos θ) or = g sin θ r or 2 (1 − cos θ) = sin θ or [2 (1 − cos θ)]2 = 1 − cos2 θ or (5 cos θ − 3) (cos θ − 1) = 0 As cos θ = 1, i.e., θ = 0° is not possible. 3 cos θ = 5 −1 3 i.e., θ = cos 5 Option (c) is correct.
5. At point P (for the circular motion)
mv2 R If at point P skier leaves the hemisphere. N =0 mv2 ∴ mg cos θ = R m R or mg cos θ = 2g h + R 4 m R or mg cos θ = 2 g R (1 − cos θ) + R 4 1 ⇒ cos θ = 2 (1 − cos θ) + 4 −1 5 i.e., θ = cos 6 Option (c) is correct. mg cos θ − N =
6. Velocity at B
v = 2 gh = 2 gR (cos θ2 − cos θ1) = 2 gR (cos 37 ° − cos 53 ° ) 4 3 = 2 gR − 3 5 =
2 gR 5
For the circular motion at B, when block just leaves the track v2 = g cos θ2 R1
2
v 2R = g cos θ2 5 cos θ2 2R = 5 ⋅ cos 37 ° 2R = 4 5⋅ 5 R = 2 Option (c) is correct. mv2 7. mg cos θ = a a 42 + 2 ⋅ g ⋅ 4 or g cos θ = a a a g− u2 + 2 ⋅ g ⋅ 4 4 or g = a a 3 a or g ⋅ a = u2 + 2 g 4 4 ag ⇒ u= 2 Option (c) is correct. v2 4 8. = 2 r r 2 ⇒ v= r or
R1 =
∴ Momentum = mv 2m = r 9.
N cos θ = mrω2 and N sin θ = mg ∴ N = m g2 + r2ω4 2π = m g2 + r2 T
4
2π = 10 10 + (0.5) 1.5 a 2
= 128 N Option (b) is correct. 1 10. θ = ω 0 t + αt2 2 1 = ω 0 + α t t 2
2
4
153
154 | Mechanics-1 v 1 1 v2 4 πR = + − ⋅ t R 2 R 4 πR v v v 4πR = − R 2R v
2H g
2π g = 2π T 2H 2g =π H 2π 12. ω (minute hand) = rad s −1 3600 2π rads −1 ω (second hand) = 60 For second hand to meet minute hand for the first time. 2π + Angle moved by minute hand in t second = Angle moved by second hand in t second 2π 2π or 2π + t= t 3600 60 t t 1= − 60 3600 t 59 1= × 60 60 3600 s ⇒ t= 59 Option (d) is correct. ω=
13.
( PR)2 + (QR)2 = ( PQ)2 (v A ⋅ 2)2 + (vB ⋅ 2)2 = 302 v2A + v2B = 225
60°
30°
B'
PR = cos 30° PQ vA ⋅ 2 3 = 30 2 ⇒ v A = 7.5 3 ms −1 Substituting value of v A in Eq. (i), vB = 7.5 ms −1 mv2 14. mg cos θ − N = r When breaks off N = 0 mv2 ∴ mg cos θ = r Further,
R
T=
vB
30 m
H
⇒
t=2s
vB = AD
= 4π = 2 rev. 1 H = gT2 2
11.
R
…(i)
N
θ
sin mg
O
or
g cos θ =
θ
Acceleration = g sin θ
mg
2 gr (1 − cos θ) r 3
√5
θ 2
or
cos θ = 2 (1 − cos θ) 2 or cos θ = 3 Acceleration of particle when it leaves sphere = g sin θ g 5 = 3 Option (b) is correct.
Circular Motion | 15. For minimum velocity (v) :
tan θ = tan 45° = 1 µ =1 As µ = tan θ θ is the angle of repose. N
∴ (given)
N cos θ
µN cos θ µN
µN sin θ
θ
Therefore, the automobile will be at the point of slipping when its velocity is zero For maximum velocity (v′ ) mv ′2 N sin θ + µ N cos θ = r Also N cos θ = mg + µ N sin θ i.e., …(ii) N (cos θ − µ sin θ) = mg Dividing Eq. (i) by Eq. (ii), sin θ + µ cos θ v ′2 = cos θ − µ sin θ gr
aT = g anet = a2c + a2T = (3 g)2 + g2 = g 10
Option (a) s correct. 17.
and ∴
N sin θ
N = mrω2 µN = mg g µ= rω2 10 = 0.2 = 0.2(5)2
19. At point A : Velocity is zero and as such its
vertical component will also be zero. At point B : Velocity is completely horizontal and as such its vertical component will again be zero. T θ
u = 5 gR Particle's velocity (v) when it is at B i.e., when its velocity is vertical would be given by the relation v R
u = √5gR
ac =
v2 = 3g R
mg
= 31.62 m/s
P
the value of
At B :
N
R
16. For the particle to just complete the circle
v2 = u2 + 2 ( − g) R = 5 gR − 2 gR = 3 gR
µN
Option (c) is correct. gra 18. vmax = h 10 × 200 × 075 . = 1.5
A
Now, as θ = 45° and µ = 1 v′ = ∞ Option (d) is correct.
155
v v1
B
In figure, TP = R cos θ At point P : v2 = 2 g ( R cos θ) v = 2 gR cos1/ 2 θ ∴ v⊥ = v sin θ = 2 gR cos1/ 2 sin θ For v⊥ to be maximum d v⊥ = 0 dθ d i.e., 2 gR cos1/ 2 θ sin θ = 0 dθ 1 or cos1/ 2 θ cos θ + cos −1/ 2 θ( − sin θ) sin θ = 0 2 sin2 θ or cos θ cos θ − =0 2 cosθ
156 | Mechanics-1 2 cos2 θ − sin2 θ = 0 3 cos2 θ = 1
or or
θ = cos −1
or
21. At highest point B B
1 3
T mg
Option (b) is correct. 20. At any time
at = ac dv v2 i.e., = dt R dt −2 ∴ ∫ v dv = ∫ R + k v −2 + 1 t i.e., = +k −2 + 1 R 1 1 or − = +k v R At t = 0, v = v0 (given) 1 0 ∴ − = +k v0 R 1 i.e., k= − v0 1 t 1 Thus, − = − v R v0 1 tv − R or − = 0 v Rv0 Rv0 or u= R − tv0 v0 or ω= R − tv0 dθ v0 i.e., = dt R − tv0 v0 or dθ = dt R − tv0 2π T v ∫ 0 dθ = ∫0 R −0tv0 dt or or or or
2π = − [log e ( R − tv0 )]T0 = log e R − log e ( R − Tv0 ) R − Tv0 log e = = − 2π R T 1 − v0 = e−2 π r R T= (1 − e−2 π ) v0
Option (c) is correct.
R u
A
mv2 l (where v = velocity at point B) Thus, T =0 mv2 if mg = l 2 i.e., v = gl If u = velocity at the lowest point A T + mg =
v2 = u2 + 2 ( − g) (2l) or
gl = u2 − 4 gh
⇒
u2 = 5 gl
i.e.,
u = 5 gl → →
(If T = 0, T ⋅ a = 0) Option (b) is correct. 22. For any value of u at the lowest point both →
→
T and a will be towards the centre of the circle and thus → →
T ⋅ a will be positive. Option (d) is correct. 23. Change in PE of the system m M
l l/2
M
l + mgl 2 M = + m gl 2 = Mg
m
Circular Motion | 2
24. Decrease in KE = Increase in PE
1 M mu2 = + m gl 2 2
i.e.,
∴ Initial speed given to ball, M + 2m u= gl m 25. Maximum and minimum velocities will be
respectively at the lowest and the highest points. v
or i.e.,
= (2 v) − 4 gL 3v2 = 4 gL 4 gL v = gL = 2 3 3
157
(Qu = 2 v)
Option (b) is correct. gL 26. u = 2 v = 4 3 ∴ KE at the lowermost position =
(Min. velocity)
=
1 mu2 2
8 mgL 3
27. Let ω = velocity of the particle when
ω u (Max. velocity)
v2 = u2 + 2 ( − g) 2L = u2 − 4 gl
moving downwards. ∴ ω2 = v2 + 2 gL 4 gl = + 2 gl 3 10 gl or ω= 3 Option (a) is correct.
More than One Correct Options 1. At point C : vmin
v'
A aT = 0 g ac(min) θ
B T mg
B aT = g sin θ
l
C aT = g ac(max)
v
g 2
vmax
or
D aT = 0
or aT is maximum and ac is somewhere between maximum value (at D) and minimum value (at A). ∴ Option (b) is correct. At point D : ac is maximum while be aT is minimum. ∴ Option (d) is correct. 2. At point B :
T + mg =
mv ′2 l
i.e.,
2mg + mg =
mv ′ l
v ′2 = 3 gl v ′ = 3 gl
Option (d) is correct. Now,
v2 = v ′2 + 2 g (2l)
∴
v2 = 3 gl + 4 gl = 7 gl
i.e., v = 7 gl Option (b) is correct.
158 | Mechanics-1 4. Particle
can’t have uniform motion because of change in direction of motion i.e., its velocity value. [Option (a)] Particle can’t have uniformly accelerated motion as acceleration changes direction even if speed is constant. [Option (b)] Particle can’t have not force equal to zero as centripetal force would be required for the circular motion. [Option (c)]
3. N sin α = mg N r
N sin θ
N cos θ mg h α
5. For this see figure in answer 3.
N cos α = mrω2 g g = tan α = r tan θ rω2 2π g = T r tan α r tan α T = 2π g
⇒ i.e., or
If ω is increased N cos α will increase. Thus, N will increase (as α is constant) [Option (b)] And as such net force = N 2 + m2 g2 + 2 N cos (90° + α)
If α is increase r will also increase and as such T will increase. Option (c) is correct. h tan2 α T = 2π g
on the block will increase. ∴ Option (a) is correct. As N increases, the value of N sin α (acting opposite to mg) will increase and the block will upwards i.e., h will increase.
Thus, if h is increase, T will increase. Option (a) is correct. Match the Columns 1. At point B : 2
v
2
v = u − 2 gl = 12 gl − 2 gl i.e., v = 10 gl = 10 × 10 × 1 = 10 ms −1 b (a) → (p). A Acceleration of bob : v2 10 g ac = = = 100 ms −2 l 1 aT = g = 10 ms −2 anet = a2c + a2T = 1002 + 102 ms −2 (b) → (s). Tension in string : mv2 1 × 100 T= = = 100 N l 1
B
(c) → (r). Tangential acceleration of bob : aT = g = 10 ms −2 (d) → (p). 2. v = 2 t
dv =2 dt
∴
i.e., a = 2 ms −2 At t = 1 s : v = 2 ms −1 v2 22 ∴ ac = = = 2 ms −2 r 2 aT = 2 ms −2 ∴
anet = 22 + 22 = 2 2 ms −2 →
→
Angle between a net and v = 45° [As ac = aT ]
Circular Motion | → →
→
→
F=f F =1 f
(a) a ⋅ v = |a ||v|cos 45° 1 2 1 = 2 2 ⋅2 2 →
= a net ⋅ v
= 4 unit. Thus, (a) → (r). →
→
→
→
(b) |a × ω|= |v||ω|sin 90° →
[ω will be perpendicular to the plane of circle] = anet ω v = anet r 2 =2 2⋅ r = 2 2 unit Thus, (b) → (p) → →
(c)
→ →
v⋅ ω = |v||ω|sin 90° v = vω = v ⋅ r 2 (2) = 2 = 2 unit Thus, (c) → (q) →
→
→
i.e.,
∴ (c) → (r). If v is increased F will increase which will automatically increase the value of f . ∴ (d) → (p). 4. Speed of particle is constant
= 4 unit Thus, (d) → (r) M v2 h 3. N1 = g − 2 ra On increase v, the value of N1 will decrease. Thus, (a) → (q). M v2 h N2 = g + 2 ra On increasing v, the value of N2 will increase. ∴ (b) → (q). As the centripetal force ( F ) would be provided by the frictional force (f )
→
→
|a |=
∴
|v|2 →
|r | ( 4)2 + ( − a)2
( −6)2 + ( b)2 =
(3)2 + ( − 4)2
36 + b2 = → →
→
16 + a2 5
…(i)
→
(A) r ⋅ v = |r ||v|cos 90° = 0 (3 ^i − 4 ^j) ⋅ ( 4 ^i − a ^j) = 0
i.e.,
12 + 4 a = 0 a= −3 ∴ (a) → (s). (B) Substituting value of a ( = −3) in Eq. (i) 36 + b2 = 5 36 + b2 = 25
→
(d) |v × a |= |v||a |sin 45° 1 = v anet ⋅ 2
159
b = 25 − 36 ∴ (b) → (s). →
(C) r = 3 ^i − 4 ^j →
r = |r |= 5
∴ ∴ (c) → (r). →
→
→
→
→
→
^
(D) ( v × a ) = |v||a |sin 90° k ^
= |v||a | k →
→
→
Thus, r ⋅ ( v × a ) → → →
^
= r| ⋅ v||a |k →
→ ^
= (3 ^i − 4 ^j) ⋅|v||a|k = 0 ^
^
[as ^i ⋅ k = 0, ^j ⋅ k = 0] ∴ (d) → (s).
160 | Mechanics-1 5. (A) As speed is constant
Average speed = 1 ms → v2 → v1
=
−1
B → ∆v mg
R 2 R 2 2 2 = ⋅1 = AB arc πR π 2 speed
∴ (b) → (q) (C) Modulus of average acceleration
→ v1
→
= |a c | [as speed is constant] A
→
→
→
|v − v1| |∆v| v 2 = 2 = = R t AB t AB π 2 = 2 [ Q v = 1 ms −1 and R =
∴ (a) → (s) (B) Modulus of average velocity =
AB t AB
∴ (c) → (r)
2 ] π
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision
8
Introductory Exercise 8.1 1. If
a body is placed in a uniform gravitational field, the CG of the body coincides with the CM of the body. n
→ rCM
→
Σ mi ri
=
i =1 n
Σ mi
i =1 n
→
while
→
Σ mi g i ri
r CM =
i =1 n
Σ mi g i
i =1
If a body is placed in a uniformly → increasing gravitational field ( g ) in the upward direction the CG of the body will be higher level than the CM. And, if the body is placed in a uniformly decreasing gravitational field in the upward direction the CG will be at a lower level the CM.
2. The centre of mass of a rigid body may lie
inside, on the surface and even outside the body. The CM of a solid uniform sphere is at its centre. The CM of a solid ring is at the centre of the ring which lies outside the mass of the body thus, the statement is false. (For further details see answer to 1 Assertion and Reason JEE corner). 3. Centre of mass always lies on the axis of symmetry of the body, if it exists. The statement is thus true. 4. Statement is true. 5. As more mass is towards base. r Distance < . 4 6. If
two equal masses are kept at co-ordinates (R, 0) and ( −R, 0), then their centre of mass will lie at origin.
7.
3 kg (1/2, 3/2)
CG CM g increasing
g decreasing
CM CG
CG shifts from CM according to the magnitude and direction of the gravitational field (by some other agency eg, earth) in which the body is placed. In zero gravitational field CG has no meaning while CM still exists, as usual.
1 kg (0, 0)
2 kg (1, 0)
m1 X1 + m2 X2 m1 + m2 m1 y1 + m2 y2 = m1 + m2
X CM = YCM 2 r = X 2CM + YCM
= Distance of centre of mass from A
162 | Mechanics-1 y
8.
YCM =
A1 y1 − A2 y2 A1 − A2
9. A1 = 4 a2 , x1 = a, y1 = a 3a 3a A2 = a2 , x2 = , y2 = 2 2 x
O
Introductory Exercise 8.2 1. Method 1 t=0s (10 – x) CM (i)
a
1 kg
2 kg y
2m t=1s
7–y CM (f)
T
a
T
°
1x = 2 (10 – x) 20 x= cm 3
60
⇒
2.
in
At t = 0s
A
2 kg
+v e
x
B
A
mg s
1 kg A t=0s
60°
mg s
in
30°
30
°
1m t=1s
1 y = 2 (7 − y) 14 cm ⇒ y= 3 Displacement of CM = Position of CM ( f ) − Position of CM ( i) = ( y + 2) − ( x) 14 20 = + 2 − 3 3 =0 Method 2 M vCM = m1v1 + m2v2 = 1 ( 2 ms −1) + 2 ( − 1 ms −1) =0 As velocity of CM is zero, there will not be any change in the position of the CM.
mg sin 60° − T = ma T − mg sin 30° = ma Adding above equations, mg (sin 60° − sin 30° ) = 2 ma 3 − 1 a= g 4
⇒ → → a1 and a2 →
∴ a CM = or
are at right angle →
→
m a1 + m a2 1 = ( a1 + a2 ) 2m 2 a g( 3 − 1) → |a CM|= = 2 4 2
…(i) …(ii)
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision |
163
Introductory Exercise 8.3 1. vCM
20 v + 60 ( + 3) = =0 20 + 60
vCM = 60 kg + ve
$ 20 × 20 $i + 30 × 20 $i + 40 × 20 k 20 + 30 + 40 y (10i + 20k) cms–1
20 kg
30 g v
j
3 m/s 20 kg
→ v
60 kg
[As there is no force along horizontal direction]. ⇒ Velocity of trolley (v) = − 9 ms −1 Total energy produced by man = KE of man + KE of trolley 1 1 = × 60 × 32 + × 20 × ( − 9)2 = 1.08 kJ 2 2 2. On streching and then releasing the
spring the restoring force on each block at instant will be same (according to Newton’s 3rd law of motion). Now, as force is same momentum p of each block will also be same (∆t being same) [As according to Newton’s second law of motion rate of change of momentum of a body is directly proportional to the net force applied on the body.] p2 Now, as KE = , 2m KE of blocks at any instant will be inversely proportional to their respective masses. 3. As no external force acts on the system of
particles, the velocity of CM shall not change. Thus, y 20 cms–1 30 g 20 cms–1 j x 20 cms–1
k
i
20 g
k
i
x
20 g
z →
^
vCM =
20 × 0 + 30 (10 ^i + 20 k) + 40 v 20 + 30 + 40 ^
^
300 ^i + 600 ^j+ 40 k = 400 ^i + 600 ^j+ 800 k →
^
40 v = 100 ^i + 600 ^j + 200 k
⇒
→
^
v = ( 2.5 ^i + 15 ^j + 5 k ) cms −1
i.e.,
4. Velocity of projectile at the highest point
before explosion = 10 2 ^i ms −1 → ^ v = 10 √2 i
20 45°
x
20 cos 45° = 10√2 ms–1
R/2
As no extra external force would be acting during explosion, the velocity of CM will not change m → m→ ⋅0+ v 2 2 = 10 2 ^i m →
v = 20 2 ^i R Range of rest half part = ×2 2 (as the velocity of the projectile has doubled at the highest point) u2 (as α = 45°) = R= g ⇒
= z
Rest
(20)2 = 40 m 10
164 | Mechanics-1 Therefore, the rest half part will land at a 40 distance of 40 + m i.e., 60 m from the 2
= Height of point P + Extra maximum height attained by second half part
point of projection.
1 ( 2000)2 sin2 α = 20 × 1 − g × 12 + 2 2g 2000 1 = 15 + × 2 × 10 5
20 √2 i u
R
= 15 + 20 = 35 m
5. At point P
6. Momentum of platform + boy + stone
along x-axis after throwing stone = before throwing stone 1 ∴ (60 + 40) v + 110 = 0 2
u = √(40)2 + (20)2 P 20 ms–1 t=0s
20√2 ms
–1
–1 α = √2000 ms t = 15
15 m
⇒
45°
v=−
5 2 ms −1 100
60 kg
Horizontal velocity = Initial horizontal velocity −1 = 20 2 cos 45° ms = 20 ms −1 Vertical velocity = 20 2 sin 45° − g ⋅ 1 (as t = 1s) = 70 ms −1
40 kg
→
10 cm
1 kg X
Time of flight of stone
→
u = 2 v = 40 ^i + 20 ^j ms −1
Angle of projection (α) of 2nd half part after explosion at point P. 20 α = tan −1 √5 40 = tan
−1 1
2
1
α 2
X
10 ms–1
∴ v (velocity of projectile at point P i.e., at t = 15 just before explosion) Now, as the projectile breaks up into two equal parts and one part comes to rest, the velocity of other half part after explosion will be
45° 1 kg
→
= 20 ^i + 10 ^j
10 ms–1
T = √2 s
2 × 10 sin 45° = 2s g
∴ Horizontal displacement of platform (+ boy) =vT 5 2 1 m = − 10 cm =− × 2=− 100 10 7. Thrust due to the upward component of the velocity of the bullet will rotate the movable end of the barrel and thus the bullet leaving the barrel will travelling at an angle greater than 45° when it comes out of the barrel.
1 sin α = 5 Maximum height attained by second half part
Bullet Bullet
θ = 45°
θ' > θ
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision |
165
Introductory Exercise 8.4 1. To just lift rocket off the launching pad
Thrust force = Weight dm v − = mg dt dm mg − = dt v (20 + 180) × 9.8 = 1.6 × 103
or or
i.e.,
180 kg =9s 20 kgs −1
v = − 9.8 × 9 + 1.6 × 103 ln
200 20
= 3.6 kms −1 2. Mass at time t, m = m0 − µ t
dm = −µ dt ma = thrust force − mg
∴
= 1.225 kgs −1 (i) Rate of consumption of fuel = 2 kgs −1 ∴ Time required for the consumption of fuel 180 kg t= = 90 s 2 kgs −1 Ultimate speed gained by rocket m v = u − gt + v log e 0 m
t=
∴
…(i)
Substituting u = 0, v = 1.6 kms −1, m0 = (20 + 180) kg and m = 20 kg in Eq. (i). 200 v = − 9.8 × 90 + 1.6 × 103 ln 20 −1 = 2.8 kms (iii) Rate of consumption of fuel = 20 kgs −1
dm ma = v − − mg dt
or or
ma = µ v − mg d2 x or ( m0 − µt) 2 = µ u − ( m0 − µ t) g (Qv = u) dt m 3. v = u − gt + v ln 0 m m0 = 0 − gt + u ln t m0 1 − 3 3 (at t = 1 s) = − g 1 + u ln 2 3 = u ln − g 2
Introductory Exercise 8.5 →
^
^
1. u (at t = 0 s) = ( 10 3 i + 10 j ) ms −1 20 m/s 60°
At t = 1 s Horizontal velocity = 10 3 ^i ms −1 Vertical velocity = (10 − g ⋅ 1) ^j ms −1 = 0 ^j ms −1 →
→
→
−1 →
Change in velocity, ∆ v = v − u = ( 10 3 ) ^i − (10 3 ^i + 10 ^j) = − 10 ^j ms −1 i.e.,
→
= Change in momentum in the time interval t = 0 s to t = 2 s = m [(Velocity at t = 2 s) − (Velocity at t = 0 s)] ^ ^ −1 = 2 kg [( 4 i + 4 j) ms − ( 4 ^j ) ms −1 ] = 8 ^i Ns
(Q 1 kg ms −1 = 1 Ns)
3. Spring will become taut when the ball
would go down by 2 m.
^
v = 10 3 i ms
∴
→
2. Impulse ( J ) imparted
∆ v = 10 ms −1, downwards.
v2 = u2 + 2 × g × 2 or v = 2 10 ms −1 ∆p = m ∆v = 1(2 10 − 0) kg ms −1 = 2 10 kg ms −1 ∴ Impulse imparted = 2 10 Ns.
2m
166 | Mechanics-1 Introductory Exercise 8.6 1. At maximum extension of the spring both
blocks will move with same velocity v. + ive 1.0 ms–1
2.0 ms–1
3 kg
of
v
Rest
V
m
m
m m
or or
1 2 K K = m⋅ + EPE max 4 m K EPE max = 2
v1'
m2
m1
m2
m1
∆ K (Transfer of KE of particle of mass m2 ) 1 1 = m2 v22 − m2 v ′22 2 2 2
From above we conclude that ∆ K will be maximum when m1 − m2 = 0 m2 i.e., =1 m1
During collision the EPE of the system would be at its maximum value when both the particles move with same velocity V given by the relation mv + m × 0 = ( m + m) V (Law of conservation of momentum) v ⇒ V = 2 1 2K = 2 m Applying law of conservation mechanical energy 1 K + 0 = ( m + m) V 2 + EPE max 2 or K = mV 2 + EPE max
v2'
m − m2 1 1 m2 v22 − m2 1 v2 2 2 m1 + m2 2 m − m2 1 = m2 v22 1 − 1 2 m1 + m2
1 mv2 = K 2 2K v= m
∴
v1 = 0
=
2. Kinetic energy of particle = K
⇒
v
m − m2 2 m2 v2 ′ = 1 v2 + v1 m1 + m2 m1 + m2 m − m2 (as v1 = 0) = 1 v2 m1 + m2
6 kg
∴ ( − 1.0) × 3 + ( + 2.0) × 6 = (3 + 6) v ⇒ v = + 1 m/s Applying law of conservation mechanical energy 1 1 1 1 3 ( − 1)2 + 6 (2)2 = 9 × 12 + x2 2 2 2 k −1 Substituting k = 200 Nm x = 0.3 m = 30 cm
3. Let us consider the following case
of
4. Continuing from the previous answer
∆K =1− K =
m1 − m2 m1 + m2
2
( m1 + m2 )2 − ( m1 − m2 )2 ( m1 + m2 )2 4 m1m2 = ( m1 + m2 )2
5. Substituting m2 = m and m1 = 2m in the
result of the previous question no. 4. u
Rest
m
2m
∆ K 4 (2m) ( m) = K (2m + m)2 8 = 9
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision |
∴ Total number collisions between the balls = 2. m − m1 2m1 7. v ′2 = 2 v2 + v1 m1 + m2 m1 + m2
6. As the collision is elastic u
Rest
v1
v2
m1
m2
m1
m2
v1 = and
m1 − m2 u m1 + m2
v2 m2
v2 = v1 + u m − m2 = 1 u+u m1 + m2 2 m1 = u ( m1 + m2 ) B
4m
4m
4m Rest
8. In elastic collision of two bodies of equal
masses, the velocities get interchanged. Before collision of A with B velocity of A = + v velocity of B = zero Therefore, after collision of A with B velocity of A = zero velocity of B = + v After collision of B with wall
2v 5
velocity of B = − v Before collision of B with A velocity of A = zero
m − 4m 3 − v m + 4m 5 9 = v 25 9 3 6 = v + − v = − v 5 25 25
vB =
velocity of B = − v
2
vA
2
Therefore, after collision of B with A velocity of A = − v velocity of B = zero
After second collision : of B with A A 6v 25
4m
m1
Ans. No.
4m
4m
After
v2′ = v1 and v ′1 = v2
C
m
3v 5
m2
However, if m1 = m2
After collision of B with C m − 4m vB = v 1 m + 4m 3 =− v 5 3 2 and vC = − v + v = v 1 5 5 Before second collision of B with A B
m1
v1'
We see that v ′2 ≠ v1 and v ′1 ≠ v2 .
C v
v2'
Elastic collision
Rest
A
Before
v1
m − m2 2m2 v ′1 = 1 v1 + v2 m1 + m2 m1 + m2
Before first collision of B with C A
167
B 4m
C 9v 25
4m
2v 5
As velocity of C is greater than that of B, no further collision will take place.
Now, no more collision will take place as A will move towards left with speed v leaving B at rest to its right side. Thus, we see that speeds of the balls remains unchanged after all the possible collisions have taken place.
168 | Mechanics-1 Introductory Exercise 8.7 1. Applying
law
of
conservation
of
momentum v
Rest
v'
2v 3
m
m
m
m
A
B
A
B
∴ vmin = 2 m / s vmax = 2 2 m / s 3. A ( m) hitting B (2m) B
2v mv = mv ′ + m ⋅ 3 v v′ = 3 velocity of separation e= velocity of approach 2v 2v v − v′ − 3 3 =1 = = 3 v−0 v 3
i.e.,
1 + e 1 − e v and v ′2 = v 2 2
2. v ′1 = v
Rest
v2'
v1'
m
m
m
m
A B Before collision
A B After collision
KE after collision = 0.2 J (given) 1 1 2 2 ∴ mv2′ + mv ′1 = 0.2 2 2 or (Q m = 0.1 kg) v ′22 + v ′21 = 4 2 2 1 − e 2 1 + e 2 or v + v =4 2 2 or
v2 (2 + 2e2 ) = 4 4
Extreme cases If collision is perfectly in elastic i.e., e = 0 v2 (2) = 4 4 i.e., v = 2 2 m/s If collision is perfectly elastic i.e., e = 1 v2 (2 + 2) = 4 4 i.e., v = 2 m/s
2m m
v A
r
A collision is elastic 2 ( m) 2 vB = v= v 2m + m 3 m − 2m v and vA = v=− 2m + m 3 ∴ Velocity of B w.r.t. A = vB − v A 2 v = v − − = v 3 3 ∴ Next collision between the balls will take place after time 2π r t= v 4. As collision is elastic v2'
v'1
m
m
m
A B Before collision
A
B
p m
Rest
After collision
1 − e p v 2′ = 2 m Now, Momentum of A before impact − Impulse given by B on A = Momentum of A after impact. ∴ p − J = m v ′2 1 − e p or p− J = m 2 m 2J or e= −1 p
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision |
169
Introductory Exercise 8.8 V −0 v−0 V = v m = M
1. v0 sin α = v cos (α + β) v0
v0
co
sin
=
α
v0 s in ( α+
s α v0
β)
α β v co
4.
s (α
A
α + β)
u
α
Rest ∠PRX = 30° ∠QRX = 30°
P R
v0 cos α = sin (α + β) (Impact being elastic) ∴ tan α = cot(α + β) π or cot − α = cot(α + β) 2 π or −α =α +β 2 π or β = − 2α 2 2. Speed after n impacts
Speed before first impact = u ( = 2 gh) Speed before one impact = e u Speed after 2 impacts = e ( eu) = e2u ........ ......... ........ ........ ........ ......... Speed after n impacts = enu Height ( H ) upto which the ball rebounds after nth rebound ( enu)2 H= 2g 2n
=
e
v
A
Rest 30° X Angle OA
30° C B
Applying law momentum
conservation
of
mu = Resultant momentum of A and B balls along the axis of X. or
M
m
M
A B Before collision
A
B After collision
m v M Velocity of separation e= Velocity of approach
of
V
Rest
V =
As the balls are of same size, the centres of the balls P, Q and R will be at vertices of an equilateral triangle when ball C just strikes balls A and B symmetrically and as such the balls A and B will follow the path as shown below
v
Rest
⇒
B
u 2g
3. mv + 0 = 0 + MV m
Rest Q
2
= e2 n ⋅ h
v
X
C
or or
mu = 2 mv cos 30° 3 u = 2v 2 u=v 3
As the ball C will strike ball A (and as well as ball B) with velocity ucos 30°
170 | Mechanics-1 Velocity of approach of ball C towards ball A = ucos 30° − 0 = ucos 30° =u 3 6. Velocity of separation of ball A away from ball C = v u sin θ v ∴ e= 3 u θ A 2 v = 3 (v 3 ) 2 2 = 3
∴ after impact = 2 ^j ∴ Velocity after impact = − ^i + 2 ^j At A, u|| = u cos θ u⊥ = u sin θ v
e u sin θ H1 T1 B
u cos θ
u sin θ
R1
T2
H2
R2
∴ At B u|| = u cos θ u⊥ = e u sin θ 2u sin θ T1 1 g = = T2 2 e u sin θ e g
5. x-component of velocity
∴ 45° 45°
2u⊥ Using T = g 2 ^j
45° v = 2i^ + 2j^ X
2 ^i
before impact = 2 ^i ∴ after impact = − e 2 ^i 1 = − × 2 ^i 2 = − ^i y-component of velocity before impact = 2 ^j
2 (u sin θ) (u cos θ) 2 u⊥ u|| R1 2 = Using R = R2 2 ( eu sin θ) u cos θ g g 1 = e (u sin θ)2 H1 u2 2g = Using H = ⊥ 2 H2 ( e u sin θ) 2g 2g 1 = 2 e
AIEEE Corner Subjective Questions (Level 1) 1. D 4kg
1m
C 3kg
P (CM) 1m
yCM
1m
B 2kg A 1kg
1⋅0 + 2⋅1 + 3 ⋅1 + 4 ⋅0 1+2+3 + 4 5 m = 10 1⋅0 + 2⋅0 + 3 ⋅1 + 4 ⋅1 = 1+2+3 + 4 7 m = 10
xCM =
y
1m
x
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 2
5 7 AP2 = x2CM + y2CM = + 10 10 2. xCM
2
4 V ⋅ 0 + − πa 3 b 3 xCM = 4 3 V − πa 3 4 − π a 3b 3 = 4 4 πR 3 − πa 3 3 3 a3 =− 3 b 3 R − a
4.
= 0.74 m 2 A x + A2 x2 = 1 1 A1 + A2 A1 = a2 A2 = π
a
O
171
a2 4
a 2
By symmetry yCM = 0 a
a
5. In Fig. 1, C is CM.
πa2 ⋅ ( a) 4 = 2 πa a2 + 4 π = a 4+π a2 ⋅ 0 +
Fig. 2
C
m1
m2
(a4 , b4 ) X
m2
C'
…(i) ∴ m1a = m2 ( l − a) In Fig. 2, C′ is CM. m1( a + b − l1) = m2 ( l − a + l2 − b) …(ii) Substituting Eq. (i) in Eq. (ii), m1( b − l1) = m2 ( l2 − b) or m1b − m1l1 = m2 l2 − m2 b or ( m1 + m2 ) b = m1l1 + m2 l2 or
b=
a
A a A ⋅ 0 + − 4 4 xCM = A A + − 4 A 4 a =− ⋅ ⋅ 4 3A 4 a =− 12 A b A ⋅ 0 + − 4 4 yCM = A A + − 4 b =− 12 b a Centre of mass − ,− 12 12
l2
b m1
Y
O
a
l1
3. Let A = area of rectangle
b
l
Fig. 1
6. xCM =
m1l1 + m2 l2 m1 + m2
∫ x dm ∫ dm dx
O
x
where, dm = mass of element of length dx. ∫ x ρ dx A = ∫ ρ dx A x2 x ρ0 2 dx l = l x2 ∫ 0 ρ0 l2 ax l
∫0
=
∫0
l
x 3dx
l
x2 dx
∫0
(Qρ = ρ0
x2 ) l2
172 | Mechanics-1 l
7. xCM
3 x4 = 3⋅ 4 0 x 3 = l 4 1 × 10 + 2 × 12 = 1+2
New position of CM
10 m 10 m
m
6ms–1 1 kg 2 kg 4ms–1
+ ive
34 m 3 1 × ( − 6) + 2 × ( + 4) 2 = = + ms −1 1+2 3
= vCM
1 kg 2 ms
1 ms
→
70 ^i + 105 ^j m 3 0.10 kg
⇒ m = 0.3 kg (b) Momentum of system = Momentum of CM = ( m + 0. 1) kg × 6 ^j ms −1 = 2.4 ^j kg ms −1 →
→
v=u+at (1)(0) + (2)(10$i + 10 $j) = − 10 $j 3 20 10 $ = $i − j m/s 3 3 →
New position vector ( r1) of particle A → → → 1→ ( r1) = s 0 + u t + a t2 2 → 1 = (10 ^i + 20 ^j) + 0 + ( − 10 ^j) ⋅ 12 2 = 10 ^i + 15 ^j m New position vector ( r2 ) of particle B → 1 r2 = (20 ^i + 40 ^j) + (10 ^i + 10 ^j) ⋅ 1 + ( − 10 ^j) ⋅ 12 2 ^ ^ = 30 i + 45 j
→
v (0.3) + 0 (0.10) (c) v CM = m 0.3 + 0.1 → 4→ v m = v CM 4 4 = × 6 j^ ms −1 3 →
+ ive
^
→
=
m ⋅ 0 + 0 ⋅ 1 × 12 m + 0. 1
(a) 3 =
9. Acceleration ( a ) = − 10 j ms −2 →
10 ^i + 15 ^j + 60 ^i + 90 ^j 3
12 m
= 0 ms −2 ∴ Displacement of CM in 1 s = 0 m. →
=
6 ^j ms–1 m kg 3 m CM O
2 kg –1
→
r1 (1) + r2 (2) 1+2
10.
x′ CM (new position of CM) 34 2 = m + 2 s ms −1 3 3 38 m = 12.67 m = 3 1 × 2 ( + 2) + 2 × ( − 1) 8. vCM = 1+2 –1
→
→
R=
= 8 ^j m s −1 m
11.
A • t = 0s 2m
B • ( t = 100 ms) Position of 1st particle ( A) at t = 300 ms 1 s1 = × 10 × (300 × 10−3)2 2 = 0.45 m Position of 2nd particle ( B) at t = 300 ms (B is at the position of A at t = 100 ms) 1 s2 = × 100 × (200 × 10−3)2 2 = 0.20 m 2 m × 0.2 + m × 0.45 ∴ Position of CM = 2m + m = 28.3 cm
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | Velocity of 1st particle ( A) at t = 300 ms v1 = 10 × 300 × 10−3 = 3 ms −1 Velocity of 2nd particle at t = 300 ms v2 = 10 × 200 × 10−3 ms −1 = 2 ms −1 2m × 2 + m × 3 ∴ vCM = 2m + m 7 = = 2.33 ms −1 3 m A ⋅ 0 + mB ⋅ 80 12. 24 = m A + mB or 24 ( m A + 0.6) = 80 × 0.6 or m A = 1.4 kg ∴ Total mass of system = 1.4 kg + 0.6 kg = 2.0 kg
14. Let x = displacement of wedge (30 kg)
towards right. ∴ Displacement of block A towards right (along x-axis) when it arrives at the bottom of the wedge A
30°
5 kg
B (50 kg) R
Q
X
x
= QR − x = 0.5 − x Now, as net force on the system (wedge + block) along x-axis is zero, the position of CM of the system, along x-axis, will not change
vCM = 6.0 t2 ^j
1
aCM = 12 t ^j ms −2
∴
Net force acting on system (at t = 3 s) = Total mass of system × (aCOM at t = 3 s) = 2.0 kg × 36 ^j ms −2
A
5 (0.5 − x) = 30x 0.5 m x= 7 = 71.4 mm
∴
^
i.e.,
= 72 N j 13. L O
173
15. As no external force acts on the system,
x
dx
xCM =
=
∫ x dm ∫ dm L ∫ 0 x ( Ax dx) L
∫o L
=
∫0
L
∫0
A
x2 dx x dx
B FA
A
i.e.,
Ax dx
L3 / 3 L2 / 2 2 = L 3
=
the velocity of CM will be zero.
i.e.,
FB B
m Av A + mBvB =0 m A + mB vA m 2 =− B =− vB mA 1
(a) ∴ Ratio of speeds = 2 p m v m m (b) A = A A = A − B = − 1 pB mBvB mB m A (c)
p2 / 2 m A KA p2 m 2 = 2A = 2A × B = KB pB / 2 mB pB m A 1
174 | Mechanics-1 16. While man travels from P to Q
or
Negative sign shows that the balloon will move downwards.
CM P
m
vCM
m v ( M + m)
V =−
Q
3 v m − v + mv 3v 2 = = m+m 4
v
M
L . 2 ∴ Time taken by man to reach point Q starting from point P L /2 2L tPQ = = 3v/ 4 3v
+ ive
Displacement of CM (along horizontal) =
While man travels from Q to P 3v m − − v + mv 2 vCM = m+m 3v =− 4 1 Displacement of CM = − 2 ∴ Time taken by man to reach point P starting from point Q −L / 2 2 L t QP = = − 3v / 4 3 v ∴ Total time = tPQ + t QP 2L 2L + 3v 3v 4L = 3v =
4L ×v 3v 4L = 3
∴ Net displacement of trolley =
17. (a) While the man climbs in the rope, no
extra external net force acts on the system (balloon along with rope + man). Force applied by man to gain velocity to climb up is an internal force and as such the velocity of the CM of the system will remain stationary. mv + ( M + m) V = 0
v
m
(b) If the man slops climbing i.e., v = 0, then according to the above relation the value of v will also be zero. Thus, the balloon will also slop. 18. m
vi
M
⇒
v2
v3
…(i)
mv1 = Mv1 + mv3 m = 4 × 10−3 kg M = 1 kg v3 = 100 m/s v1 = 500 m/s Find v2 2 (a) 0 = v2 − 2as = v22 − 2(µg) s v2 ∴ µ= 2 2 gs
Here s = 0.30 m (b) Decrease in kinetic energy of bullet 1 = m(v12 − v23 ) 2 1 (c) KE of block = mv22 2 19. m
v1
m1
v2
µ = 0.5 ⇒
m2
v3 ⇒
( m + m1)v2 = mv1
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | mv1 ∴ v2 = m + m1 mv1 Common velocity v3 = m + m1 + m2
…(i) …(ii)
V2 =
or
2 m2 g l (1 − cos α) M ( M + m) α 2 m2 g l 2 sin2 2 2 V = M ( M + m)
or or
20. Let v = Velocity of pendulum bob when the
pendulum is vertical α
PE = mgh h
M
KE = 1 mv 2 v
2
V = 2 m sin
α 2
gl M ( M + m)
21. Let the track shift by x (to the right) when the cylinder reaches the bottom of the track. ∴ x′ + x = R − r or x ′ = ( R − r) − x (x′ = horizontal displacement of cylinder w.r.t. ground) As no force would be acting along horizontal direction, for no shift in CM along horizontal. We would have mx ′ − Mx = 0 or m [( R − r) − x ] − Mx = 0 or ( M + m) x = m ( R − r) m ( R − r) ⇒ x= M+m Now, as the PE of the cylinder would change into the kinetic energies of the
KE = 1 2
(R – r) m
V
V = Velocity of wagon when pendulum is vertical ∴ mv = MV M or v= V m KE of wagon + KE of bob = PE of bob 1 1 MV 2 + mv2 = mgh 2 2 2 1 1 M or MV 2 + m V = mgh 2 2 m 1 M or MV 2 1 + = mgh 2 m 1 or MV 2 ( M + m) = m2 gh 2
2 m gh M ( M + m)
V2 =
or
For m1 : v23 = v22 − 2a1s1 and Find s1 For m2 : v23 = 2a2 s2 and Find s2 µm1 g Here : a1 = = µg m1 µm1 g and a2 = m2
175
2
x
x'
x
M
x
O
cylinder and the track we have, 1 1 …(i) mg ( R − r) = mv2 + MV 2 2 2 where, v = velocity of cylinder and V = velocity of track. (when cylinder just reaches bottom of the track)
176 | Mechanics-1 Applying law of conservation of momentum mv + MV = 0 M or v=− V m Thus, Eq. (i) becomes 2 1 M 1 mg ( R − r) = m − V + MV 2 2 m 2 1 M or mg( R − r) = MV 2 + 1 2 m 2 2 m g ( R − r ) or V2 = M ( M + m) or
V =m
2 g ( R − r) M ( M + m)
22. No change in momentum of ball and also
that of wall along horizontal direction. √2 m/s
√2 m/s
2 m/s
45 °
+ ive
2 m/s
√2 m/s
50 g
45° 45°
√2 m/s
Wall
Along perpendicular direction : Momentum of ball after reflection
→
p f = − 50 × 10−3 × 2 kg ms −1 →
|p f |= 50 × 10−3 × 2 kg ms −1
∴
∴ Change in the magnitude of the → → momentum of the wall = |p f |− |p i | =0 23. m0 = 40 kg
m = ( 40 + 160) kg = 200 kg vi = 2 km/s = 2 × 103 kg −1 Rate of consumption of fuel = 4 kgs −1 ∴ Time (t) required to completely brunt out 160 kg of the fuel = = 40 s. 4 kgs −1 Ultimate vertical speed gained by the rocket m = − g t + vi ln m0 200 = − 10 × 40 + 2 × 103 × ln 40 = − 400 + 3218 = 2818 m/s = 2.82 kms −1 24. When y length of rope has fallen on table
top A y
→
( p f ) = − 50 × 10−3 2 kg ms −1 Momentum of ball before reflection
L
A v = √2gy
→
( p i ) = + 50 × 10−3 2 kg ms −1 →
→
→
∆p = p f − p i
∴
= − 100 × 10−3 2 kg ms −1 = − 0.14 kg ms −1 i.e.,
→
|∆p|= 0.14 kg ms −1
Momentum of wall (when ball strikes wall) →
p i = 50 × 10−3 × 2 kg ms −1
∴
→
|p i|= 50 × 10−3 × 2 kg ms −1
Momentum of wall (when ball rebounds from the wall)
dm dt dm dy =v ⋅ dy dt dm M = v2 = v2 dy L M W = g L Fnet = W + Fth Fth = vrel
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | M M ∴ Fnet = yg + v2 L L M M = yg + 2 gy L L M =3 yg L = Weight due to 3 y length of the rope. 25. Horizontal velocity of incoming sand y
Fth
v = 2 ms–1
O
→
i
x
→
v=0
Horizontal velocity of the conveyer belt →
^
= ( v) (say) = v i ∴
→
→
→
u rel = v1 − v2 →
^
= 0 − (v i) = − v ^i ∴
dm dt dm ^ = − v ⋅ i dt →
Fth = v rel
dm > 0, the falling sand particles exert dt thrust force which decelerates the conveyer belt. As
Force required to keep the belt moving →
→
F = − F th dm ^ =v i dt = (2 ms −1 × 5 kgs −1) ^i = 10N ^i Power delivered by motor to drive belt at 2 m/s →
→
= |F||v|= 10 N × 2 ms −1 = 20 W
26.
177
ma = Fth − mg or ma = Fth (neglecting mg as compare to Fth ) dm or ma = u − dt dm a or = − dt m u dm a or ∫ m = ∫ − u dt a or log e m = − t + K u Now, as t = 0, m = m0 (given) log e m0 = K a Thus, log e m = − t + log e m0 u m a or log e =− t m0 u or
m = m0 e
−
a t u
27. u = 100 ms −1, v = 0 ms −1, s = 6 cm
v2 = u2 + 2as u2 ⇒ a=− 2s (a) Now, v = u + at u u 2s ∴ t=− =− 2 = a u u 2s 2 × 6 × 10−2 = = 1.2 ms 100 (b) Impulse on log = − Change in momentum of bullet = − m (0 − u) = mu = 5 × 10−3 × 100 = 0.5 Ns (c) Average force experienced by the log Impulse 0.5 = = time 1.2 × 10−3 = 416.67 N 28. Let us consider right direction as positive.
Impulse of the spring on the block = Change in momentum of block = m (v − u)
178 | Mechanics-1 32. m1 = 1 kg and m2 = 1 kg
= 3 [( + 40) − ( − 50)] = + 270 Ns
v2 = + 4ms–1 v1 = – 6ms–1
(to the right) Impulse Average force on block = ∆t 270 = 0.02 = 270 Ns
m2 1 kg
A B Before collision
(to the right)
= 13.5 kN →
^
→
dp = 4 ^i dt → or dI = 4 ^i dt
∴
or
→
(where, I = Impulse) →
dI =
∫
2
∫0
4 ^i dt
→
I = 4 ^i (2 − 0)
or
30. m ∆v = F ∆t
or or ∴
m ∆v = Area under F-t graph 16 + 8 m ∆v = × 20000 2 m ∆v = 240000 240000 ∆v = = 200 m/s 1200
33. v1 ′ =
v − u = 200 ∴
v2
v = 200 ms −1 as u = 0
31. v1 = v2 ′ = − 2 ms −1
v2 = v1 ′ = + 3 ms v2
v1
m2
m1
A
+ ive
B After collision
1 − 2 2 × 2 = ( +4) + ( − 6) 2 + 1 2 + 1 4 = − −8 3 28 28 =− ms −1 = ms −1 3 3 (in − ive x direction).
= 8 ^i kg-ms −1 or
v1'
2 − 1 2 × 1 = ( − 6) + ( 4) 2 + 1 2 + 1 8 = −2+ 3 2 = + ms −1 3 2 = ms −1 (in + ive x direction) 3 m − m1 2m1 v2 ′ = 2 v2 + v1 m + m 1 m1 + m2 2
^
p = m v = 2 (2t i − 4 j)
29.
v2'
m − m2 2m2 v1 ′ = 1 v1 + v2 m1 + m2 m1 + m2
= 13500 N →
m1 2 kg
(given)
m
1
−1
1+ e 1− e v2 and v2 ′ = v2 2 2 v2'
v1 = 0 m 2
m
1
v1' = 2v2' m
2
Rest
v2' = – 2ms–1 v1' = + 3ms–1
m
m
m
m
A
B
A
B
+ ive
(When two bodies of equal masses collide elastically they interchange their velocities)
⇒ or or or
v1 ′ 1 + e = v2 ′ 1 − e 1+ e 2= 1− e 2 − 2e = 1 + e 1 e= 3
(Qv1 ′ = 2 v2 ′)
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 34. mv = mv1 + mv2 v m
m
Rest Before collision
⇒ Also,
or or or or
or
or or
or
v1
v2
m
m
v22 + v12 − 2 v1v2 v2 (v + v2 )2 − 4 v1v2 e2 = 1 v2 v2 v2 − 2 e2 = v2
e2 =
1 2 1 or e= 2 2m2 35. v1 ′ = v2 m1 + m2 or
After collision
v = v1 + v2 1 1 mv12 + mv22 3 2 2 = 1 2 4 mv 2 3 (v12 + v22 ) = v2 4 3 2 (v1 + v2 ) − 2v1v2 = v2 4 3 2 2 v − 2 v1v2 = v 4 1 2 2 v1v2 = v 4 Velocity of separation e= Velocity of approach v2 − v1 e= v−0 v2 − v1 e= v 2 v − v1 e2 = 2 v
179
v2 = 2 ms–1
e2 =
v1 = 0 ms–1
v2'
v1'
B
A
B
A m2 = 3 kg
m1 = 2 kg
…(i)
=
2×3 ×2 2+3
12 = 2.4 ms −1 5 m − m1 v2 ′ = 2 v2 m1 + m2 =
and
=
3−2 ×2 2+3
=
2 = 0.4 ms −1 5
Distance between blocks when they stop sliding v ′21 v ′22 = − 2µ k g 2µ k g =
1 (v ′21 − v ′22 ) 2µ k g
=
1 [(2.4)2 − (0.4)2 ] 2 × 0.3 × 10
=
2.8 = 0.933 m 3
180 | Mechanics-1 Objective Questions (Level 1) Single Correct Option 1. Momentum remains conserved. Decrease
in momentum of the ball is transferred to sand while KE does not remain conserved as it gets used up in doing work against friction. net 2. Fext = M × aCM net ∴ If Fext = 0,
aCM = 0 d i.e., v =0 dt CM or vCM = constant Option (a) is correct. 3. The forces acting on the blocks would be equal and opposite as per Newton's 3rd law of motion. Acceleration of the blocks will depend upon their masses as per Newton’s 2nd law of motion. Accelerations being different velocities will be unequal. M1
F
F
M2
Option (c) is correct. 4. While colliding the balls will apply equal
and opposite impulsive force on each other. Impulsive forces will change the momentum of the balls but the total momentum of the system of 2 balls will remain conserved impulsive forces being internal ones. Change in momentum of the system will definitely be due to external gravitational forces on the balls but as the time of impact shall be very less the impulsive force will over shadow the weak gravitational force. 5. External force acting on the cannon shell
before explosion is the gravitational force. Now, as no extra net external force would be act on the shell during collision the momentum of the system shall remain conserved and the CM of the system (now broken in pieces) will also keep on following the path which the shell would have followed had the explosion not taken place. Further, as the explosion would never be super-elastic, the KE of the system can’t increase after explosion. Option (d) is correct.
6.
Velocity of separation =e Velocity of approach
As in an elastic collision e < 1 Velocity of separation < velocity of approach (when e = 0, the velocity of separation in zero and the colliding bodies do not separate from each other.) Further, whether the collision in elastic or inelastic the law of conservation of momentum always hold gord. ∴ Option (d) is correct. → → 7. p = M v CM , Option (a) is correct. →
→
→
→
p = p1 + p2 + p3 + ....,
Option (b) is correct. Further, we define momentum for every type of motion. ∴ Option (d) is correct. 8. Let us consider a system of two masses as
shown in figure. → v1
→ vCM
m1
CM
→ v2 m2
Momentum of system about CM →
→
→
→
= m1 ( v1 − v CM ) + m2 ( v2 − v CM ) →
→
→
= m1 v1 + m2 v2 − ( m1 + m2 ) v CM →
→
= ( m1 + m2 ) v CM − ( m1 + m2 ) v CM →
=0
Option (c) is correct. 9. Option (a) If collision is inelastic. A
m
v
v m B
v C
m
Option (b) If collision is perfectly inelastic Option (c) If the dimensions of the particles → 0 ∴ Option (d) would be the answer.
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 10.
⇒
14. Horizontal velocity of the leaving coal :
F ∆t = m ∆v m ∆v F= ∆t 5 × (65 − 15) = 2 = 125 N
system
O
v
→
v1 = + v ^i
d C
Horizontal velocity of the system
mo O
CM d1
→ v2
mo d mo + mc 8 = d 8+6 4 = d 7 4 = × (1.2 × 10−10 m) 7 = 0.64 × 10−10 m
d1 =
⇒
Rest
v
V
M
m
M–m
9 kg
3 kg
6 kg
→
→
second n × [change in momentum per second] ≤ F 40 n × (1200 − 0) ≤ 144 100 144 or or n ≤ 3 n≤ 48 ∴ Option (a) is correct. i.e.,
m v ( m − m) 3 =− × 16 9−3
16. Change in momentum along x-axis v
v sin θ – P +
= − 8 ms −1 1 ∴ KE of 6 kg mass = × 6 × ( − 8)2 2 = 192 J 2 m2 13. v1 ′ = v2 m1 + m2 v1 = 0
v2'
v1'
m2
m1
m2
m1
After collision
(as m1 < < m2 )
Q
θ v cos θ y
v2
= 2 v2 Option (b) is correct.
→
15. If n be the number of bullet shots per
V =−
Before collision
→
= + v ^i
U rel = v1 − v2 = 01 → → dm → ∴ Fth = U rel =0 dt As, the leaving coal does not exert any thrust force on the wagon, the speed of the wagon won’t change. Option (a) is correct. ∴
12. M × 0 = mv + ( M − m) V
⇒
x
^i
11. mc × d1 = mo × ( d − d1) mc
181
v cos θ x θ
v cos θ
v
= m (v cos θ − v cos θ) = 0 ∴ Net change in momentum = Change in momentum along y-axis = m[( + v sin θ) − ( − v sin θ)] = 2 mv sin θ (as θ = 45°) = mv 2 Option (a) is correct. 17. Velocity of ball before first impact i.e.,
when it reaches point Q of the horizontal plane
182 | Mechanics-1 50 kg
P u=0 450 kg O 50 kg
+ eg Q
1st
450 kg
2nd
v = 0 + ( g) ⋅ 1 = g ∴ Velocity of ball after 1st impact = ev = eg time elapsed between 1st and 2nd impact with the horizontal plane ( + e g) = ( + e g) + ( − g) t ⇒ t = 2e 4 2 (as, e = L) = s 3 3 Option (c) is correct.
∫ x dm = ∫ x m dx ∫ dm ∫ m dx
∴
CM (boat)
50 × 10 + 450 × 5 50 × ( x) + 450 × ( x + 5) = 50 + 450 50 + 450 (Initially)
⇒ x = 1m Option (b) is correct. 20. As discussed in the answer to previous
question no. 19. M M/3
O
Smooth
L
O
x
x
Smooth
(where m = mass per unit length) Ax2 ∫ x L dx Ax2 = Q m = 2 L Ax ∫ L dx L
x 3dx
L
x2 dx
∫0
M 3
(Initially)
(Finally)
M x M + = ML 3 3 x= L 4
⇒
i.e., the distance that plank moves =
L4 3 × 3 4 L 3 = L 4 Option (a) is correct.
∴ The distance that the man moves 3 =L− L 4 L = 4 Option (b) is correct.
=
19. As there is no net external force acting on
the system in the horizontal direction, the CM of the system shall not shift along x-axis.
X
L M L x + ML + 2 3 2 = M M M+ M+ 3 3
Mx +
L
∫0
X
M/3 L
x
=
(Finally)
M
dx 0
X
x
– eg
18. xCM =
X
CM (boat) 10 m
21. v
m
Rest
Rest
3m
m
Before collision
u
3m
After collision
mv + 3 m ⋅ 0 = m ⋅ 0 + 3 mu
3 L 4
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | v ∴ u= 3 Velocity of separation e= Velocity of approach u−0 u = = v−0 v 1 = 3 Option (d) is correct.
Impulse on ball = Change in momentum of ball 9 4 = − mv0 − + mv0 20 5 5 = − mv0 4 25. If a ball dropped from height h rebounds
u
to a height h′, then speed of ball just before 1st impact, u = 2 gh Just after 1st impact u′ = 2 gh′
m
∴
22. Change in momentum of A or B = mu u m
Rest
Rest
m
m
Before collision
Option (b) is correct. F'
F'
m B
+
Option (c) is correct.
h′ = e2 h
Height attained after nth impact = e2 n h = (0.8)2 n ⋅ 1
(as h = 1 m)
2n
= (0.8)
→
^
or
→
^
475 v = 500 ( ^i − k)
20 ^ ^ ( i − k) 19 Option (c) is correct. or
→
v=
27. While force is increasing with time
F = kt (where k is + ive constant) or X
5
53° Y
i.e.,
500 × 1 ^i = (500 − 25) v + 25 (20 k)
v0 cos 37° = v0 4 5 v0
3 v cos 53° 4 0 = 9 v0 20 3v 4 0
= 0.8
before throwing block = Momentum of car after throwing block + Momentum of block
F
F′ m ⇒ F ′ = ma0 Acceleration of block B : F − F′ aB = m F − ma0 F = = − a0 m m
37°
h′ 64 = h 100
26. Momentum of car (+ block)
a0 =
24.
u′ = u
Option (d) is correct.
23. Acceleration of block A A
e=
After collision
(As collision is elastic) Impulse = Change in momentum Ft0 = mu 2 2 mu ⇒ F= t0
m
183
37° 4
53°
3
or ∴ or
ma = kt dv k = t dt m k t2 v= +C m 2 k t2 v= ⋅ m 2 (If at
t = 0, v = 0)
184 | Mechanics-1 Thus, graph between v and t would be
30. As no external force is acting along the
horizontal direction on the system (wedge + block). The CM of the system shall not change along horizontal when the block moves over the wedge but would change along vertical as net force (= gravitational force) is acting on the block. Further, as no non-conservative force is acting on the system, its total energy will not change. Option (d) is correct.
v0 Graph would be parabolic in nature
v
t0
t
While force is decreasing with time. F = − kt (where k is + ive constant) k t2 ⇒ v=− + C′ m 2 At t = 0, v = v0 k t20 ∴ C ′ = v0 + m 2 k 2 Thus, v = v0 + ( t0 − t2 ) 2m Thus, graph between v and t would be
Parabola
v v0
31.
v' m
m
M v
Rest Before collision
Just after collision
or 8 π R2 x1 = 2πR2 R or 8 x1 = 2 R R or x1 = 4 Option (c) is correct. A x + A2 x2 + A3x3 29. xCM = 1 1 A1 + A2 + A3 0 = [ π ( 4 R)2 − π R2 − π R2 ] x1 + [ − π R2 ] 3 R + [ − π R2 ] ⋅ 0 or 14 x1 = 3 R 3 or x1 = R 14 Option (d) is correct.
Final position
mv = ( M + m) v ′ From final position, v ′ = 2 gh From Eq. (i), m or v = 2 gh M+m
…(i)
M v = 1 + 2 gh m
t
∴ Option (a) is correct. A x + A2 x2 28. xCM = 1 1 A1 + A2 [ π (3 R)2 − π R2 ] x1 + [ − πR2 ] 2R 0= π (3 R)2 − π R2
Rest h
or t0
m
M
Option (c) is correct. 32. As no net extra external force is acting on
the system the CM of the gun and the bullet system remains at rest. The force exerted by the trigger of the gun on the bullet is an interval one. 33. m1 g − T = m1a
T
∴
T
T T
a m2g
m1g
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | T − m2 g = m2 a ∴ ( m1 − m2 ) g = ( m1 + m2 ) a m − m2 or a= 1 g m1 + m2 m a + m2 ( − a) aCM = 1 m1 + m2 m1 − m2 = a m1 + m2 2
m − m2 = 1 g m1 + m2 Option (b) is correct. m 34. v = u − gt + vi ln 0 m m0 m (neglecting gravity as given) m or v = v2 ln 0 (Taking u = 0) m m0 or (as vi = v) v = v ln m m ∴ log e 0 = 1 m m0 1 ⇒ = e = 2.718 m Option (a) is correct. or
v = u + v2 ln
36. Velocity of separation would be zero as the
collision is completely inelastic. A
v cos θ 2 v
θ v sin 2 θ v sin 2
θ/2
2m
V
θ/2 v v cos θ 2
B
Velocity of approach = Velocity of A w.r.t. B or velocity of B w.r.t. A θ θ = + v sin − − v sin 2 2 θ = 2 v sin 2 Common velocity ( V ) after collision θ θ mv cos + mv cos = 2 mV 2 2 θ ∴ V = v cos 2 Option (d) is correct. 37. v sin 53 ° = u sin 37 °
35. Velocity of water after striking the plate
would be almost zero as then it flows parallel to the plate.
185
u sin 37° u cos 37°
u
v cos 53°
v v sin 53°
53°
37°
Floor
Jet of water Water flow
Force exerted on plate = − [Rate of change of momentum of water] m ⋅ 0 − mu =− t mu m = = ⋅u t t kg = 0.5 × 1ms −1 s = 0.5 N Option (c) is correct.
4 3 =u× 5 5 3 i.e., v= u 4 Impulse exerted on floor = − [Change in momentum of ball] = − [( − mv cos 53 ° ) − ( + mu cos 37 ° )] = m [v cos 53 ° + u cos 37 ° ] v×
3 3 4 = m u× +u× 5 5 4 5 = mu 4 5 = × 1 × 5 = 6.25 Ns 4
186 | Mechanics-1 x = Displacement of boat x ⋅ 40 + (1 + x) 15 = 2 × 40 + 1 × 15 i.e., x = 1.46 m (The frictional forces on the boat by the boy and that by the boy on the boat are internal forces).
38. As the CM will not change along x-axis,
for no net force acting on system (wedge + block) along x-axis. m
h
M x'
a = cot θ h
θ
X
a
x
41.
Mg M
h
m X Mg a
mx + M ( x + x ′ ) = Ma + Mx ′ m mx + Mx = ma or x = a M+m m h cotθ = M+m
∴
2 cm
A
D M
M
B
2 cm
2 cm 2 cm
M
C
M A
→
M falls very close to 5 mortar, its velocity after explosion must be − u. As piece of mass
Shell A u (say) M or ta r
M
Y M
→
v=
42. The shell explodes at A, the highest point.
Option (b) is correct. 2(4M ) + 3 (− M ) + 3 (+ M ) 39. xCM = = 2 cm 4M + (− M ) + (+ M ) Y
→
m v1 + m v2 + m v 3 m+m+m → → 1 → = [ v1 + v2 + v 3 ] 3 1 ^ = [v0 ^i + ( − 3v0 ^j ) + (5v0 k)] 3 v ^ = 0 [ ^i − 3 ^j + 5 k ] 3 Option (d) is correct. →
R/2 After explosion path of M/5 part
2M D+C
R/2
Shell 4M 5
M
(Velocity of 4M/5 man part of shell after explosion) 3/2 u After explosion path of 4M/5 part
B R/2
O
2 cm
2 cm
O
X
2 cm
2 cm
D
X
3R/4
yCM
2(4M ) + 1(− M ) + 3 (+ M ) = = 2.5 cms 4M + (− M ) + (+ M )
Option (c) is correct. 40. As explained in the answer to question no.
20 and 19.
Thus, Mu = i.e.,
M 4M ( − u) + (v) 5 5 3 v= u 2
As the velocity has increased 40 kg 15 kg 2m 15 kg
x
2m
range of this part will be 3 R 3R of i.e., 2 2 4 R 3 R 5R ∴ D= + = 2 4 4 Option (c) is correct.
3 times the 2
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | m − m1 43. v2 ′ = 2 v2 m2 + m1 v2
v1 = 0
A m
1 1 1 mv22 − m − v2 2 3 2 = 1 2 mv2 2 1 1 − = 2 18 1 2 8 = 9
v2' m
B 2m
m2 = m
187
2
2m
m1 = 2m
=
m − 2m v2 m + 2m
=−
1 v2 3
Fraction of KE lost by colliding particle A KE lost = Initial KE
JEE Corner Assertion and Reason 1. To answer this question, let us find the
centre of mass of an L-shaped rigid body (as shown in figure) of uniform thickness.
The centre of mass of a uniform plate lies at its centre. The CM of a uniform sphere is at its centre.
Y
Y L
4L A2
xCM
L P
A1
2L O
L
2L
4L
A x + A2 x2 = 1 1 A1 + A2 =
(6L × 2L) 3 L + ( 4 L × 2L) L (6L × 2L) + ( 4 L × 2L)
=
36L3 + 8 L3 12L2 + 8 L2
=
44 L3 20L2
These examples show that assertion is false. It is correct that centre of mass and centre of gravity of a body coincide if the body is placed in a uniform gravitational field. Option (d) is correct. F − F1 F 2. a1 = 1 , a2 = 2m m
Thus, we see that CM is lying outside. Centre of mass of the rigid body of uniform thickness as shown in figure would be at point P which is neither outside nor inside.
a1
a2
m
2m F1
= 2.2 L Similarly yCM = 2.2 L
X
L
X
∴
F
ma1 + 2ma2 m + 2m F1 + ( F − F1) = 3m F = 3m
aCM =
F
188 | Mechanics-1 As F is constant, the value of aCM will remain constant (Reason). As CM is accelerated the velocity of CM will obviously increase (Assertion). Option (a) is correct. 3. As per assertion if force is applied on a
system the linear momentum of the system must not remain conserved. But, it will not be true if we apply two equal and opposite external forces on the system as then net external force on the system will be zero and the linear momentum will remain conserved as given under Reason which is correct. Option (d) is correct. 4. A rocket moves forward due to the thrust
force produced on it as per Newton’s 3rd law of motion (as given under Reason) when gas inside (not the surrounding air as given under assertion) it is pushed backwards. Option (d) is correct. 5. Linear momentum of a system remains
conserved when no net external force acts on the system i.e., only a net external force on a system can change its linear momentum. Inside a system internal forces are always in pairs and as such can't change linear momentum. If two blocks connected by a spring placed on a smooth surface are stretched apart the internal restoring forces acting on blocks will definitely increase the KE of the system but this argument under Reason has nothing to do with the nothing under Assertion that internal forces can't change linear momentum. Option (b) is correct. p2 6. KE = 2m 1 i.e., KE ∝ (if momentum p is constant). m ∴ Nothing under reason is correct. When bullet is fired and comes out of the gun. Linear momentum of gun = Linear momentum of bullet = p
(In magnitude) (KE) gun Mass of bullet ∴ = (KE) bullet Mass of gun (as given under Reason) This is what is given in Assertion. Thus, both Assertion and Reason are true and also reason is the correct explanation of the assertion. Option (a) is correct. 7. As no net external force is acting there on
the system along horizontal direction the momentum of the system remains conserved along horizontal direction but as gravitational force (a net external force) acts on block in the vertical direction (downwards) the momentum of the system does not remain constant along vertical direction. As overall momentum of the system does not remain constant, Assertion is true. As wedge is at rest, Reason is false. Option (c) is correct. 8. In any collision, there is not change in the
momentum of the system as given in reason, which is true. ∴ i.e.,
→
→
→
∆p1 + ∆p2 = 0 →
→
∆p1 = − ∆p2
…(i)
∴ Assertion is false. Option (d) is correct. 9. Reason is true as explained in the answer
to question no. 6 and also in Eq. (i) in the answer to question no. 8. As KE is inversely proportional to mass, K the KE of the block of man 2m will be 2 when the KE of the block of mass m is K . ∴ Assertion is true. Further, as Reason is the correct explanation of the Assertion. Option would be (a). 10. Assertion is false as for example heat energy can be given to a system without any increase in momentum of the system while KE given to a system increases its momentum.
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 2
p KE = as given is Reason which is 2m true. Option (d) is correct. 11. As no external force would be acting on
the system of electron and proton along the line joining electron and proton the CM of electron and proton will remain at rest. Therefore, Assertion is false. Further, as proton is heavier than electron the reason is true. Option (d) is correct. m − m2 2m2 12. v1 ′ = 1 v1 + v2 m1 + m2 m1 + m2 v2 m2
v1
m1 A B Before collision
and v2 ′ =
v2'
v1'
m2
m1 A B After collision
m2 − m1 2m1 v2 + v1 m1 + m2 m1 + m2
v2 ′ − v1 ′ m − m1 − 2m2 2m1 − m1 + m2 = 2 v2 + v1 m1 + m2 m1 + m2 = − v2 + v1 = − (v2 − v1) i.e., v ′21 = − v21 i.e., relative velocity of A w.r.t. B after collision = − (relative velocity of A w.r.t. B before collision) ∴ Reason is true and Assertion is false. Option (d) is correct. 13. As explained in the answer to question no.
11, the CM of the objects will remain at rest. Therefore, assertion is false.
x1
189
x2
m1
CM
m2
m1x1 = m2 x2 x2 m1 = x1 m2
∴
∴ x2 > x1 as m2 > m1 Reason is true. Option (d) is correct. →
14. F =
→
dp dt
(Newton's second law of motion)
→
→
a=
F m
(outcome of the above)
∴ Reason is true. →
First equation tells that if same force F is applied on different masses the rate of →
p change of momentum i.e., d of each mass dt will be same. Second equation tells that if →
same force F is applied on different → masses the a produced in each will be different. ∴ Assertion is true. Further, as Reason is the correct explanation of the Assertion. Option (a) is correct. 15. Assertion is false as explained in the
answer to question no. 12. In every type of collision the linear momentum of the system remains conserved. Therefore, Reason is true. Option (d) is correct.
190 | Mechanics-1 Objective Questions (Level 2) Single Correct Option 1. m1v1 + m × 0 = m 5 g l + m1
v1 3
α = cos −1
or
3 4
Option (c) is correct. →
→
3. Here |v2 |= |v1|= v say → dp
→ v2
l
→ v1
dm
m1
m Rest
v1
u = √5gl
v1/3
2 v1 = m 5g l 3 3 m v1 = 5g l 2 m1
Net momentum of the two elements as → shown in figure = dp = |dp|
l α
2m H m v
m
⇒
⇒
m
u 2 2 0 = v2 + 2 ( − g) H v2 H= 2g v=
=
or or or
= 2v dm sin θ M = 2v ( R dθ) sin θ π R 2M v = sin θ dθ π π /2 2Mv p= ∫ sin θ dθ 0 π 2M v =− [cos θ ] π0 / 2 π 2M v 2Mv =− [0 − 1] = π π Option (b) is correct. 1 1 4. (2 m) v2 = kx20 2 2
(u / 2)2 2g
u2 8g 2g l H= 8g =
l l (1 − cos α) = 4 1 1 − cos α = 4
→ v1 dl
2. mu = 2mv
u = √2gl Rest
θ θ dθ
Option (b) is correct.
m
2θ
R
m1 ⋅ ∴
dm
dθ
m
kx0
kx0
or or or or
v=
k x0 2m
2mv = 2 mk x0 Fav × ∆t = 2 mk x0 Fav =
Option (b) is correct.
2 mk ∆t
x0
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision |
4 p = 1(1 − p) 1 ⇒ p= m 5 Displacement ( x) of bar when pendulum becomes vertical x = sin θ p
5. As the collisions of the striker and the
walls of the carrom are perfectly elastic the striker will follow the path OPQROP ... Q
R
45°
P
x = p sin θ 1 1 m = 0.1 m = sin 30°= 5 10 When the ball reaches the other extreme end the bar will further shift to the left by distance x and as such the net displacement of the bar will be 2x i.e., 0.2 m. Option (b) is correct.
45° A
O
Change in KE = work done against friction 1 mv2 − 0 = µ mgs 2 (m = mass of striker, s = displacement of the striker) v2 ⇒ s= 2µ g (2)2 = 1m = 2 × 0.2 × 10 1 1 PQ = OP = OA 2 = 2= m 2 2 2 1 1 OP + PQ = + = 1 m 2 2 ∴ Striker will stop at point Q where 1 1 co-ordinates are , . 2 2 2 Option (a) is correct. 6. As no force is acting on the system along
horizontal, the CM of the system will not shift horizontally. 4 kg p
θ
x
7. p
ep
Floor During first collision
ep
e2p
During Second collision
e 2p
e 3p
During third collision
Floor
Momentum imparted to the floor in 1st collision = p − ( − ep) = p (1 + e) 2nd collision = ep − ( − e2 p) = ep (1 + e) 3rd collision = e2 p − ( − e3 p) = e2 p (1 + e) As theoretically there will be infinite collision, total momentum imparted to floor = p (1 + e) + ep (1 + e) + e2 p (1 + e) +.... ∞ = p (1 + e) [1 + e + e2 + K ∞ ] 1 + e 1 = p = p (1 + e) 1− e 1 − e Option (d) is correct. 8. Let F be the frictional force applied by
1–
p
CM
191
1 kg x
x 4 kg
CM
plate when bullet enters into it 1 …(i) ∴ mu2 = Fh 2 If plate was free to move mu + 0 = ( M + m) u′ m ∴ u′ = u M+m 1 1 New KE of bullet = mu2 − ( M + m) u′2 2 2 (entering plate)
192 | Mechanics-1 2
m 1 1 mu2 − ( M + m) u 2 2 M + m 1 m 1 M 2 = mu2 1 − = mu 2 M + m 2 M + m …(ii) = Fh′ Dividing Eq. (ii) by Eq. (i), h′ M = h M+m M i.e., h′ = h M + m
3 m1 = m2 m1 1 = m2 3
∴
=
or
12. As the resultant of the velocities of 1st
and 2nd are just opposite to that of 3rd, the 4th particle will travel in the line in which 3rd is travelling. v 2nd
Option (a) is correct. 9. v2 = 16 ms–1
45°
2
v1 = 2gs = 10 ms–1
v2' Rest
m
m
A
B
Now, v1 ′ =
u
A
1+ e v2 2
v
B
Let the velocity of 4th particle is u as shown in figure. ∴ u cos 45° + v cos 45° = v i.e., u = v ( 2 − 1) Total energy released 1 1 1 1 = mv2 + mv2 + mv2 + m [v( 2 − 1)] 2 2 2 2 2 1 2 2 = mv [3 + ( 2 − 1) ] = mv2 [3 − 2 ] 2 Option (a) is correct.
(entering plase)
10 =
13.
b l2 bL l L2 b yCM = l + + 2 2 6 according l= bL bl + to question 2 mg L l2 L l L2 or l l + = + + 2 2 2 6 l2 L2 mg or = 2 6 A L or l= 3 11.
m1 v m1 1 A B ⇒ v2
∴ ∴
v
3rd
1+ e ⋅ 16 2 1 i.e., e= 4 Option (b) is correct. l 1 L ( l ⋅ b) + L ⋅ b l + 2 2 3 10. yCM = 1 ( l ⋅ b) + L ⋅ b 2 ∴
1st
4th
mg
A
B
v2
m1v1 = m2v2 − m1v2 e=1 2 v2 = v1 m1(2 v2 ) = m2v2 − m1v2
…(i)
mg
2mg
2mg
2mg
2mg C
mg B
A
B
u
C A
B
mg C
u
mg
mu = 2mv1 C will increase u the tension in i.e., v1 = 2 the string. mu = 2mv 2 u i.e., v 2 = 2
2 mg
mg
2 mg
C will also increase the weight of B when collision takes place. Thus, mu = (2m + m ) v 3 u i.e., v 3 = 3
u
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | ∴ v1 : v2 : v3
193
m A
u u u = : : = 3 :3 :2 2 2 3 Option (b) is correct.
14.
Place of collision
time = t (say)
g
h
CM
3m w
h/2
h/2
v
v
2m
30°
B
v cos 30°
Motion of B : h h 1 h =v − g 2 g 2 g
vCM = v cos 30° 3 =v 2 Option (a) is correct. →
^
15. r CM = i
i.e.,
h=v
∴
v=
→
vB =
r1 (position vector of lighter piece) ^
= 3 ^i + 2 ^j − 4 k →
r CM = →
→ m1 r1
r2 =
+ m1 + m2
( m1 + m2 ) r CM − m2
→ m1 r1
^
= − ^j + 2 k ∴ The heavier part will be at (0, − 1, 2). Option (d) is correct. 16. Motion of A :
i.e.,
t=
h g
h =0 g
gh − g ⋅
gh
Rest
A
B
m
2m
ω A
B
m + 2m
m gh = 3 m ω gh ∴ ω= 3 Velocity of the combined mass when it reach ground v ′2 = ω2 + 2 gh gh = + 2 gh g 19 gh i.e., v′ = 3 Option (d) is correct.
2 ^ 2 i − (3 ^i + 2 ^j − 4 k) 3 = 4 3 1 ^ ^ = [6 i − 2 (3 ^i + 2 ^j − 4 k)] 4 1 ^ = [ − 4 ^j + 8 k ] 4 ^
h 1 2 = gt , v A (at time t) = g t = 2 2
gh
Collision of A and B at time t :
→ m2 r2
→
h g
gh
17. u = velocity of man w.r.t. cart
Let v = velocity of cart w.r.t. ground ∴ Velocity of man w.r.t. ground = u + v m (u + v) + 2 mv = 3 m ⋅ 0 u ∴ v=− 3 Work done = KE gained by man and cart 1 1 = m (u + v)2 + 2 mv2 2 2
194 | Mechanics-1 2
1 u 1 u m u − + 2 m − 3 2 3 2 1 4u2 1 u2 = m⋅ + 2m 2 3 2 9 2mu2 mu2 = + 3 9 7 = mu2 9 Option (d) is correct. 30 m + 50 m 18. vCM = m+m
2
1 mv2 2
= 2 mv2 −
=
3 mv2 2 3 p2 = 2m Option (c) is correct. =
20. According to question 4m
P
4m CM
= 40 m/s upwards. 30 m/s
4m CM
x
x a
a
m
a 2 = ( 4 m) x + ( m) a 4m + 4m 4m + m
( 4 m) x + ( 4 m) Initial position of CM
40 m 20 m
x a 4x + a + = 2 4 5 a i.e., x= 6 Option (b) is correct. or
50 m/s
m
If the velocity of CM becomes zero at displacements 02 = 402 + 2 ( − 10) s ⇒ s = 80 m ∴ Maximum height attained by CM = 20 m + 80 m = 100 m Option (c) is correct.
21. xCM of Fig. 1 will as that of Fig. 2. y a
40
a
20
10
a
a a Fig. 1
O
19. As the masses are equal and the collision
is elastic, the particles will exchange their velocities as shown in figure. p
50 O
A B Before collision 2v
2v
v
A B After collision
Gain in KE of 1st particle 1 1 = m ( − 2 v)2 − mu2 2 2
a x
a
a Fig. 2
a 3a + 50 ⋅ 2 2 = 40 + 50 20a + 75a = 90 95a 19 a = = 90 18 40 ⋅
xCM
In terms of velocity A B Before collision
40
– 2p
In terms of momentum
v
x
y a
20
m R
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | yCM
3a a 40 ⋅ + 50 ⋅ 2 2 = 40 + 50
There is no need to find the value of yCM
17 a 18 Option (a) is correct.
195
24. Velocity of 2nd ball when 1st with velocity
v strikes 2nd at rest
=
22. Conservation of momentum along y-axis
v m0 0 sin 45° = 2 m0 v sin θ 2 v i.e., …(i) 2 v sin θ = 0 2 Conservation of momentum along x-axis. v0/√2 Rest
m/22
m/2n – 1
1
2
3
n
1st
2nd
3rd
nth
2m0
x
θ 2m0 v
v m0 0 cos 45° + 2 m0 v cos θ = m0v0 2 v i.e., …(ii) 2 v cos θ = 0 2 Squaring and adding Eqs. (i) and (ii), v 2v = 0 2 2 v ∴ v= 0 2 2 Option (b) is correct. 1 1 23. kx20 = m2v22 2 2 k i.e., v2 = x0 m2 m × 0 + m2v2 vCM = 1 m1 + m2
r
v2 = v
Rest
v2'
v1'
1
2
1
2
m2 = m
m1 = m/2
45° x
m0
m/2
v1 ′ =
m0 v0
m
2m2 2m 4 v2 = v= v m m1 + m2 3 +m 2
Velocity of 3rd ball when 2nd with velocity 4 v strikes 3rd at rest 3 v2 = 4 v 3 2 m2 = m 2
Rest
v2'
3 m1 = m 4
m 2⋅ 2m2 2 4v v1 ′ = v2 = m m 3 m1 + m2 + 4 2 4 4 = ⋅ v 3 3 2 4 = v 3 3−1 4 Velocity of 3rd ball = v 3 m2 4 As in every collision = m1 + m2 3 n−1 4 The velocity of nth ball = v 3
(When wall just breaks off the velocity of mass m1 would be zero) m2 = v2 m1 + m2 m2 k = x0 m1 + m2 m2 x0 = k m2 m1 + m2
Now, this must be equal to 5 gr for it to complete the circle n−1 4 ∴ v = 5 gr 3 n−1 3 i.e., v = 5 gr 4
Option (b) is correct.
Option (a) is correct.
v1' 3
196 | Mechanics-1 25. Impulse given to the block will also
release it from abstraction besides giving and then imparting the restoring force on it due to velocity to it 5cm of the spring expansion will accelerate it. Impulse = 4 kg ms −1 4 kg ms –1 Initial velocity (u) = = 2 ms −1 2 kg
5cm 6.7 cm F
F 6.7 cm 3 m/s 6.7 cm
Average Acceleration ( a) kx = = 2m
5 4000 × 100 2×2
KE of the block moves blocks ahead and the spring stretches by 6.7 cm but the block does not return due to same region. Block is now at rest for the second time. ∴ Distance travelled by block when it comes to rest for the second time = (5 + 6.7 + 6.7 + 6.7) cm = 25 cm approx. Option (b) is correct.
= 50 m/s2
5 m 100 v2 = u2 + 2as
Displacement ( s) = x = ∴
= 22 + 2 × 50 ×
5 100
= 4+5 =9 ∴ v = 3 ms −1 Option (b) is correct. 26. Compression in spring Velocity gained
by block when the spring is at its natural length will compress the spring. 1 2 1 kx = mv2 2 2 m x=v k 2 =3 4000 2×9 ×5 = 4000 × 5 45 10000 6.7 m = 100 = 6.7 cm
=
Spring at its natural length. Block at rest first time due to some reason. Restoring force brings the spring to its natural length and block attains a velocity of 3 ms −1.
M1= 8 m 2v
27.
(M + s) 48 m
A
P
M2= 16 m 2v B
6L 12 L
8 m × AP = 16m × (12L − AP) AP = 24 L − 2 AP 3 AP = 24 L AP = 8 L As the CM of M and S does not change, the CM of the bar shall also not change i.e., the displacement of bar will be zero. Let x be the displacement of rod. ∴ x × 8 m + ( x + 6 L) 48 m + ( x + 12 L) 16m = (6L × 48 m) + (8 L × 24 m) i.e., x = 0 m i.e., no displacement of bar. Option (d) is correct. 8 m (2 v) + 16 m ( − v) 28. vCM of M and S = =0 8 m + 16 m There CM of M and S will not change while they move i.e.,the point P (where they meet) is at the edge of the table supporting the end B. Option (b) is correct.
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 29. When the spider eats up the moth and
v relative 2
travels towards A with velocity to rod.
197
20L 20L = = = 20T v L/ T = 20 × 4 = 80 s 31. Form CM not to shift
v 24 m + vR + 48 m vR = 0 2
8L x'
[vR = velocity (absolute) of rod] v ⇒ vR = − 6 Option (c) is correct. 30. Time taken by spider to reach point A
starting from point B 4L 8L = + v v/2
M+S 48 m
48 m
( x ′ + 8 L) 24 m + ( x ′ + 6L) 48 m = 64 × 48 m 8L i.e., x ′ = − 3 Option (a) is correct. ⇒
More than One Correct Options
v 2
1. Along vertical : 2 mV sin θ = m ⋅ sin 45°
i.e.,
2 V sin θ =
v 2 2
m v m
Rest 2m
…(i) v/2
45° θ 2m L
Along horizontal : v mv = m cos 45° + 2mV cos θ 2 1 i.e., …(ii) 2 V cos θ = v 1 − 2 2 Squaring and adding Eq. (i)and (ii), 2 2 v v 8 V 2 = + v − 2 2 2 2 2 v2 v v = + v2 + − 2⋅v ⋅ 8 7 2 2 5v2 v2 = − 4 2 Dividing Eq. (ii) by Eq. (i) 1 2 2 tan θ = 2 2−1 2 2
=
1 < 1 2 2−1
∴ θ < 45° Thus, the divergence angle between the π particles will be less than . 2 Option (b) is correct. 1 Initial KE = mv2 2 2 1 v 1 Final KE = m + 2 mV 2 2 2 2 1 1 5 1 = mv2 + 2 ⋅ − 2⋅ 2 32 8 2 4 As Final KE < Initial KE Collision is inelastic. Option (d) is correct. m − m1 2. v2 ′ = 2 v2 m1 + m2 v2
v1 = 0
m2 = m
m1 = 5m
v2'
m − 5m v2 5m + m 2 = − v2 3 2 =− 2g l 3 =
v1'
198 | Mechanics-1 mv ′22 = mg l mv ′2 or T= + mg l m8g = + mg 9 17 mg = 9 Option (a) is correct. Velocity of block 2m2 v1 ′ = v2 m1 + m2 2m = ⋅ 2g l 5m + m 1 = 2g l 3 Option (c) is correct. Maximum height attained by pendulum bob v ′2 8 g l / 9 4 l = 2 = = 2g 2g g
u = mv sin θ e + v
T−
= mu sin θ (1 + e) Option (d) is correct. u 1 − (1 − e2 ) sin2 θ = u 1 − sin2 θ + e2 sin2 θ = u cos2 θ + e2 sin2 θ = u cos2 θ +
= v2 cos2 φ + v2 sin2 φ =v Option (c) is correct. cos2 θ + e2 sin2 θ tan2 φ = cos2 θ + sin2 θ tan2 θ = cos2 θ (1 + tan2 φ) = cos2 θ sec2 φ cos2 θ = cos2 φ
v sin φ v v cos φ
Option (d) is correct. →
^
^
4. u = (3 i + 2 j) ms −1
u sin θ θ
φ
v sin φ u sin θ cos θ sin φ = × cos φ sin θ tan φ = tan θ Option (b) is correct. Change in momentum of particle = ( − mv sin φ) − ( + mu sin θ) ∴ Impulse delivered by floor to the particle = mv sin φ + mu sin θ sin φ u = mv sin θ + sin θ v and
Final KE v2 = 2 Initial KE u
=
u
+
v2 sin2 φ u2
= u2 cos2 θ + v2 sin2 φ
Option (d) is correct. v cos θ 3. v cos φ = u cos θ ⇒ = u cos φ
u cos θ
u v
e=
u mM
→
v = (–2 ^i + ^j) m/s m
Impulse received by particle of mass m →
→
= − mu + mv = − m (3 ^i + 2 ^j) + m( − 2 ^i + ^j) = − m (5 ^i + ^j) unit Option (b) is correct. Impulse received by particle of mass M = − (impulse received by particle of mass m) = m (5 ^i + ^j) Option (d) is correct.
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | 5. T = m1a
and
m2 g − T = m2 a m2 Solving, a= g m1 + m2 a m1
T
T T
a
T
x
m2 y m2g
m1a + m2 0 m1 + m2 m1 = a m1 + m2 m1m2 = g ( m1 + m2 )2
( aCM ) x =
Option (b) is correct. m 0 + m2 a ( aCM ) y = 1 ( m1 + m2 ) m2 = a m1 + m2
3 mv2 16 Option (c) is correct. =
2
m = g m1 + m2
8. As the mass of the system keeps on
Option (c) is correct. 6. As the block comes down, the CM of the
system will also come down i.e., it does not remain stationary. mg aCM = ≠g m+M aCM is downwards and also aCM < g. Option (d) is correct. As no force acts along horizontal direction, the momentum of the system will remain conserved along horizontal direction. Option (c) is correct. 7. Velocity of B after collision : v2
v1 = 0
v2'
v1'
m2
m1
m1
m2
A B Before collision
199
1 + e v1 ′ = v 2 2 3 1 = v [as e = and v2 = v (given)] 4 2 v ≠ 2 Impulse given by A to B = change in momentum of B 3 = m v − m ⋅ 0 4 3 = mv 4 Option (b) is correct. Velocity of A after collision 1 − e v2 ′ = v 2 2 v = 4 Loss of KE during collision 1 1 = mv22 − m(v ′21 + v ′22 ) 2 2 2 2 1 v 3v = m v2 − − 4 4 2
A B After collision
decreasing momentum of the system does not remain constant. Thrust force is developed on the rocket due to Newton’s 3rd law of motion. Option (b) is correct. dv vi dm As, a= = − −g dt m dt The value of a will remain constant if vi dm and − are constant. dt Option (c) is correct. Fnet = Ft (Thrust force due to gas ejection) − W (weight of rocket) F a = net m Thus, Newton’s 2nd law is applied. Option (d) is correct.
200 | Mechanics-1 Match the Columns 1. If x0 is the compression made in the
spring, the restoring force on B will decrease from kx0 to zero as the spring regains its original length. Thus, the acceleration of B will also decrease from kx0 to zero. mB
2. Initial aCM =
m ( + g) + m ( + g) m+m u=0 2nd particle
180 m +
mA
mB
k
A
20 ms–1 1st particle
B m x0
kx0
A
B
kx0
So, the aCM will also decrease from kx0 to zero. m A + mB ∴ (a) → (r) When spring is released after compressing it, the restoring on B will accelerate it towards right while the reaction force on A will apply a force on the wall which in turn will apply equal and opposite force on A and consequently A will travel towards right. As both travel towards right the velocity of CM will be maximum in the beginning. After this A will start compressing the spring and at a certain instant when the spring is compressed to maximum value both the blocks will travel towards right with a constant velocity and then the velocity of CM will become constant. ∴ (b) → (q) As the blocks will never move along y-axis, the y-component of the CM of the two blocks will not change. ∴ (d) → (p) As the two blocks will keep on moving towards right (surface below being smooth) the x-coordinate of the CM of the blocks will keep on increasing. ∴ (c) → (s)
=+g = + 10 SI unit ∴ (a) → (q) Initial vCM =
m ( − 20) + m × 0 m+m
= − 10 ∴ |vCM|= 10 SI unit ∴ (b) → (q) For the time taken by the first particle to return to ground 1 s = ut + at2 2 0 = ( − 20) t + 5t2 ⇒ t=4s Now, as the collision of the first particle with the ground is perfectly inelastic, the first particle will remain on ground at rest. Now, let us find the position of 2nd particle at t = 5 s 1 s = (0) 5 + (10) 52 2 = 125 m The particle (2nd) will still be in space moving downwards. m ⋅0 + m ⋅ g aCM = m+m g (SI unit) = =5 2 ∴ (c) → (p) Velocity of 2nd particle at t = 5 s v = 0 + 10 × 5 = 50 ms −1
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | ∴ At t = 5 s vCM
m ⋅ 0 + m ⋅ 50 = m+m (SI unit)
= 25 ∴ (d) → (s) 3. Initial KE of block B = 4 J A
B
m
m = 0.5 kg
1 ∴ × 0.5 × u2 = 4 2 ⇒ u = 4 ms −1 ∴ Initial momentum of B = 0.5 × 4 = 2 kg ms −1 ∴ (a) → (r) Initial momentum pCM = pA + pB =0+2 = 2 kg ms −1 ∴ (b) → (r) Velocity given to block B will compress the spring and this will gradually increase the velocity of A. When the spring gets compressed to its maximum both the blocks will have the same velocities i.e., same momentum as both have same mass. pA = pB (at maximum compression of the spring) But, pA + pB = initial momentum of B. ∴ pA + pA = 2 i.e., pA = 1 kgms −1 ∴ (c) → (q) After the maximum compression in the spring, the spring will gradually expand but now the velocity of block A will increase and that of B will decrease and when the spring attains maximum expansion the velocity of B will be zero and so will be its momentum. ∴ (d) → (p) 4. If collision is elastic, the two blocks will
interchange there velocities (mass of both balls being equal). Thus, velocity of A after collision = v
201
∴ (a) → (r) If collision is perfectly inelastic, the two balls will move together (with velocities V). ∴ mv = ( m + m) V v ⇒ V = 2 ∴ (b) → (s) 1 If collision is inelastic with e = , 2 1+ e v1 ′ = ⋅ v2 2 1 1+ 2 ⋅ v [Qv = v (given)] = 2 2 3 = v 4 ∴ (c) → (p) 1 If collision is inelastic with e = , 4 1 1 + 4 v = 5 v v1 ′ = 2 8 ∴ (d) → (q). 5. If A moves x towards right A
B 50 kg C 30 kg
60 kg
Smooth
Let plank (along with B) move by x′ to the right. x × 30 + x ′ (60 + 30) ∴ =0 30 + (60 + 30) x i.e., x′ = − 3 x = , towards left. 3 ∴ (a) → (r) If B moves x towards left Let plank (along with A) move x′ to the left x ⋅ 60 + x ′ (30 + 30) ∴ =0 60 + (30 + 30) i.e.,
x′ = − x = x, towards right
202 | Mechanics-1 ∴ (b) → (p) If A moves x towards right and B moves x towards left. Let plank moves x′ towards right 30 × x + 60 ( − x) + 30 ( x ′ ) ∴ =0 30 + (60) + (30) i.e., x ′ = x = x, towards right ∴ (c) → (p) If A and B both move x towards right. Let plank moves x′ towards right (30 + 60) x + 30x ′ ∴ =0 (30 + 60) + 30 i.e.,
x′ = − 3 x = 3x, towards right
∴ (d) → (s) 6. For man to be in equilibrium
1 T1
T1
T1 N
T1
T2 T2
m1 N
2
T2 T2 m2g
m1g
…(i) N + T1 = W For the block of mass m1 to be in quilibrium …(ii) T2 = N + m1 g For the block of mass m2 to be in quilibrium …(iii) T2 = m2 g For the equilibrium of pulley 2 …(iv) T1 = 2T2 Solving Eqs. (i), (ii), (iii) and (iv) W = 3 m2 g − m1 g = (3 m2 − m1) g
= (60 − 10) 10 = 500 N ∴ (a) → (r) For the equilibrium of man N = W − T1 = W − 2T2 = W − 2 ( N + m1 g) i.e., 3 N = W − 2m1 g N = 100 N ∴ (d) → (s) Force exerted by man on string to accelerate the centre of mass of the system upwards Centre of mass of the system will move upwards if man move upward. i.e., when T1 > W − N > 500 − 10 > 400 N Options are 500 N and 600 N. (b) → (r) and (s). ∴ Force ( T1) exerted by man on string to accelerate the centre of mass of the system. Centre of mass of the system will move downward if man moves downward. i.e., when T1 < W − N < 400 N Options are 100 N and 150 N. ∴ (c) → (q). 2×3 +0×6 7. vCM = 3+6 2ms–1
3 kg
6 kg
2 ms −1 3 When both the blocks move with same velocity (say v) deformation in the spring will be maximum 2 × 3 = 3v + 6 v 2 i.e., v = ms −1 3 = velocity of A = velocity of B =
Centre of Mass, Conservation of Linear Momentum, Impulse and Collision | vCM
2 2 3 × +6× 3 3 = 3+6 2 = ms −1 3
When both the blocks move with same velocities, each will be at rest w.r.t. the other ∴ (a) → (p), (r), (s) (b) → (p), (r), (s) Minimum speed of 3 kg block will be 2 ms −1 and at that moment velocity of CM 3 2 will be ms −1. 3 ∴ (c) → (p) Initial velocity of 6 kg block is zero. When the spring is compressed to maximum value the velocity of 6 kg will be maximum and the velocity of CM will 2 be ms −1 (as explained above). 3 ∴ (c) → (p).
8. vCM =
203
2 × 5 + 1 ( − 10) 2+1
= 0 ms −1 + ive Rough 1 kg 5 ms–1
10 ms–1
2 kg Smooth
∴ (a) → (r) Momentum of CM = 2 ( + 5) + 1 ( −10) = 0 kg ms −1 (b) → (r) ∴ Velocity and so the momentum of 1 kg block will decrease to zero as the surface below is rough. ∴ (c) → (q) Velocity and so the KE of 2 kg block will decrease to zero when the velocity of 1 kg block becomes zero (according to law of conservation of momentum). ∴ (d) → (q).
