Physics Sheet Solutions 2nd Dispatch CLASS : XI
Contents Preface 1.
Newton's Laws of Motion Exercise
2.
96 - 113
Rigid Body Dynamics Exercise
8.
079 - 95
Centre of Mass Exercise
7.
059 - 078
Circular Motion Exercise
6.
044 - 058
Work, Power & Energy Exercise
5.
026 - 043
Gravitation Exercise
4.
001 - 025
Friction Exercise
3.
Page No.
114 - 156
Unit and Dimensions Exercise
157 - 161
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TOPIC : NEWTON'S LAWS OF MOTION EXERCISE-1 PART - I SECTION (A) A-1.
Gravitational, Electromagnetic, Nuclear.
A-4.
Newton's IIIrd Law
A-6.
Vertical wall does not exert force on sphere (N' = 0).
A-8.
action reaction pairs
(1) and (2) (3) and (4) (5) and (6) (7) and (8)
SECTION (B) B-1.
N = F + mg N = mg + mg N = 2mg
[equilibrium]
B-3.
If is obvious that block can`t accelerate in y direction N – mg cos = 0 N = mg cos
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 1
B-5.
Due to symmetry normal reactions due to left and right wall are same in magnitude W – N cos 60 – N cos 60 = 0 [Equilibrium in vertical direction] W–
B-7.
N1 cos300 N1
N N – =0 2 2 = 50 +
N 3 – 2 = 2 2
N1 sin300
=
N=W
N2 2
50 ................ (1)
N2 2
N1 = 2 N2 ..............................(2) Solving equation (1) & (2) N1 = 136.57 N N2 = 96.58 N
SECTION (C) C-1.
Since string 2 is massless net force on it must be zero. T2 = F = 10 N T1 + mg = T2 [Equilibrium of block] T1 + 1 × 10 = 10 T1 = 0
C-3.
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 2
TC – 10 = 0 TC = 10 N TB – TC – 5 = 0 TB – 10 – 5 = 0 TB = 15 N TA – TB – 5 = 0 TA = 20 N
C-5.
[Equilibrium of block] [Equilibrium of 2]
[Equilibrium of 1]
For finding distance travelled we need to know the acceleration and initial velocity of block. m 2g – T = m 2a [Newton’s second law for m2] T – m 1g = m 1a [Newton’s second law for m1] m2g – m1g = (m2 + m1)a [adding both the equation]
a=
(m2 – m1)g m2 m1
a=
6–3 ×g 63
a=
g 10 = m/s2 3 3
s=ut+ S=
1 at2 2
=0×2+
1 10 × × 22 2 3
20 m 3
T – m1g = m, a
g
T = m1 g 3
=3×
40 3
T = 40 N
Force exerted by clamp on pulley is 2T 2 × 40 = 80 N C-7.
VA = VP2 = 10m/s For pulley
V1 V2 VP = 2
P P
VB – VC VP2 = 2
VB 2 VB = 22 m/s 2
and
VP1
VA 10 2
VP P2
10 =
0=
P1
VP
V1 1 2 V2
VA VP2
V A. A
Vp2 = 10m/s .
C 2m/s = VC
VB. B
2
VA = 10 m/s
SECTION (D) D-1.
Since string is inextensible length of string can’t change rate of decreases of length of left string = rate of increase of length of right string
V1 cos 1 = V2 cos 2
RESONANCE
V1 cos 2 V2 = cos 1
SOLN_NEWTON'S LAWS OF MOTION - 3
D-3.
Since rod is rigid, its length can’t increase. velocity of approach of A and B point of rod is zero. u sin – v cos = 0 v = u tan at any angle x and y coordinates of center of mass are
X
cos 2
...............(i)
sin 2 from (i) and (ii) Y
...............(ii)
2 4 equation of circle. X2 Y 2
D-5.
V1 =
10 – 20 2
[constrained relation of P1 ] V1 = – 5 m/s 10 =
–5 V2 2
V2 = 25 m/s VC = V2 = 25 m/s upwards VP1 = V1 = – 5 m/s
VP = 5 m/s downward [because we have assumed upward direction as +ve for V1]
SECTION (E) E-1.
Since point A is massless net force on it must be zero other wise it will have acceleration. F1 – 60 cos 45 = 0 F1 = 30 2 N F2 – 60 cos 45 = 0
F2 = 30 2 N W – 60 sin 45 = 0 W = 30
E-3.
2 N
F = ma
ˆ ˆ a = ax i + a y j
=
d2 x dt 2
d2 y ˆi + ˆ dt 2 i
= (10) ˆi + (18 t) ˆj
at t = 2 sec t = 2 sec
ˆ a = 10 ˆi + 36 j
ˆ ˆ F = 3 (10 i 36 j )
= 30 ˆi 108 ˆj
F =
302 108 2 = 112.08 N
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 4
E-5.
R4 – mg = ma
R4 – 1 = 0.1 × 2
R4 = 1.2 N
R3 – mg – R4 = ma R3 – 1 – 1.2 = 0.1 × 2
R3 = 2.4 N
Similarly R2 = 3.6 N R1 = 4.8 N F =6N Fnet = ma = 0.1 × 2 = 0.2 N
m
B m
> T
2a
B
2x 2a
> >
(a) When the block m is pulled 2x towards left the pully rises vertically up by x amount. aB = 2aA F.B.D. of blocks
T x
T
T = m2a ............. (1)
2T
>
E-7.
A 2m a
F.B.D.
2T FBD of A
FBD
A 2m a
2mg – 2 T = 2ma
2mg mg – T = ma ................(2) (1) + (2) mg = 3ma a = g/3 aB = 2g/3
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 5
(b) = xB + 3xA 2
+3
aB
d2 x A
dt dt 0 = – aB + 3aA
B m
2
xB
T T
aB = 3aA ........... (1) For B
^
0=
d2 x B
A
^
^ T
^
2T^ 3m
aA
T = maB ....................... (2) 3mg
For A
3mg – 3T = 3maA ............... (3) mg – T = maA
By (1) , (2) & (3) aB = 3g/4 Ans.
SECTION (F) F-1.
Reading of weighing machine is equal to the normal reaction Normal reaction is not affected by velocity of lift, it is only affected by acceleration of lift. For I, II and III a = 0 N – mg = 0 [Equilibrium of man] N = mg = 600 N For IV, VI and VII a = +2 m/s2 N – mg = ma [Newtons II law] N = 60 × 2 + 60 × 10 = 720 N For V and VIII a = – 2 m/s2 N – mg = ma [Newtons II law] N = 60 × (–2) + 60 × 10 = 480 N
F-3.
(a) For A For B
(b)
T AB = 2mg, T BC = mg 2mg + mg = maA T AB – mg – T BC = maB 2mg – mg – mg = maB T BC – mg = mac T AB = 2mg T AB – mg = maB 2mg – mg = maB aB = g () aA = 0 & aC = g().
TAB
aA = 3g A m
maB = aB = 0 ac = 0.
TAB
m B m
Cm
mg
mg mg
TAB
B m
TBC
TBC
aB
mg
SECTION (G) G-1.
If we take both A and B as a system then there is no external force on system. mAaA + mBaB = 0 [Newton’s II law for system ] 60 aA + 75 × 3 = 0
aA =
–15 m/s2 4
–ve sign means that acceleration is in direction opposite to the assumed direction.
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 6
G-3.
4F – (M + m)g = a =
G-5.
(M + m)a
4F 4F – (M m)g = –g Mm Mm aA
tan 45º = a B
(wedge constrained relation)
N sin 45 = ma For Rod A mg – N cos 45 = ma From equation (1) & (2) a =
...........(i) ...........(ii)
g 2
SECTION (H) H-1.
Pseudo force depends on mass of object and acceleration of observer (frame) which is zero in this problem. Pseudo force is zero.
H-3.
F.B.D. in frame of lift It is obevious that block can accelerate only in x direction. ma is Pseudo force. mg sin + ma sin = max [Newton`s II law for block in x direction] ax = (g + a) sin
PART - II SECTION (A) A-1.
Experimental fact.
A-3.
Force exerted by string is always along the string and of pull type. When there is a contact between a point and a surface the normal reaction is perpendicular to the surface and of push type.
SECTION (B) B-1.
F – N = 2 ma, N = ma1
[Newton`s II law for block A] [Newton`s II law for block B]
F 3
N=
N = 2 ma2 [Newtons II law for block A] F – N = m 2a [Newtons II law for block B] N = 2F/3 so the ratio is 1 : 2
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 7
B-3.
F – N = Ma N – N’ = ma N’ = M’a
N’ = M’
[Newtons law for block of mass M] [Newtons law for block of mass m] [Newtons law for block of mass M’]
F M m M'
N = (m + M’)
F M m M'
N > N’
SECTION (C) C-1.
Point A is mass less so net force on it most be zero otherwise it will have acceleration. F – Tsin = 0 [Equilibrium of A in horizontal direction] F
C-3.
T = sin
10 – T2 = 1 a [ Newton’s II law for A ] T2 + 30 – T1 = 3 a [ Newton’s II law for B ] T1 – 30 = 3a [ Newton’s II law for C ]
a=
T2 =
C-5.
g 7 6g 7
Mg – T = Ma [ Newton’s II law for M] T – mg = ma [ Newton’s II law for m]
T=
2 m Mg mM
If m << M than m + M M
T=
2 m Mg M
T = 2 mg Total downward force on pulley is 2T = 4 mg.
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 8
SECTION (D) D-1.
The length of string AB is constant.
speed A and B along the string are same u sin = V V
u sin = V
D-3.
u = sin
By symmetry we can conclude that block will move only in vertical direction. Length of string AB remains constant
Velocity of point A and B along the string is same. V cos = u
D-5.
Let
u
V = cos
AB = , B = (x , y) v B = vx ˆi + vy ˆj v B = 3 ˆi + v y ˆj 2
2
x +y =
(i)
2
2x vx + 2y vy = 0
3 +
y v =0 x y
3 + (tan600) vy = 0 vy = – 1 Hence from (i) v B = 3 ˆi – ˆj Hence vB = 2 m/s
D-7.
V = (velcoity of B w.r.t ground)
V–4 =2 2
V = 8 m/s (velcoity of B w.r.t ground) V' = 6 m/s (velcoity of B w.r.t lift )
D-8.
u cos 45° = v cos 60°
or v =
2u
SECTION (E) dv a dt
E-1.
F ma
E-4.
In free fall gravitation force acts.
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 9
E-5.
M2 g sin – T = M2a T – M1g sin = M1a By adding both equations
[Newton’s II law for M2] [Newton’s II law for M1]
M2 sin – M1 sin g M1 M2
a= E-6.
Case 1
T1 – mg = ma1 2 mg – T1 = 2 ma1 a1 =
[Newton’s II law for m] [Newton’s II law for 2m]
g 3
Case 2
F – mg = ma2 2 mg – mg = ma2
[Newton’s II law for m] a2 = g
a2 > a1
E-7. F = m1 4 F = m 26 F = (m1 + m2)a F
F
1
1 = 4 6a
a = 2.4 m/s2.
F 6 ˆi – 8 ˆj 10 kˆ
2 2 2 as
x=
–m 2F
O2 = 32 +
F m a
F ma
E-11.
1
F = 4 6a
E-10.
[Newton’s II law for m1] [Newton’s II law for m2] [Newton’s II law for (m1 + m2)]
62 82 102
0 2 12 2
=m1
m = 10
2 kg.
F x m
2 2 2 as
2F1 x m
RESONANCE
2F1 – m
0 = 9 + m 2 F
F1 = 9F
SOLN_NEWTON'S LAWS OF MOTION - 10
E-12.
Mg sin – T = Ma T = Ma By dividing both equations 2 T = Mg sin
T=
[Newton’s II law for block 1] [Newton’s II law for block 2] Mg sin 2
SECTION (F) F-1.
T – mg = 0 [ Equilibrium of block] T – 10 = 0 T = 10 Reading of spring balance is same as tension in spring balance.
F-2.
F-4.
F – k x = m1 a1 kx = m2a2 By adding both equations.
[Newton’s II law for M1] [Newton’s II law for M2]
F = m1a1 + m2a2
a2 =
F – m1 a1 m2
Weight of man in stationary lift is mg. mg – n = ma [Newton’s II law for man] N =m (g – a) Weight of man in moving lift is equal to N.
mg 3 m (g – a) 2
a=
g 3
SECTION (G)
G-2.
F = m1a1 [Newton’s II law for m1] 180 = 20 a1 a1 = 9 m/s2 Net force on m2 is 0 therefore acceleration of m2 is 0.
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 11
G-3.
30 – T2 = 3 a T2 – T1 = 6 a T1 – 10 = 1 a By adding three equations 30 – 10 = 10 a
[Newton’s II law for 3 kg block] [Newton’s II law for 6 kg block] [Newton’s II law for 1 kg block] a = 2 m/s2.
SECTION (H) H-1.
FBD of block is shown w.r.t. wedge and FBD of wedge is shown w.r.t. ground. FP is pseudo force. mg sin 37 – ma cos 37 = mab = 10 ×
3 5
4 5
= 2 m.s2 w.r.t. wedge
ab = g sin 37 – a cos 37
block is not stationary w.r.t. wedge N – ma sin 37 – mg cos 37 = 0 [Newton’s II law for block]
N = 1 × 10 ×
–5×
4 3 +1×5× 5 5
N = 11 N. Net force acting on block w.r.t. ground. F=
(mg sin 37 )2 (mg cos 37 – N)2 2
= =
3 4 10 10 – 11 5 5
2
62 32
F = 3 5 N.
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 12
H-3.
F.B.D. of wedge is w.r.t. ground and F.B.D. of block is w.r.t. wedge. Let a is the acceleration of wedge due to force F. FP is pseudo force on block mg sin 30º – ma cos 30º = 0 [Equilibrium of block in x direction w.r.t. wedge] a = g tan 30º F = ( M + m)a [Newtons II law for the system of block and wedge in horizontal direction] F = (M + m) g tan 30º.
H-4.
acceleration of point A and B must be some along the line to the surface a sin = g cos a = g cot
H-5.
F.B.D. of block is w.r.t. wedge and F.B.D. of wedge is w.r.t ground. FP is pseudo force on block . mg sin – ma cos = 0 [ Equilibrium of block w.r.t. wedge along x direction ] a = g tan
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 13
EXERCISE-2 PART - I
1.
mg – NAB = maA [Newton’s II law for block A in vertical direction] mg sin + NAB sin = maB [Newton’s II law for block B in x direction] aA = aB sin [Constrained relation for contact surface between block A and B] Solving above three equations we get NAB =
mg cos2 1 sin2
mg cos + NAB cos – nBC = 0 NBC =
mg cos +
mg cos2 cos
NBC sin – NWC = 0 NWC =
[Equilibrium of block B in y direction]
2
1 sin
NBC =
2 mg cos 1 sin2
[Equilibrium of block in horizontal direction]
2 mg sin cos 1 sin2
NBC cos + mg – NFC = 0
[Equilibrium of block C in vertical direction ]
2
3.
NFC =
NFC =
2 mg cos 1 sin2
mg
mg (2 cos2 ) 1 sin2
mg – Ncos 37 = maB [Newton’s II law for block B in vertical direction] N sin 37 = maA [Newton’s II law for block A in horizontal direction] aB cos 37 = aA sin 37 [constrained relation for contact surface between block A and B] By solving above three equations we get aA =
12 g 25
N=
4mg 5
aB =
9g 25
NBW = N sin 37 [Equilibrium of block B in horizontal direction]
NBC =
12 mg 25
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 14
6.
aB + a = 2aA [constrained relation for pulley 1] O + a = 2aB [contrained relation for pulley 2] From above two equations 3aB = 2aA
3 a 2 B F – 2T = 2maA [Netwon's II law for block A] 3T = 4 m aB [Netwon's II law for block B] From equation I, II and III aA =
aB = 8.
..........I ..........II ...........III
3F . 17m
mAg – 2T = mAaA [Newton's II law for block A] T – mBg = mBaB [Newton's II law for block B] aB + O = 2aA [constrained relation for pulley P1] mA = 4mB [Given in question] From above four equations aA =
g = 2.5 m/s2 4
aB =
g = 5 m/s2 2
h=
1 2 a t [Equation of motion for block A] 2 A
2 sec. 5 H is the distance travelled by block
t=
B in vertical direction till
H=
1 a t2 2 B
1 2 5 2 5
2 second 5
[Equation of motion for block B]
2
H = 0.4 m H´ is the distance travelled by block B due to gained velocity. v 1 = a Bt = 5 × 0.4 v1 = 2 m/s v22 = v12 + 2a H´ 02 = 22 + 2 (–10) H´ H´ =
2 = 0.2 m 10
Total distance = H + H´ = 0.6 m = 60 cm
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 15
9.
4F1 – F2 = ma [Newtons II law for block]
4F1 – F2 m For t = 0 to 2 sec. F1 = 30N F2 = 10N a=
4 30 – 10 = 2.75 m/s2 40 For t = 2 to 4 sec F1 = 30N
a=
F2 = 20N
a=
4 30 – 20 = 2.5 m/s2 40
a=
4 10 – 40 = 0 m/s2 40
For t = 4 to 6 sec. F1 = 10N F2 = 40N
For t = 6 to 12 sec F1 = 0 , F2 = 0 a = 0 m/s2 V12 – V0 = a0–2(2 – 0) + a2–4(4 – 2) + a4–6(6 – 4) + a6–12(12–6) V12 – 1.5 = 2.75 × 2 + 2.5 × 2 + 0 × 2 + 0 × 6 V12 = 12 m/s
11.
All the forces shown are in ground frame. aw is the acceleration of wedge w.r.t ground and a is the acceleration of blocks w.r.t wedge. mAg sin45º – T = mA (a – aw cos 45º) [Newton's II law for block A along the wedge in ground frame] mAgcos – N = mA awsin45º [Newton's Ii law for block A in direction to the wedge in ground frame.] T – mBg sin 45 = mB (a – awcos 45) [Newton's II law for block B along the wedge in ground frame.] NB – mBg cos 45º = mB (awsin45) [Newton's II law for block B in direction to the wedge in ground frame] NAsin45 + T cos 45 – NB sin 45 – T cos45 = mwaw [Newton's II law for wedge in horizontal direction in ground frame]. After solving above five equations we will get aw =
2 m/s2 5
acceleration of B w.r.t ground in
RESONANCE
2 13 m/s2. 5
SOLN_NEWTON'S LAWS OF MOTION - 16
13.
16.
2aA = a + aB 2aA = 3 + aB 2T – 100 = 10aA 50 – T = 5aB aB + aA = 0 2aA – 3 + aA = 0 aA = 1m/s2 aB = – 1m/s2 .
Fnet = mg – 2F cos
anet = g –
17.
2k L2 x 2 – L m
=
x L2 x 2
2
Fs = K 2mg 2 < mg
<
Fs
T + Fs = mg T
= mg –
K 2
If it is So Fs > mg i.e. 19.
20.
<
string unstretched & T = 0. 2
N sin = mb N sin = m(a cos – b) 2mg – N cos = ma sin
4g sin
a=
1 h = a sin t2 2
1 sin2
t=
h(1 sin 2 ) 2g sin 2
.
acceleration of bead along rod is
ma cos = a cos m 1 a cos t2 = 2 t= N=
2 a cos
(ma sin )2 (mg)2 .
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 17
22.
By newtons law on system of (A, B, C) along the string. (a) (M + m – M) g = (2M + m) a
mg 2M m (b) free body diagram ‘C’ block FBD
a=
mg – N = ma
N=m N=
2M gm 2M m
(c) T – Mg = M
gm g 2M m
mg for A block 2M m
T = Mg +
Mmg 2M m
for pulley P = 2T + Mg = 2Mg +
6M 3m 2m 2Mmg + Mg = Mg 2M m 2M m
6M 5m P = 2 M m Mg
23.
mg – T = ma
a=
2g 29
a 3a 6g = = 2 2 58
S=
1 a t2 2 rel
58 3g
t = 1.4 sec.
0.2 mg = 2.9 ma
arel = a + 1=
2T – 1.8 mg = 1.8 m
1 6g 2 t 2 58
t=
RESONANCE
a 2
SOLN_NEWTON'S LAWS OF MOTION - 18
PART - II 1.
Slope of vrel – t curve is Constant. arel = Const. = a1 – a2 0 Inference that at least one reference frame is accelerating both can’t be non - accelerating simultaneously.
3.
T1cos45º = T2cos45º (T1 + T2) sin45º = mg
T1 = T2
2 T1 = mg
mg T1 =
.
2
T sin = mg +
mg 2
T sin = mg +
T1
mg 2 2 dividing (i) and (ii) T cos =
tan = 6.
=
2
.........(i)
.........(ii)
M m/2 2M =1+ Ans. m/2 m
w – f = ma
T1
w – ma = g
m
m
w 1 – mg a = f
w 1 – w a = f
a
w 1 – g = f
8.
a1 =
10.
2mg – mg m
a2 =
a1 = g
a2 =
2m – m g
a3 = 0
2m m g 3
So,
a1 > a2 > a3
By setting string length constant L
= 31 + 22
3v0 = 2vA vA vAB
3 v 2 0 = vA– vB
=
=
v0 towards right. 2
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 19
12.
T 2
=
2m 2m 3 g m 2 m3
m1g 2
=
2m 2m 3 g m 2 m3
m1
=
4m 2m 3 m 2 m3
1 1 + m2 m3 13.
T sin
=
4 m1
= m (g sin + a0)
T cos = mg cos
16.
g sin a 0 = g cos
g sin a 0 = tan–1 g cos
tan
T1 + T2 = mg If upper spring is cut
mg – T2 = m × 6 .....(i) If lower spring is cut :
mg – T1 = ma
......(ii)
adding (i) and (ii) 2mg – {T1 + T2} = m (a + 6) 2mg – mg = m (a + 6) mg = m (a + 6) g=a+6 a = 4m/s2. 18.
A+B+C+D+E = 300 i B + C + D + E = – 100 i A + C + D + E = 2400 j equation (1) - equation (3) give B = 300 i – 2400 j equation (1) - equation (2) give A = 400 i Adding equation (4) and (5) A + B = 700 i – 2400 j a(A + B)
=
..........(1) ........... (2) ........... (3) .............(4) ............. (5)
A B 100
= 7 i – 24 j
RESONANCE
= 25 m/s2
SOLN_NEWTON'S LAWS OF MOTION - 20
20.
For first case tension in spring will be Ts = 2mg just after 'A' is released. 2mg – mg = ma a = g In second case Ts = mg 2mg – mg = 2mb b = g/2 a/b = 2
22.
(Force diagram in the frame of the car) Applying Newton’s law perpendicular to string mg sin = ma cos a tan = g Applying Newton’s law along string
T – m g2 a 2 = ma
T = m g 2 a 2 + ma Ans.
24.
F.B.D. of mass m is w.r.t. trolley T sin ( – ) + mg sin – FP = 0 [Equilibrium of mass in x direction w.r.t. trolley] T sin ( – ) + mg sin – mg sin = 0 T sin (– ) = 0 since T cant be zero , sin (– ) must be zero =
25.
Maximum acceleration of block is 10 m/s2 . S= =
26.*
1 2 at 2
1 × 10 × 0.22 = 0.2 m = 20 cm. 2
T = m 1g when thred is burnt, tension in spring remains same = m1g. (m1 – m2 ) g= m2
m 1g – m 2g = m 2a
for m1
30.*
a = upwards
a=o
By string constraint aA = 2aB ................................(1) equation for block A. 10 × 10 ×
1
2 equation for block B. 2T –
400
– T = 10 aA ......(2)
= 40 aB .........................(3)
2
Solving equation (1) , (2) & (3) we get aA =
5 2
m/s2
RESONANCE
aB =
5 2 2
m/s2
T
=
150 2
N
SOLN_NEWTON'S LAWS OF MOTION - 21
31.*
Apply NLM on the system 200 = 20 a + 12 × 10 80 =a 20
= 4 m/s2
spring Force = 10 × 12 = 120 N 32.*
There is no horizontal force on block A, therefore it does not move in x-directing, whereas there is net downward force (mg – N) is acting on it, making its acceleration along negative y-direction. Block B moves downward as well as in negative x-direction. Downward acceleration of A and B will be equal due to constrain, thus w.r.t. B, A moves in positive x-direction.
Due to the component of normal exerted by C on B, it moves in negative x-direction.
EXERCISE-3 1.
(A) Let the horizontal component of velocity be ux. Then between the two instants (time interval T) the projectile is at same height, the net displacement (uxT) is horizontal uxT average velocity = = ux (A) p, r T (B) Let ˆi and ˆj be unit vectors in direction of east and north respectively.. V DC 20 ˆj , V BC 20 ˆi and V BA 20 ˆj V AD VDC V CB VBA = 20 ˆj 20 ˆi 20 ˆj = 20 ˆi VAD 20ˆi Hence V AD VBC (B) p, r (C) Net force exerted by earth on block of mass 8 kg is shown in FBD and normal reaction exerted by 8 kg block on earth is 120 N downwards.
Hence both forces in the statement are different in magnitude and opposite in direction. (C) q, s (D) For magnitude of displacement to be less than distance, the particle should turn back. Since the magnitude of final velocity (v) is less than magnitude of initial velocity (u), the nature of motion is as shown.
Average velocity is in direction of initial velocity and magnitude of average velocity = less than u because v < u. (C) q, r 2.
uv is 2
Let a be acceleration of two block system towards right
F2 F1 a = m m 1 2 The F.B.D. of m2 is F2 – T = m 2 a
m1 m2 Solving T = m m 1 2
F2 F 1 m 2 m1 (B) Replace F1 by – F1 is result of A m1 m 2 T = m m 1 2
RESONANCE
F2 F 1 m 2 m1 SOLN_NEWTON'S LAWS OF MOTION - 22
(C) Let a be acceleration of two block system towards left
F2 F1 a = m m 1 2 The FBD of m2 is F 2 – N2 = m 2 a
m1 m2 N = m m 1 2
F1 F2 m1 m 2 (D) Replace F1 by –F1 in result of C Solving
m1 m 2 N = m m 1 2 3.
F2 F 1 m m 1 2
FBD of Block in ground frame : applying N.L. 150 + 450 – 10 M = 5M
150 N 450 N
600 15 M = 600 M = 15 M = 40 Kg Ans. Normal on block is the reading of weighing machine i.e. 150 N.
4.
5 m/s2
Mg = 10 M
If lift is stopped & equilibrium is reached then T = 450 N
450 + N = 400
N
Mg = 400 M
N = – 50 So block will lose the contact with weighing machine thus reading of weighing machine will be zero. T = 40 g So reading of spring balance will be 40 Kg.
T
40 g T = 450 N
5.
40 Kg
N = 400 N a
a=
950 400 40
Mg = 400 N
a=
6.
ap =
450 45 = m/s 2 40 4
Ans.
10 t =t 10 v
t
dv =t dv t dt dt 0 0 Putting v = 2 we have t = 2 sec.
dx t 2 Now dt 2 xB = 2 × 2 = 4 m
RESONANCE
v=
t2 2
2
t3 xp = 6 0
Hence relative displacement = 4 –
=
4 3
4 8 = m 3 3 SOLN_NEWTON'S LAWS OF MOTION - 23
7.
From above t3 t2 = 12 t = 2 3 sec. 6 after 4 seconds VB =2 m/s
2t = 8.
a=t=4
42 = 8 m/s Vrel = 8 – 2 = 6 m/s. 2 Inertia is the propety to resist change in state of motion or rest. Vp =
9. 10.
The FBD of block A is The force exerted by B on A is N (normal reaction). The forces acting on A are N (horizontal) and mg (weight downwards). Hence statemt I is false.
11.
If the lift is retarding while it moves upward, the man shall feel lesser weight as compared to when lift was at rest. Hence statement1 is false and statement 2 is true.
12.
Newton's third law of motion is valid in all reference frames. Hence statement-1 is incorrect.
13.
(i) (True) (ii)
14.
(True)
(i) Earth (ii) 4N (iii) No (v) 4N , hand , book , downward
(iv) 4N , Earth, book , upward (vi) nd (vii) rd
EXERCISE-4 PART - I 1.
2mg cos =
2 mg
1 cos = 2.
2
= cos 45° = 45°
After string is cut, FBD of m
mg = g m FBD of 2m (when string is cut tension in the spring takes finite time to become zero. How ever tension in the string immediately become zero.)
2m
3mg 2mg g a= = 2m 2
2mg
a=
3.
F = 2T sin a=
T cos m x
F cos F a = 2m sin = 2m 4.
3mg
2
a x2
ma cos = mg cos (90 – )
a tan g
a dy g = dx
a d (kx2) = g dx
x = 2gk = D
a
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 24
PART–II 1.
F Mm FBD of m asystem =
T = masystem = 2.
3.
mF Mm
2 dx dx V1 = 1 V2 = 1 dt 2 1 dt 2 Impulse = |P| = |m(V2 – V1)| = |0.4 (–1 –1)| = 0.8 Ns
Vertical component of acceleration of A a1 = (g sin ). sin = g sin 60º . sin 60º = g .
3 4
That for B a2 = g sin 30º . sin 30º = g
4.
A =
(aAB)=
3g g g – = = 4.9 m/s2 4 4 2
3 2 , B = 5 5
2 K = K A 5
5.
0
t
F dv 0 e bt dt m
v=
0
F0 1 e bt mb
K = KAA = KBB KA =
F = ma = F0 e–bt v
1 4
5K 2
KB =
5K . 3
dv F0 bt e dt m
F0 v= m
t
e bt b 0
RESONANCE
SOLN_NEWTON'S LAWS OF MOTION - 25
TOPIC : FRICTION EXERCISE-1 PART - I SECTION (A) A-1.
A-4.
Friction is kinetic because their is relative motion. Direction of friction is such that it opposes the relative velocity.
A-5.
a = – µmg/m = – µg = – 1 m/s2 V 2 – Vi2 = 2as (V = 0 V = 5 m/s) f
f
i
25 s= = 12.5m. 21
SECTION (B)
B-4.
R = mg + 60 = 160 N
f = 80 N ( No sliding )
angle of friction B-5.
ablock =
= tan–1
f = tan–1 R
80 160
= tan–1
1 2
Ans.
mg = g = 0.15 × 10 = 1.5 m
aT = 2 ST – Sb = 5
1 1 aT t2 – a t2 = 5 2 2 B
1 2 t [2 – 1.5] = 5 2 t2 = 20
ST = =
1 a t2 2 T
1 × 2 × 20 = 20 m. 2
RESONANCE
SOLN_FRICTION - 26
B-6.
N = mg – F sin F cos = N = [mg – F sin ]
mg cos sin F is minimum when cos + sin is max F=
d (cos + sin ) = 0 d – sin + cos = 0 = tan or ;k = tan–1 also vr% cos + sin
=
1 2 for = tan-1
d sfy ,
mg
thus
Fmin =
1 2
SECTION (C)
C-1.
30 = smg
30 = s × 5 × 10 s = 0.6.
Again, kN kmg S=
a=
1 2 at 2
2S t
2
=
2 10 = 0.8. 25
30 – kmg = m × 0.8 C-2.
(i)
aA =
k =
30 m 0.8 = 0.52. mg
15 F = =3 5 m
0 =0 10 fAB = 0, fBG = 0.
aB =
(ii) fBG 75 Since fAB can’t be greater than fBG therefore acceleration of B will be zero. 30 25 = 1m/sec2 5 fAB = 25 N, fBG = 25 N.
and aA =
RESONANCE
SOLN_FRICTION - 27
(iii)
fAB 25
aA
25 5
or
aA 5
Let there is no sliding between A and B then common acceleration of A and B. 200 75 = 8.33 15 Since aA 5 Hence, there will be sliding between A and B in that case.
=
200 100 = 10 m/sec2 10 fAB = 25 N, fBG = 75 N.
aA = 5 m/sec2, aB =
(iv) aA 5 Let A and B move together then common acceleration. 90 75 = 1m/sec2 15 As common acceleration is less than aA hence A and B will move together aA = 1m/sec2, aB = 1m/sec2 fAB = mA × 1 = 5N, fBG = 75 N.
=
PART - II SECTION (A)
A-1.
Let acceleration in Ist case is a1 and that in second case is a2 Now ,
1 2 1 a1t = a2(2t)2 2 2
a2 =
a1 4
............(i)
Clearly a1 = mg sin = g sin ............(ii) m mg sin mg cos and rFkk a2 = = g sin – g cos ............(iii) m From (i), (ii) and (iii), we get A-2.
= 0.75.
The normal reaction on the block is N = mg – F sin Net force on block is Fcos – µN = Fcos – µ mg + µFsin or acceleration of the block is a=
F F(cos µ sin ) µmg = (cos + µsin) – µg m m
RESONANCE
SOLN_FRICTION - 28
A-3.
N = 50 – 40 sin30 = 30 a=
40 cos 30 º 0.2 30 = 5.73 m/sec2 5
A-4. Apply Newton’s law for system of m1 and m2 a=
(m1 m 2 )g sin 37 º [m1g cos 37 º m 2 g cos 37 º ] m1 m 2
= g[sin37º – cos37º] Now apply Newton’s law for M1
m1g sin37º + T – m1gcos37º = m1a = m1g[sin37º – cos37º] T = 0 and a = 4m/sec2
SECTION (B) B-3.
Solving from the frame of cart , we get N = ma, mg = N B-4.
mg = ma
a=
g
Solving from the frame of truck f mg = 6
B-7.
f = 5N.
Fpseudo = 5×1 = 5N
Rest
Apply Newton’s law for system along the string mB g = (mA + mC) × g
mC =
mB 5 – mA = – 10 = 15 kg 0 .2
RESONANCE
SOLN_FRICTION - 29
SECTION (C) : C-2.
fs fk ( s k ) mg = = (S – k) g m m = (0.5 – 0.4)10 = 1 m/sec2 a=
C-3.
When F is less than µsmg then tension in the string is zero. When µsmg F < µs2mg then friction on block B is static. If F is further increase friction on block B is kinetic.
C-4.
Solving from the frame of elevator
g = 12.5 4
geff. = g + f=
1 × 2 × 12.5 = 5N 5
a2 =
5 = 2.5 m/s2 2
a1 =
30 5 25 = m/sec2 8 8
EXERCISE-2 PART - I SECTION (A) 2.
a1 =
mg k mg g = (1 – k) 2m 2
a2 =
k mg k g = 4m 4
s1 =
1 a t2 2 1
s2 =
1 a t2 2 2
s1 – s2 =
s2 =
7 8
7 1 g 1 g (1 – k)t2 – k t2 = 8 2 2 2 4
7 7 t2 = 2g(1 ) g = g(2 3 ) k k k
7 k 7 k g 1 1 a2 t2 = × × g(2 3 ) = 8(2 – 3 ) 2 2 4 k k
RESONANCE
SOLN_FRICTION - 30
4.
Solving from the frame of rod. ab =
Now,
ma cos ma sin = a [cos – sin ] m =
5.
2 a[cos sin ]
f1 = 3 × 0.25 × 10 = 7.5
If
F = 17.5 + 25 + 37.5 = 80 N F = 200 then aB = aC T – f 1 – f 2 = m Ba F – T – f2 – f3 = mCa from equation (1) and (2) F – f1 – 2f2 – f3 = (mB + mC)a
a 6.
2 ab =
1 a t2 2 b
.........(1) .........(2)
F f1 2f2 f3 200 7.5 35 37.5 = = 10 m/sec2 mB m C 12
The F.B.D. of A and B are (force of friction)
For sliding to start between A and B, the frictional f = µ N =
1 × 2 × 10 = 5 N = fmax 4
Applying Newton’s second law to system of A + B F = (mA + mB) a = 6a .....................(1) Applying Newton’s second law to A
fmax 5 f = mA a amax = m = = 2.5 m/s2 .......................(2) 2 A 7.
from (1) and (2) Fmin = (mA + mB) 2.5 m/s2 = 6 × 2.5 = 15 N (i) The F.B.D. of A and B are
For A to be in equilibrium ; F = N sin .....................(1) For B to just lift off ; N cos = mg + µs N .....................(2) For horizontal equilibrium of B ; N = N sin ................(3) From (2) and (3) N (cos – µs sin ) = mg From equation (1)
RESONANCE
or
F=N×
3 5
4 2 3 N = mg 5 3 5
F=
or N =
5 mg 2
...............(4)
3 mg 2 SOLN_FRICTION - 31
(ii) The acceleration of the block A be a and B be b F – N sin = 2ma ...............(1) N cos – mg – µkN = mb ...............(2) N = N sin ................(3) From constraint = a sin = b cos ................(4) Solving (1), (2) , (3) and (4) we get 10.
b =
3g 22
Considering the forces on the chain for the given situation we have
F – k ( – x)g = a
F k ( x )g dv – = .v.. dx
0
F dx –
0
k ( x ) g dx =
v
dv v 0
x 2 v2 F x – g k x 2 2 0 0
v
0
F v2 – g k = 2 2 2F k g = v..
PART - II 3.
On smooth surface a1 = g sin v2 = u2 + 2a1s1 = 0 + 2 g sin .m On rough surface a2 = g sin – g cos v´2 = v2 + 2a2s2 O = 2mg sin + 2g (sin – µ cos )n
m n = tan n
6. If acceleration of the car is a0, acceleration of the block 2a0 = 2 × 2 = 4 m/s2 ()
F = N = 0.3 × 50 × 10 = 150 T – F = ma T – 150 = 50 × 4 T = 350 N.
RESONANCE
SOLN_FRICTION - 32
8.
FBD of A
N T
8m
N
a
C
a mg 8mg If the acceleration of ‘C’ is a For block ‘A’ N = 8 ma .... (1) 8 mg – N = 0 .... (2) and acceleration a can be written by the equation of system (A + B + C) m 1 g = (10 m + m 1) a .... (3)
m1g 8 mg = 8 m 10m m 1 10 m + m 1 = m 1 10 m = ( – 1) m 1
m1 =
10 m Ans. 1
10.
(i)
(ii) (iii)
11.
Let the blocks does not move then T1 = 20 – 4 T2 = T1 – 8 = 20 – 4 – 8 = 8 Since T2 < max possible friction force for 6 kg block hence it will be at rest and this assumption is right. Therefore tension in the string connecting 4kg and 6 kg block = 8N friction of 4 kg block = N = 0.2 × 4 × 10 = 8N friction force on 6 kg block = 8N
So block ‘Q’ is moving due to force while block ‘P’ due to friction. Friction direction on both +Q blocks as shown. P
fmax=8 fmax=9
Q
4
fmax=8
5
F
First block ‘Q’ will move and P will move with ‘Q’ so by FBD taking ‘P’ and ‘Q’ as system F–9=0 F=9N When applied force is 4 N then FBD Q
0
0 P
4
4
4 kg block is moving due to friction and maximum friction force is 8 N.
8 = 2 m/s 2 = amax. 4 Slipping will start at when Q has +ve acceleration equal to maximum acceleration of P i.e. 2 m/s 2. F – 17 = 5 × 2 F = 27 N. So acceleration =
RESONANCE
SOLN_FRICTION - 33
13.
Applying Newton’s law for the system of m and 3m along the length of the string we get 3mg sin45 – 3mg cos 45 – mg sin45 = (3m + m)a =
2 5
as
a=
g 5 2
now making the F.B.D. of m we get – T – mg sin 45 = m a
T= T=
mg 5 2
+
mg 2
6 mg 5 2
Now from F.B.D. of pulley we get
Force exerted by string on pulley
6mg 6mg = T 2 = × 2 = (downward) 5 5 2 14.
F.B.D. for A block
F.B.D. for B block
RESONANCE
SOLN_FRICTION - 34
for block A mg sin – f1 = ma .........(1) for motion w.r.t. block B mgsin – µmg cos = ma .........(2) for limiting case a=0 and a=b=0 mg sin = µmg cos µ = tan = tan–1 µ for block B mgsin + f1 – f2 = mb for motion w.r.t wedge f2 = 2µmg cos mgsin + f1 2µmg cos = mb ..........(3) for no relative motion between A and B block from equation (1) & (3) : a = b 2mg sin – 2 µmg cos = 2ma for limiting case a = 0 = tan–1 (µ) for motion tan–1 (µ) when block B is moving w.r.t wedge mgsin + f1 – 2µ mgcos = mb But f1 = µmg cos mg sin – µmg cos = mb for block A mg sin – µmgcos = ma a = b. 16. *
The free body diagram of the block is N is the normal reaction exerted by inclined plane on the block.
Applying Newton’s second law to the block along and normal to the incline. mg sin 45° = T cos 45° + N ............... (1) N = mg cos 45° + T sin 45° ............... (2) Solving we get = 1/2 so any value of greater than 0.5 is answer 18.*
Applying NLM on the part that moves through slit. T2 – f – T1 =0 For 4 kg mass 40 – T2 = 4a For 2 kg mass T1 – 20 = 2a On solving 10 = 6a
T1
T1
f m1
5 a= m/s2 3 70 Force exerted on 2kg mass by string = T1 = N. 3
T2
40
m1 20
T2
Tension in the string will not be same throughout, due to the friction force exerted by the slit.
RESONANCE
SOLN_FRICTION - 35
19.*
The breaking force is insufficient, so the block will not slide. So friction force = 100 N and acceleration will be 20 m/sec 2 only
( 200)2 (100)2 = 100 5 N All mechanical interactions are electromagnetic at microscopic level. Net contact force on the block = 20.* There are two possibilities (i) 100 kg block slides down the incline (ii) 100 kg block slides up the incline
case-(i)
we get, 100 g sin 37 – ×(100 g) cos 37 – mg = 0 case (ii)
100 3 4 – 0.3 × 100 × = 36 kg 5 5 mg = 100 g sin 37 + g cos 37 × 100
m=
100 3 100 4 + × 0.3 5 5 = 84 kg
m=
To remain in equilibrium, m [36, 84] kg therefore, m can be 37 and 83 kg. N 20 N
2
2
8 +6 21.*_
mg = 50 N N = 50 – 20 = 30 N Limiting friction force = µN = 12 N and applied force in horizontal direction is less than the limiting friction force, therefore the block will not slide. For equilibrium in horizontal direction, friction force must be equal to 10 N.
f 53° 6i + 8j From the top view, it is clear that = 37° i.e. 127° from the x-axis that is the direction of the friction force. It is opposite to the applied force. Contact force =
N2 f 2 = 10 10 N
RESONANCE
SOLN_FRICTION - 36
22.* F1 = mgsin + mg cos . F2 = mgsin – mg cos . But q mg = w = tan
23.*_
sin cos) cos
F1 = w (sin +
Now F1 = 2 F2 mg sin+ mg cos = 2 (mg sin – mg cos) sin+ cos = 2 sin – 2 cos 3cos = sin tan = 3tan.
F1 = w sin( + )sec
tan = 3
mgsin + mg cos = ma a = g sin + g cos 3 4 = 10 = 14 m/sec 2. 5 5
If f r = mg sin = mg ×
3 3mg = < f r max 5 5
f r < f r max =
v=
3mg 4mg < hence insect can 5 5
mg
move with constant velocity.
n co
n sta
t
fr
sin
EXERCISE-3 PART - I 1.
(i)
FBD in (case (i))
{1 = 0, 2 = 0.1}
2N
O
1 kg
N = 10
A
1 kg
N = 10
B mg
2N mg
While friction’s work is to oppose the relative motion and here if friction comes then relative motion will start and without friction there is no relative motion so both the block move together with same acceleration and friction will not come. A
B
mg
RESONANCE
aA = aB = 10 m/s2
mg
SOLN_FRICTION - 37
1
0 A 1 kg
(ii)
B 1 kg
10
10
1
10 0 10 Friction between wall and block A oppose relative motion since wall is stationary so friction wants to stop block A also and maximum friction will act between wall and block while there is no friction between block. Note : Friction between wall and block will oppose relative motion between wall and block only it will not do anything for two block motion. 1
A
B
10 aA = 9 m/s2 ; 1
10 aB = 10 m/s2 f
A (iii)
B
10
f
10 10 Friction between wall and block will be applied maximum equal to 1N but maximum friction available between block A and B is 10 N but if this will be there then relative motion will increase while friction is to oppose relative motion. So friction will come less than 10 so friction will be f that will be static. 1 f
A
B f
10
10
by system (20 – 1) = 2 × a a = 10
1
A
(iv)
19 = 9.5 m/s2 2
B
10 10
1
aA =
11 10 = 1 m/s2 1
aB =
10 1 = 9 m/s2 1
RESONANCE
10
SOLN_FRICTION - 38
2.
The acceleration of two block system for all cases is a = 2 m/s 2 In option (p) the net force on 2 kg block is frictional force Frictional force on 2 kg block is f = 2 × 2 = 4N towards right In option (q) the net force on 4 kg block is frictional force Frictional force on 4 kg block is f = 4 × 2 = 8N towards right In option (r) the net force on 2 kg block is 2 × 2 = 4N Friction force f on 2 kg block is towards left. 6–f=2×2 or f = 2N In option (s) the net force on 2 kg block is ma = 2 × 2 = 4N towards right. Friction force on 2 kg block is 12N towards right. (A) 4.2 m/s2
(B*) 3.2 m/s2
(C) 16/3 m/s2
(D) 2.0 m/s2
3. & 4. First, let us check upto what value of F, both blocks move together. Till friction becomes limiting, they will be moving together. Using the FBDs a1
F
F f
F F
f
15 kg F a2
10 kg block will not slip over the 15 kg block till acceleration of 15 kg block becomes maximum as it is created only by friction force exerted by 10 kg block on it a1 > a2(max) Ff f = for limiting condition as f maximum is 60 N. 10 15 F = 100 N. Therefore for F = 80 N, both will move together. Their combined acceleration, by applying NLM using both as system F = 25a
a= 5.
80 = 3.2 m/s2 25
If F = 120 N, then there will be slipping, so using FBDs of both (friction will be 60 N) For 10 kg block 120 – 60 = 10 a a = 6 m/s2 For 15 kg block 60 = 15a a = 4 m/s2
6. & 7. In case 80 N force is applied vertically, then 80 f
F f
F F
F
For 10 kg block 80 – 60 = 10a a = 2 m/s2 For 15 kg block in horizontal direction. F – f = 15a a = 4/3 m/s2, towards left.
RESONANCE
SOLN_FRICTION - 39
8.
F sin + f = mg and Fcos = N for minimum ; f = N = Fcos mg Fmin. = sin cos
9.
As f = 0
10.
If F < Fmin. ; block slides down due to mg
11.
Friction always opposes relative motion.
12.
Due to pseudo force, the person observes the block to move back. Also the accelerating person does not observe any relative motion between body and the rough surface. The minimum force required to pull the block of mass m lying on rough horizontal surface is mg F= = 60 N, inclined at an angle tan–1 with horizontal (where is the coefficient of friction). Hence 2 1 statement 1 is true and statement 2 is false.
13._
F sin = mg mg F= sin
14._
There is no tendency of relative motion between the blocks. Hence Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
15._
R = f 2 N2 = mg if body does not move. But if it moves then f < mg sin
16.
2
2
A f
f B
Fext
R=
f (mg cos )
< mg
(i)
Since the initial velocity of block is along positive x-axis. So the direction of frictional force will be in – ˆi at that moment – ˆi ... Ans.
(iii) The block begins to slide if F cos 37° = µ (mg – F sin 37°) 5t [cos 37° + µ sin 37°] = µ mg 4 3 5t = 70 5 5
or ;k
t = 10 second
EXERCISE-4 PART - I
1.
N = mg + F sin 60 =
3
× 10 +
F 3 2
...... (i)
F cos 60 = N ................. (ii)
1
F 2
=
F 2
=5+
2 3
RESONANCE
F 4
× (10
3 +
F 3 ) 2
F =5 4
F = 20 N SOLN_FRICTION - 40
2.
8
aA = g [sin 45 – A cos 45] =
2
, aB = g [sin 45 – B cos 45] =
aAB = aA – aB = g (B – A) cos 45 = Now
1 2
sAB =
Again sA =
1 2
aAB t2 aA t2 =
1 2
(
2
8
=
1 2
, sAB =
1 2
×
) 4
2
1 2
7 2
2
t2
t = 2 sec.
sA = 8 2 m
3.
Solving from the frame of disc
Let accleration of the block relative to the disc is a then 25 m cos – N = m a ..........(i) Now, there will be two normal as there are two contacts (i) Horizontal and (ii) vertical NH = 25 m×sin = 25 × m ×
3 = 15 m 5
NV = mg = 10 m f = NH + NV from (i) we get 4.
=
2 2 (15 m) + (10 m) 5 5
a=
= 10 m
m (25 cos – 10) = 10 m/sec2 m
Statement-1 is also practical experience based; so it is true. Statement-2 is also true but is not the correct explanation of statement-1. Correct explanation is ''there is increase in normal reaction when the object is pushed and there is decrease in normal reaction when object is pulled".
5. P1 = mgsin – mgcos P2 = mgsin + mgcos Initially block has tendency to slide down and as tan > , maximum friction mgcos will act in positive direction. When magnitude P is increased from P1 to P2, friction reverse its direction from positive to negative and becomes maximum i.e.mgcos in opposite direction.
RESONANCE
SOLN_FRICTION - 41
6.
mg F1 =
2
mg F2 =
mg 2
mg
2 2 F1 = 3F2 1 + = 3 – 3 4 = 2 1 2 N = 10 N = 5 Ans. =
7.
f = 0, If sin = cos f towards Q, sin > cos f towards P, sin < cos
RESONANCE
= 45° > 45° < 45°
SOLN_FRICTION - 42
PART - II 1.
Force, F = R = Mg weight of block = R = 0.2 ×10 = 2N
2.
F = ma
3.
Now ,
v = u + at or
or
0 .6 = a = – 0.6 10
a g
=
0 = 6 + 10a
a 0.6 = g 10 0.06
so
Let the mass of block be m. Frictional force in rest position F = mg sin 30º 10 = m × 10 ×
R
1 2
mg sin 30º
m=
s1 =
1 2 a1t1 2
........ (i)
a2 = g sin – g cos
s2 =
1 at2 2 22
........ (ii)
a1 = g sin
F
mg cos30º
mg 30º
2 10 = 2 kg 10 When fiction is absent
4.
mg = ma
When friction is present
From Eq. (i) and (ii)
1 2 1 2 a1t1 a 2 t 2 2 2
5.
or
a1t12 = a2 (nt1)2 ( t2 = nt1)
or
a 2 g sin g cos 1 2 or a1 g sin n
or
1 – k =
1 n2
or
or
a1 = n2a2
1 g sin 45 º g cos 45 º 2 g sin 45 º n k = 1 –
1 n2
According to work-energy theorem, W - K = 0 (Initial and final speed are zero) Work done by friction + work done by gravity = 0
+ mgl sin = 0 2 cos = 2 sin
– (mg cos)
= 2 tan
v2 100 100 100 100 = = 1000 m 2 k g 2 0.5 10 52
6.
s=
7.
F1 = mg sin + mg cos F2 = mg sin – mg cos
F1 sin cos F2 = sin cos 3 tan 2 = = = 3. tan 2
RESONANCE
SOLN_FRICTION - 43
TOPIC : GRAVITATION EXERCISE-1 PART - I SECTION (A) A 2.
mass of each sphere m = Volume × =
4 3 r 3
m.m (2r )2
F = G
2
4 G r 3 3 = 4r 2 = A 3.
tan =
4 G22r4 N. 9
Ans.
8 4 = 6 3 = 53°
F =
GmM r2
= G a =
0.260 0.01 (0.1) 2
2F cos m
0.260 3 (0.1)2 5 = 31.2 G m/s2 = 2G
SECTION (B) B 1.
Ex = –
v = – (20x + 40y) = – 20 x x
Ey = –
v = – y (20x + 40y) = – 40 x
E = Ex ˆi + Ey ˆj = – 20 ˆi – 40 ˆj
Ans.
It is independent of co ordinates Force = F = m E = 0.25 {– 20 ˆi – 40 ˆj } = – 5 ˆi – 10 ˆj magnitude of F = 5 2 10 5 = 5 5 N
RESONANCE
SOLN_GRAVITATION - 44
SECTION (C) C 1.
Potential energy at ground surface
– GMm R potential energy at a height of R is potential energy =
– GMm 2R When a body comes to ground Loss in potential energy = Gain in kinetic energy potential energy =
– GMm 1 – GMm = – mv2 R 2R 2
C 2.
GM 2 g R
gR = v2
Initial kinetic energy =
1 M V2 2 S
Initial potential energy
= –
GMm 1 = mv2 2R 2
v =
gR
GM A MS GMBMS 2GMS – = – (MA + MB) d/2 d/2 d
2GMS 1 M V2 – (MA + MB) d 2 S Finally, Potential energy = 0 Kinetic energy = 0 Limiting case Applying energy cnservation
Total initial energy =
2GMS 1 MSV2 – (MA + MB) = 0 d 2
V = 2
G(M A MB ) d
Ans.
SECTION (D) D 2.
T1 = 2
r3 (1.01r )3 , T2 = 2 GMe GMe
3/2 T2 1.01r T1 = r
T2 T1 – 1 = 0.005 × 3 D 3.
T2 3 3/2 T1 = [1 + 0.01] = 1 + 2 × 0.01
(T2 T1 ) × 100 = 0.015 × 100 = 1.5%. T1
GMm GM 2 = 2 (2R ) 4R 2
(a)
F =
(b)
Mv 2 R
=
v =
GM 4R
T =
GM 2 4R 2
2R = v
RESONANCE
2R GM 4R
= 4
R3 GM
SOLN_GRAVITATION - 45
(c)
Angular speed =
(d)
2 = T
2
=
R3 4 GM
GM 4R 3
Energy required to separate = – { total energy } = – { Kinetic energy + Potential energy }
1 2 GM2 GM2 1 2 2 = – 2 Mv 2 Mv – 2R = – Mv – 2R GM GM2 GM2 GM 2 = – M 4R – 2R = – – 4R = 4R
Ans.
(e) Let its velocity = ‘v’ Kinetic energy =
1 mv2 2
Potential at centre of mass = –
GM GM 2GM – = – R R R
Potential energy at centre of mass = –
2GMm R
For particle to reach infinity Kinetic energy + Potential energy = 0
1 2GMm mv2 × = 0 2 R v =
D 4.
(a)
4GM R
Ans.
GMmA rA UA m A rB = = GMm UB mB rA B rB rB = 19200 + 6400 = 25600 Km rA = 6400 + 6400 = 12800 km, mA = mB UA 25600 = =2 12800 UB
(b)
GMmA m A rB 2rA KA = = =2 GMmB mB rA KB 2rB
GMm , 2r Clearly farther the satellite from the earth, the greater is its total energy. Thus B is having more energy. (c)
As T.E. = –
RESONANCE
SOLN_GRAVITATION - 46
SECTION (E) : E 2.
g
Period of pendulum = 2
Let T1 be the time period at pole and T2 is time period at equator.
T1 T2
g2 g1
=
1
R 2 2 1 – e g
Re2 Re2 . Since << 1 2g g
T1 = 1 –
So ,
T1 1
R 2 g 1 – e g g
T1 = 1 –
2 1 Re 2 g
= 1–
2 6400 10 3 1 ( 2 ) × 9.8 2 (86400)2
= 0.998 second
Ans.
PART - II SECTION (A) A-2.
Net torque = F2 . – F1 . 2 2 = (F2 – F1)
2
2H2 F2 = mgH = mg 1 – 2 R 2H1 F1 = mgH = mg 1 – 1 R
= (F2 – F1)
A-4.
V=
Ans.
MV 2 R
2 Fg cos 30 =
GM2 2 2 L
mg (H1 – H2 ) = 2 R
MV 2 3 = L/ 3 2
GM L
SECTION (B) B 2.
dEnet = 2dE sin = 2
Gdm r2
= 2G . =
sin
rd r2
sin
2G sind r
RESONANCE
SOLN_GRAVITATION - 47
/2
Enet = =
dEnet =
0
2G sin d = 2G r r
m and r =
Enet =
2Gm
2 Along + y axis
B 4.
Ans.
For point ‘A’ : For any point outside, the shells acts as point situated at centre. So,
FA =
G (M1 M2 ) p2
m
For point ‘B’ : There will be no force by shell B. So, ,,
B-6.
FB =
GM1m q2
For point ‘C’ : There will be no gravitational field. So, FC = 0 Let the possible direction of gravitational field at point B be shown by 1, 2, 3 and 4(Figure 1). Rotate the figure upside down. It will be as shown in figure 2.
B
1
Figure 1 2
4 3
=
B
Figure 3
3 4
2 1
Figure 2
B
Now on placing upper half of figure 1 on the lower half of figure 2 we get complete sphere. Gravitational field at point B must be zero, which is only possible if the gravitational field is along direction 3. Hence gravitational field at all points on circular base of hemisphere is normal to plane of circular base. Circular base of hemisphere is an equipotential surface. Aliter : Consider a shaded circle which divides a uniformly thin spherical shell into two equal halves.The potential at points A,B and C lying on the shaded circle is same. The potential at all these points due to upper hemisphere is half that due to complete sphere.Hence potential at points A,B and Cis also same due to upper hemispehre
RESONANCE
SOLN_GRAVITATION - 48
SECTION (C) m
C 2.
(a)
F 120°
F
120° 120°
F m
m
Due to geometry net force is zero.
(b)
By geometry , x2 +
a2 4
= a2
x2 =
3a 2 4
x =
3a 2
F
x
a Gm2 x2
=
4 Gm2 3 a2
Gm 2 Gm 2 Gm 2 3 Gm 2 Initial potential energy = – = – 2 a a a a
Work done on system
= Final potential energy – intial potential energy = –
(d)
F2 F1
Fnet = F =
(c)
and F1 = F2
3Gm 2 3 Gm 2 3 Gm 2 – – = 2 a 2 a a
Ans.
Initial kinetic energy = 0 Initial potential energy = –
= –
Gm 2 Gm 2 – a a
2Gm 2 a
Total initial energy = – Now, kinetic energy =
Potential energy = –
2Gm 2 a
1 mv2 2 2Gm 2 Gm 2 – a/2 a/2
= –
4Gm 2 a
1 4Gm 2 mv2 – 2 a
Total energy =
1 2Gm 2 = mv2 2 a 4Gm a
v =
2
= v Gm a
RESONANCE
Ans.
SOLN_GRAVITATION - 49
SECTION : (D) D-1.
r2
v =
V
mv 2 = r
GMm
m r M
GM r2 3
3
1
2r 2
2 r 2 2r T = = = v GM
G
T
4 3 r 3
D-5*.
PE = –G m1 m2/r, ME = – G m1 m2 / 2r On decreasing the radius of orbit PE and ME decreases
D-6.
According to kepler's law applying angular momentum conservation m 1v1r1 = m 2v2r2 Vmax is (a) ans.
SECTION (E) E 1.
we = 50 × 10 = 500 N wp = 50 × 5 = 250 N Hence option A is correct
E 2*.
In case of earth the gravitational field is zero at infinity as well as the the centre and the potential is minimum at the centre .
EXERCISE-2 PART - I 2.
(a)
r < y < 2r Field due to outershell = 0 Distance from centre of solid spere = (y – r) Gravitation field intensity
y
GM = – = –
E = 0– (c)
GM r3
× distance from centre y
(y – r)
GM
GM (y – r)
2
ˆj
=
r
x
in y - direction
GM (y – r) ˆj = (y – r) (– ˆj ) r3 r3 Field due to outshell = 0 Distance from centre of solid spere = (y – r)
= –
(b)
(radius )3
GM (y – r)
2
y
y 4r
(– ˆj ) r
y > 8r For any point outside, the shells acts as point situated at centre. Distance from centre of hollow shell = (y – 4r) Field due to hollow shell = –
4GM ( y – 4r )
Distance from centre of solid spere = (y – r) Field due to solid spere = –
GM ( y – r )2
4GM GM ˆ (– j ) Total field = y – 4r ( y – r )2
RESONANCE
SOLN_GRAVITATION - 50
x
3.
(a)
Force will be due to the mass of the sphere upto the radius r In case (i) 0 < r < b ; Mass M = 0, therefore F(r) = 0 In case (ii) b < r < a ; Mass M =
(iii) a < r < ; Mass M =
r2
(b)
Uf – Ui = –
a3 b3 4 4 (a3 – b3), therefore F(r) = Gpm 2 3 3 r
Fc .dr
r1
(i) 0 < r < b ;
u(r) = - 2 Gm(a2 — b2)
(ii) b < r < a ; u(r) = (iii) a < r < ; u (r) = 5.
b3 4 4 (r3 – b3), therefore F(r) = Gpm r 2 3 3 r
2Gm (3ra2 - 2b3 - r3) 3r 4Gm 3 (a b 3 ) 3r
(a)
The gravitation field is uniform inside the cavity and is directed along OO´ . Hence the particle will strike at A.
(b)
The gravitational field at any point P inside cavity.
4 |E | = G 3
4 4 2 G = Gy OO´ = GR 3 3 3 Total workdone = m | E | . S = m . 2 GR . R 3 2 Applying work - energy theorem Workdone by all force = Change in kinetic energy
–
1 m . 2 GR . R = mv2 2 3 2
6.
v =
2GR 2 3
(a)
GMsm 1 mv2 = 2 R
(b)
1 mve2 – 2
Ans.
G m =0 2R
GmM s 1 m (V + Ve)2 = 2 R V =
2 – 1
or
V=
2G S R
or
Ve =
G R
or
V + Ve =
2GMs R
GMs R
RESONANCE
SOLN_GRAVITATION - 51
7.
Applying angular momentum conservation : mv0 = mvd v0= vd .......... (i)
1 mv02 + 0 2
Intial energy =
GMs 1 mv2 – d 2 Applying energy conservation ,
Final energy =
GMsm 1 1 mv02 = mv2 – d 2 2 2GM s .......... d From equation (i) and (ii) :
v02 = v2 –
v 022
v02 =
d2 +
d
2
2GMs v0
2
(ii)
2GM s d
–
d – 2 = 0
Solving this quadratic
d = –
GMs v0
2
2 2 GMs 1 v 0 – 1 = GM v 02
2
GMs v 2 0
+
2
Ans.
PART - II 1.
Gravitational field at ‘m’ due to hollowed - out lead sphere = { Field due to solid spere } – { Field due to mass that was removed } Field due to solid sphere =
GM d
Field due to removed mass =
2
= E1 =
GM' x2
3 M 4 R M’ = 4 × 3 2 R 3 3
And
E2 =
4R 2
= E2
=
M 8
R 2
x = d–
So ,
GM
GM R 8 d – 2
=
2
GM = 2 18 R 2 3 R 8 2
GM
Enet = E1 – E2 =
GM 1 1 – R 2 4 18
Fnet = mEnet
RESONANCE
=
=
7GMm 36 R 2
7GM 36 R 2
Ans.
SOLN_GRAVITATION - 52
4.
5.
Gm1 3 r1 = 4 Gm 2 r2
m1 + m2 = m
4R
V = KVe = K
Ve =
m1
m
2GM R
m2
=
4 r12
4 r22
m1
=
2
Initial total energy =
1 2GMm mv2 – 2 R
Final total energy =
1 2GMm m02 – 2 x
or
4 r12
=
5 Gm = Ans. 3 Gm1 R r1
2GM R
1 2GM 2GMm m.K2 – 2 R R
=
Applying energy conservation
1 2GM 2GMm mx2. – = 2 R R 1 1 x2 = – x R R 9.
Fg =
x =
0–
2GMm x
R 1– k2
Ans.
GMmr R3
pressing force
= Fg cos =
GMm
= 2 R2
GMmr cos R3
= constant
a=
Fg sin GMr sin = m R3
a=
GMy R3
10.*
In elliptical orbit sun is at one of the foci hence the distance between the planet and sun changes as planet revolves hence linear speed, kinetic energy and potential energy of planet donot remain constant
11.*
S =
2 , 1 .5
E =
2 24
1 1 – west to east = 2 1.5 24
T west to east =
2
= 1.6 hours
west to east
Similarly 1 1 east to west = 2 1 . 5 24
T east to west =
24 hours 17
RESONANCE
SOLN_GRAVITATION - 53
EXERCISE-3 1.
P.E. = –
GMm r
Total energy =
T.E. = 0
K.E. =
GMm 1 + mV2 r 2
if
GMm 1 mV 2 0 v = r 2
2GM T.E. is – ve r
For v <
1 mV2 2
2GM r
2GM , T.E. is + ve r
for v >
GM i.e. equal to orbital velocity, path is circular.. r If T.E. is negative, path is elliptical. If T.E. is zero, path is parabolic. If T.E. is positive, path is hyperbolic. If V is
2.
5.
(A) (B) (C) (D)
At centre of thin spherical shell V 0, E = 0. At centre of solid sphere V 0 , E = 0. At centre of spherical cavity inside solid sphere V 0, E 0. At centre of two point masses V 0, E=0.
4 2 3 T = GM R y = mx + c
GM R= 2 4
2
(3) Slope
=m=
log R =
2 1 log T + log 3 3
GM 2 4
20 10 11 M 3 = 18 log 4 10
M = 6 × 1029 Kg T 2 R3
RA RB
3
R = A RB
2
t=
2
= B A
rel = 80 – 0 = 70
6.
T2 / 3
2 3
GM 1 intercept c = log 2 = 6 3 4
(4) (5)
1/ 3
rel = (rel) t
R 4R
3
2
= B A
B A
1 = 8
2 = (T0) t
2 T 0
Let M and R be the mass and radius of the earth respectively. If m be the mass of satellite, then escape velocity from earth e =
( 2 Rg)
Velocity of satellite s =
e = 2
Further ]s =
GM = r
(2 R g) / 2
......... (1)
R 2g R h
R 2g Rh h = R = 6400 km
2s =
RESONANCE
SOLN_GRAVITATION - 54
7.
4 2 3 x Gm Hence time period of revolution T is T2 =
8.
T = 2
x3 Gm
T = 2
8R g
(Put x = 2R)
Now total energy at height h = total energy at earth's surface (from principle of conservation of energy)
m 1 m = m2 – GM Rh 2 R
0–GM
or
1 GM m GMm m2 = – 2 R 2R
v=
gR
( h = R)
9 to 11 Let the angular speed of revolution of both stars be about the common centre , that is, centre of mass of system. The centripetal force on star of mass m is m2
2d Gm(2m) 4 2 3 d = . Solving we get T= 3 d2 3Gm
The ratio of angular momentum is simply the ratio of moment of inertia about center of mass of system. 2
Lm LM
2d m Im 3 2 2 I M d 2m 3
Similarly, The ratio of kinetic energy is simply the ratio of moment of inertia about center of mass of system. 2
+
Km KM
2d 1 I m 2 m 3 2 2 2 1 I M 2 2m d 2 3
12.
Till the particle reaches the centre of planet, force on both bodies are in direction of their respective velocities, hence kinetic energies of both keep on increasing . After the particle crosses the centre of planet, forces on both are retarding in nature. Hence as the particle passes through the centre of the planet, sum of kinetic energies of both the bodies is maximum. Therefore statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
13._
It is minimum –ve (iii) Energy density =
15. 14.
for closed paths (circular or elliptical) the total mechanical energy is negative. (i) g’ = g – R cos 2 At equator = 0 g’ = g – R 0 = g – R
RESONANCE
B2 and B increases by a large factor.. 2 0 r
SOLN_GRAVITATION - 55
=
(ii)
g = R
9.8
= 1.24 × 10 –3 rad/s
6400 103
dA L = = constant because angular momentum of planet (L) about the centre of sun is constant. dt 2m Thus, this law comes from law of conservation of angular momentum.
(iii)
T r 3 / 2
T2 r2 T1 = r1
3/2
r2 T2 = r1
or
3/2
3.5 R T1 = 7R
3/2
(24) h = 8.48 h
EXERCISE-4 PART - I 1.
Time period of a satellite very close to earth’s surface is 84.6 minutes. Time period increases as the distance of the satellite from the surface of earth increase. So, time period of spy satellite orbiting a few hundred km, above the earth’s surface should be slightly greater than 84.6 minutes. Therefore, the most appropriate option is (C) or 2 hrs.
2.
(A) Gravitational field is a conservative force field. In a conservative force field work done is path independent. W I = W II = W III
3.
speed of particle at A VA = escape velocity on the surface of moon =
2GM R
At highest point B, VB = 0 From energy conservation. UB U A 1 – mVA2 = VB – VA = m m m 2
or
VA2 UB UA UA – GM – , also [3R2 – r2] 2 m m m 2R 3
2 GM – GM – GM R 2 1 . 5 R – 0 . 5 R – – R R h R3 100 2
or or
1 3 1 99 1 –1 – – R R h 2R 2 100 R h = 99.5 R 99R Ans rA C A
4.
mA
rB
com
Gm A m B
(rA rB ) 2 m A rA
TA2
B mB
4 2 = mArA
TA2
4 2 = m B rB
TB2
mB rB TB2
As C is com mArA = mBrB hence TA = TB 5.
(A) It is similar equation as v = ù a 2 x 2 in SHM. (B) Particle on positive x-axis move towards origin with speed decreasing as x decreasing. (C) It is spring mass system performing SHM. (D) Object moves away from Earth so its speed will decrease, since its speed is greater than escape velocity so it will never return back.
RESONANCE
SOLN_GRAVITATION - 56
6.
= 0 =0 Case I r < R FC = mg
r
R
mV 2 r
r mV 2 (g = acceleration due to gravity at surface of sphere) R r g r R
V= Case II
for r < R
r>R GMm
mV 2 r
=
r2
GM r
V=
g R r
So
7.
If only gravitational force acts on astronaut (that is in state of free fall), he shall feel weightless. Thus statement-2 is correct explanation of statement-1.
8.
W ext = U – UP W ext = 0 –
– –
Gdm .1 x
M W ext = G 2GM
=
7R 2
W ext = W ext =
9.
7R
P
2rdr 2
16R 2 r 2
=
x
rdr
2GM
7R 2
16R 2 r 2 3R
2GM zdz = [Z] z 7R 2
4R
r
dr
4R 2GM 2 2 16 R r 3R 7R 2
2GM 7R
2
4
2R – 5R
W ext =
2GM 7R 2
c.m.
A 2.2 Ms
5d 6
4
2 –5 .
B d 6
11 Ms
5d 5d dd ( 2.2Ms ) (11 Ms ) Total angular momentum about c.m. 6 6 66 = 6. Angular momentum of B about c.m. = dd (11 Ms ) 66
10.
g=
GM R2
4 (G) R 3 3 = 2 R
;
g R
g' ' R' 2 R' 6 g = R = 3 R = 11 Given,
Ve R
RESONANCE
R' 3 6 = R 22
Ve =
GM = R
4 (G) ()) R 3 3 R
; Ve = 3 km/hr. SOLN_GRAVITATION - 57
11.
12.
Ve =
2v 0
2GM = R
Ves =
KE =
1 1 mv 2e = m 2 v 0 2 2
4 2.G. R 3 3 = R
2
= mv02
4G R 3
Ves R Sarface area of P = A = 4RP2 Surface area of Q = 4A = 4 RQ2 mass R is MR = MP + MQ
RQ = 2Rp
4 4 4 3 3 3 R R = R P + RQ RR3 = RP3 + RQ3 3 3 3 = 9RP3 RR = 91/3 RP RR > RQ > RP
VR 1/3 and VP = 9
Therefore VR > VQ > VP
VP 1 VQ = 2
PART - II 1.
Electric charge on the moon = electric charge on the earth
2.
V=
3.
2GMe 10 2GMe = 10 = 110 k m/s Re Re 10 Acceleration due to gravity at leight h from earth surface.
2GM = R
g
g' = 1 h
=
2
(r x )
2
r – x = 2x x=
5.
Gm 2 (2R )2
Gm 2 4R 3
6.
1 2 = x rx
3x =
r 3
r 3
Gm G( 4m) r /3 2r / 3
=
g g 2 9 h 1 h = 2R R
G( 4m)
Gm x
R
4.
2
3Gm 6Gm 9Gm = r r r
Ans.
= m2R
= 2 Gm 4R
3
v = R
v=
Gm 4R
3
×R =
Gm 4R
GMm GMm W = 0 – R R
m = mgR = 1000 × 10 × 6400 × 103 R = 64 × 109 J = 6.4 × 1010 = gR2 ×
RESONANCE
SOLN_GRAVITATION - 58
TOPIC : WORK, POWER AND ENERGY EXERCISE-1 PART - I SECTION (A) A 1.
f = mg = F Displacement = vt (a) W mg = mg × vt cos90º = 0 (b) W N = N × vt cos90º = 0 (c) W f = –mgvt (d) W F = Fvt = mgvt.
A 6.
m = 500 g =
1 kg 2
mg sin = fk 4 1 Wfk = (mg sin ) (2) = (10) ×2=8J 5 2
A 7.
W 1 = (mg sin )4 = (20 × 10 ×
4 ) (4) = 640 J 5
WF2 + WGrav = K = 0
WF2 – (mg sin ) (4) = 0
W 2 = WF = 4 mg sin = 640 J 2
SECTION (B) B 2.
B 4.
W = Area under given graph from x = 0 to x = 35m =
1 1 × (20 + 40) × 10 – ×5×5 2 2
=
575 J. 2
F at any moment = mg
( x ) mg(
=
W=
0
x)
dx
mg . 2
SECTION (C) C 4.
Work done by resistive force = W R = K = =
1 × 20 × 10–3 (1002 – 8002) 2 – 6300 J
So, average resistive force =
RESONANCE
6300 J = 6300 N. 1m
SOLN_Work, Power & Energy - 59
C 6.
Work done by the force = F ds = F ×
1 2 at 2
1 F × × t2 2 m
=
F×
=
20 2 10 2 F2 t 2 = = 4000 J 25 2m
Now
K = =
1 m(v2 – u2) 2
1 m (2as) 2
=m×
F 1 F × × × t2 m 2 m
F2 t 2 = 4000 J 2m W F = K. =
C-10.
U = –K
1 2 kx – 2mgx = 0 2 C-11.
x=
4mg . k
(a) Since, gravitational force is conservative, So, work done by it in round trip is zero. 5 1 = 10 2 = 30º W F = mg(sin + cos) ×
(b) sin =
1 3 = 0.3 × 9.8 2 0.15 2 10 = 18.519 J (c) W f = –f.s = –mg cos (2) = – 2mg cos = – 2 × 0.15 × 0.3 × 9.8 × 10 ×
3 = –7.638 J. 2
(d) By W.E.T, Kf – Ki = W F + W f + W g Kf = (18.519 – 7.638)J = 10.880 J. C 12.
Displacement of 4kg block = 2 × 2m = 4m 4kg = 2 × 2m = 4m Final speed of 4kg block = 2 × 0.3 = 0.6 m/s 4kg = 2 × 0.3 = 0.6 m/s W f + W g = K
1 1 × 4 × (0.6)2 + × 2 × (0.3)2 2 2
– × 4 × 10 × 4 + 2 × 10 × 2 =
160 = 40 – (0.72 + 0.09) =
RESONANCE
40 – 0.81 = 0.2449 160
SOLN_Work, Power & Energy - 60
C-14.
(i) w.r.t. person in the train
Ft m (ii) w.r.t. person on ground, v1 = at =
Ft m (iii) According to person in the train, v = vc + v1 = vc +
1 F2 t 2 mv12 = 2 2m (iv) According to person on ground, K1 =
K =
1 m 2
2
Ft 1 2 v c m 2 mv c .
1 2 Ft 2 at = . 2 1 2m (vi) According to person on ground, (v) S1 =
1 F 2 Ft 2 t = + vct. 2 m 2m (vii) According to person in the train work done by F = Fs1 S = vct +
F2 t 2 2m According to person on ground, Work done by F = F.s =
Ft 2 F = 2m v c t . (viii) Comparing W g = Kg and W c = Kc . (ix) Work–energy theorem holds in moving frame also.
SECTION (D) D 5.
Let
m1 = 2m2
m2 (m1 m 2 ) a = (m m ) g = 3m g = g/3 1 2 2
So, distance travelled by each block =
Also
2m1m 2 g 4m 2 g T = m m = = 16 3 1 2
12 m2 = g
1 2 at = g/6 2
Hence, loss in gravitation P.E. during first second = (m1 – m2)gh =
(2m2 – m2) g ×
RESONANCE
12 g g = g g 6 = 2g. 6
SOLN_Work, Power & Energy - 61
D 6.
a=
32 g g= 32 5
Distance covered in fourth second =
(2n 1)a 2
( 2 4 1) g 25 Hence, work done by gravity = (m2 – m1)gh
=
= (3 – 2)g × = D 7.
=
7g 10
7g 10
7 2 g . 10
W s + W g + W f = K
1 2 kx + mgx sin37º – mg cos37º × x = 0 2
3 4 1 × 100 × (0.1)2 + 1 × 10 × 0.1 × – × 1 × 10 × × 0.1 5 5 2
=
1 . 8
SECTION (E) E 4.
Power developed by motor =
E 6.
Power P =
E 7.
P=
E 8.
mgh 400 10 120 = = 1600 W.. t 5 60
mgh t
Pt 2 10 3 60 m = gh = kg = 1200 kg. 10 10 mgh t
t=
200 10 40 mgh = sec. = 8 second. 10 1000 P
20 kg / minute = 20 kg / 60 sec =
1 kg/s 3
1 10 20 P = g (20) = watt 3 3 746 W = 1 H.P
P=
100 HP 1119
RESONANCE
SOLN_Work, Power & Energy - 62
SECTION (F) F 1.
(a) w = F.ds = ( x 2 y 2 ˆi x 2 y 2 ˆj ).(dx ˆi dyˆj ) = ( x 2 y 2 dx x 2 y 2dy ) which is not a perfect integral and hence cannot be integrated without knowing y = f(x) or x = f(y). So, work done by F depends on path. So, it is non–conservative force. (b) While moving along AB, y = 0 and along BC, x = a. a
a
0
0
2 2 2 2 W ABC = x y dx x y dy
a3 a5 = 3 3 While moving along AD, x = 0 and along DC, y = a = 0 + a2 ×
So
a
a
0
0
2 2 2 2 W ADC = x y dx x y dy
= 0 + a2 .
a3 a5 = 3 5
Along AC x=y So
a
a
0
0
a
a
0
0
2 2 2 2 W AC = x y dx x y dy
2 2 2 2 = x x dx y y dy =
F 2.
2a 5 . 5
dU (a) F(y) = dy = dU (b) F(y) = dy = – 3ay2 + 2by dU (c) F(y) = dy = –U0 cos y..
F 5.
At x = 0, total energy is in form of K.E. since U = 0 and it turns back when its K.E. = 0 So, total energy is in form of P.E. U = –K
1 2 kx = 1 2 x2 = 1 × 2 × 2 x = ± 2m
Ans.
PART - II SECTION (A) A 3.
W = (force) (displacement ) = (force) (zero ) = 0
A 6.
W = (2000 sin 15º) × 10 = 5176.8 J
RESONANCE
SOLN_Work, Power & Energy - 63
A 9.
S1 =
1 1 1 g 12 , s2 = g 22 , S3 = g 32 2 2 2
S2 – S1 =
1 1 g 3, S3 –S2 = g5 2 2
W 1 = (mg) S1, W 2 = (mg) (S2 – S1) , W 3 = (mg) (S3 – S2) W1 : W2 : W3 = 1 : 3 : 5 A 10.
T = mg + ma, S =
1 at2 2
WT = T × S =
m(g a)at 2 2
A 11.* W = K, 0 = K, k remains constant, speed remains constant. A 14.* W = K > 0 K ( = kinetic energy) increases p=
2mk , p as k.
A 15.* W f + W G + W N = K = 0 As W G = 0, W N = 0 so W f = 0.
SECTION (B) x1
B 1.
W=
cx
dx = c
x12
o
B 2.
2
F
F2
1
F = K1x1 , x1 = K , W 1 = K x2 = 2 1 1 2 K1 1 similarly
F2 2K2
W2 =
since K1 > K2 , W 1 < W 2
SECTION (C) C 2.
C 5.
a=
h=
1 F 2 t , 2 m
F , m
S=
1 gt2, 2
W = mgh = mg
1 mg2 t 2 = Kf – mu2, 2 2
Kf =
Ft2
W F = FS = F 2 m
gt 2 , W = Kf – Ki 2
1 mg2 t 2 mu2 + 2 2
Hence Ans. is (A) v
C-10.
x
V2 Kx 2 dV V = – Kx, 2 – 2 dx u 0
V2 – u2 = – Kx2 1 1 1 mu2 – mV2 = mK x2 2 2 2
Loss x2
RESONANCE
SOLN_Work, Power & Energy - 64
x
C 12.
(mg sin ) x –
mg cos
dx = 0
0
x
x dx
sin x = o cos
x tan = 0
0
x2 , 2
x=
2 tan 0
SECTION (D) D 3.
Ui + 0 = Uf + Ui – Uf =
1 mv2 2
1 mv2 2
U= m= D 4.
1 mv2 2
2U v2
1 mu2 = mgh, u2 = 2gh 2
....(i)
3h
mg 4 + K.E. = mgh K.E. =
mgh 4
mgh / 4 K.E. 1 = 3mgh / 4 = P.E. 3
D 6.
W F + W S = 0, W F – U = 0 , W F = U = E E=
1 1 K x 2 , FxA = K x2 2 A A 2 A A 2E 2F2 , K = A KA E
2F 2F K A = xA , K A =
similarly
KB =
2F2 , EB
...(i)
KA = 2KB
2F2 2F2 = 2 E E B
EB = 2E Alter : F = KA xA = KBxB EA =
1 K x2 2 A A
EB =
1 K x2 2 B B
EA K A x A EB K B x B
2
2
EA 1 1 2 EB 2 2
RESONANCE
SOLN_Work, Power & Energy - 65
D 7.
D 13.
100 =
1 1 K(2cm)2 , E = K(4cm)2 2 2
so
E =4, 100
E – 100 = 300 J
E = 400 J
1 11 ( 2m)u2 mv 2 2 22
.... (i)
1 1 (2m) (u + 1)2 = mv2 ....(ii) 2 2
From (i) and (ii) D 14.
1 2 1
W 1 = work done by spring on first mass W 2 = work done by spring on second mass W 1 = W 2 = W (say) W 1 + W 2 = Ui – Uf 2W = 0 – W= –
D 15.
u=
1 Kx2 2
Kx 2 4
W a + W c = K = 0, Wa =
W a – mg 2 – 2 cos 60 º = 0
1 mg 5 = (0.5) (10) 4 = J. 4 4
SECTION (E) E 3.
V = 0 + at, F – mg = ma , F = mg + ma, P = (mg + ma) at
E 5.
P = TV = 4500 × 2 = 9000 W = 9KW
E 6.
P1 = 80 gh/15 , P2 = 80 gh/20 P1 20 4 P2 = 15 = 3
SECTION (F) F-2.
dU dU ve, ve dx x A dx x B
So, F-5.
F-6.
FA = positive, FB = negative
WC = WC + WC = 5 + 2 = 7 PR PQ QR U = cos (x + y), x
U y = cos (x + y) F = – cos (x + y) ˆi – cos (x + y) ˆj
= – cos (0 +
) ˆi – cos (0 + ) ˆj 4 4
|F | = 1
RESONANCE
SOLN_Work, Power & Energy - 66
EXERCISE-2 PART - I F a = (m m ) 1 2
1.
F f1 = m1a = m1 (m m ) 1 2
2F – f1 – f2 = m2a – f2 = – 2F + f1 + m2a = m1a + m2a – 2F
F – f2 = (m2 + m1) (m m ) – 2F = F – 2F = – F 1 2
f2 = – F
2F – K (m2 + m1)g = (m2 + m1)a
F F 2F – (m2 + m1) (m m ) = µK (m2 + m1)g (m m )g = K 1 2 1 2 W = work done by friction force on smaller block
m1F x (m 2 m1 )
= f 1x = 2.
mg = N + F sin .......(1) N = F cos .......(2) mg = F cos + F sin F=
mg cos sin
.......(3)
ì mg cos è (10) 40000 = cos è ì sin è 5 tan
WF =
Ans.
1 ì2
F is min. if D = cos + sin is maximum and its maximum value is
Fmin. =
mg
1 2
1
WFmin =
mg
mg
=
2
1 2
10
1 ì 2 1 2 = 0.2, mg = 4000 Nt (0.2)( 4000 )10 WFmin =
4.
fK =
x 4
W=
=–
400 20 ( 1 0.04 )
2
=
8000 = 7692.307 J 1.04
Ans.
m ( – x)g
W=
=
( 1 ( 0 . 2) 2 ) 2
–ì
m( – x )g dx
2 ìmg [( x ) ] 4 × 2
ìmg 9 2 × 2 16
=–
RESONANCE
9mg 32 SOLN_Work, Power & Energy - 67
7.
(a)
Taking F = 40 N, m = 4 kg , = 53º ax = (F cos – mg sin )/m = (40 cos – 40 × ay =
4 )/4 = 10 cos – 8 5
F sin á 40 sin á = = 10 sin m 4
x=0×2+
1 (10 cos – 8) (2)2 = 20 cos – 16 2
1 (10 sin ) (2)2 = 20 sin 2 WF = (F cos ) x + (F sin )y WF = (40 cos ) (20 cos – 16) + (40 sin ) 20 sin = 800 cos2 – 640 cos + 800 sin2 WF = 800 – 640 cos WF 800 – 640 WF 160 J If W F = 160 J then 160 = 800 – 640 cos cos = 1 y = 0 and x = 20 – 16 = 4 y=
(b)
W G = (–mg sin ) (4) = (–4 × 10 × (c)
4 )4 5
= –128 J F acts along the x-axis. W G + W F = K –128 + 160 = K Kf = 32 J.
9.
Work energy theorem (Between A & C) W f + W G + W sp = K mg cos (5 + 3) + mg 2 sin = 0
2 3 tan 37o = 8 16 work energy theorem W sp + W G + W f = K
=
(bet. A & B)
mg 5 sin 37o – mg 5 cos –
1 K (0.4)2 = 0 2
16 1 4 3 3 (4 × 10) 5 (5) = × K 100 2 5 16 5
K = 9000/ 8 N/m so 10.
x=9
Work energy Theorem on “m” W G + N + W T + W f =K 2
– mg R + O + W T –
(mg sin ) R d = 0 0
W T = mgR ( + 1)
RESONANCE
SOLN_Work, Power & Energy - 68
11.
13.
W F + W Sp + W fric = K
1 Kx2 – m1g x = 0 2
Fx –
F–
F = m1g +
& Kx = m2g
1 m2g – m1g = 0 2 m 2 g 2
mg = kx K=
100 mg = = 500 N/m 0 .2 x
1 1 K (0.2)2 + mv2 = m × 10 × 0.2 2 2 1 1 × 500 × 4 × 10–2 + × 10 v2 = 10 × 10 × 0.2 2 2 10 + 5v2 = 20 v2 = 2 v=
2 m/s Since u is 4 m/s ( ) so block will compress the spring. Let x be the compression of spring.
1 1 1 1 mu 2 + K (0.2)2 + 0 = m(0)2 + Kx2 + mg (x + 0.2) 2 2 2 2 4 1 1 1 = × 10 (4)2 + × 500 × × 500 (x)2 + 10 × 10 (x + 0.2) 100 2 2 2
80 + 10 = 250x2 + 100 x + 20 25 x2 + 10 x – 7 = 0 solving this x = 0.36 m So from initial position distance is ( 0.2 + 0.36) m = 56 cm 15.
(i)
mg = T cos mg =
x 2mg 2 a x 2 a a × 2 2 a a x
x=
a 2
(ii)
1 K 2
2a
2
+ mga =
1 1 K (2a – a)2 + mv2 2 2
1 2mg 1 (2a2 – a2) + mga = mv2 a 2 2
4ga = v
RESONANCE
SOLN_Work, Power & Energy - 69
1 K 2
(iii)
2
2a
+ mg a =
2 1 2 2 K a y – mg y 2
1 1 K2a2 + mg a = K(a2 + y2) – mgy 2 2 1 2mg 1 2mg 2 1 2mg 2 2a2 + mg a = a + y – mg y a a a 2 2 2 3 mg a – mg a =
mg y 2 – mg y a
mg y 2 – mg y a 2a2 = y2 – ay y2 – ay – 2a2 = 0 y2 + ay – 2ay – 2a2 y (a + y) = 2a (y + a) 2 mg a =
16.
y = 2a
P = Fext . V Where V is the vel. of point of application Fext + m, g = T & m2g =T Fext = m2g – m,g = (m2–m1) g (a)
P = (m2–m1) g v Ans.
(b)
Fext + m,g – T = m,a T–m2g = m2a _____________________________ Fext = (m1+ m2) a (m2–m1)g = m2(g+a) – m1(g – a) P = (Fext) (0 + at) = {m2(g+a) – m1 (g – a)} at
Ans.
19.
– g [ma + m´a + M
a 1 ] = – g [ m2a + m´0 + Ma] + (M + m + m´)v2 2 2
2g[2ma ma Ma
Ma m´a] 2 =v
M m m' v=
ag
M 2(m – m´) M m m´
RESONANCE
Ans.
SOLN_Work, Power & Energy - 70
21.
U (x) = 20 + (x – 2)2 du = 2(x – 2) dx – F = 2(x – 2) F = – 2(x – 2) m (x – 2) = – 2 (x – 2) Let x = x – 2 mx = – 2 x 1x =–2x x=–2x Simple Harmonic Motion Mean position is x = x – 2 = 0 x = 2 W2 = 2 ,
Kinetic energy = =
1 mv2 2
1 (1) (2) (A2 – x2) = x – 2, x = 5 – 2 = 3 2 1 (1) (2) {A2 – 32} 2
20 =
20 = A2 – 9
A2 = 29
A = 29
Aliter : for mean position dU = – 2(x – 2) = 0 dx At x = 5 K.E. = 20 J U(x = 5) = 20 + (5 – 2)2 = 29 J Total energy, T.E. = 20 + 29 = 49 J At amplitude U(x)max = 49 J = 20 + (x – 2)2 29 J = (x– 2)2
F=–
x=2±
29
x=2+
29 , 2 –
xmin = 2 –
x=2
29
29 = –3.38
xmax = 2 +
29 = 7.38 K.E.max when U(x) is minimum at x = 2 U(x)min = 20 J KEmax = 29 J 22.
Using work energy thoerem,
3R 1 4mg R W f + mg 2 2 R 2 Wf =
2
=
1 m 2
2
3gR
1 mgR 2
f = N (as kinetic friction) Wf =
Fs cos d x Fs = k 2 R (sec 1) Wf =
f dx =
k 2 R (sec
RESONANCE
( x = 2 R tan
; dx = 2 R sec2 d)
1) cos 2 R sec2 d SOLN_Work, Power & Energy - 71
0
Wf = 4 R
2
k
(sec 2 sec ) d
0 = tan 1 (3/4)
0
= 4 R2
0
k tan n (sec tan ) = – 4 R2 k [ tan 0 ln (sec 0 + tan 0)] = 0
4 R2
3
= 23.
5
3
= R2
k n 4 4 4 mg 2 R k (3 4 n 2)
=
k [3 4 ln 2] =
mgR 2
mg R 2
1 Ans. 8 (3 4 n 2)
Velocity will be maximum when a = 0 For a = 0, F = 0 This situation occurs for ve following arrangement of springs. Natural length is c = 150 mm Now , Ui + Ki = Uf + Kf
1 1 K{ 5 c – c}2 + K{ 2 c – c}2 2 2 Ki = 0 Ui =
Uf = 2.
150
150
1 K{ 2 c – c}2 2
1 1 K{ 5 c – c}2 + K{ 2 c – c}2 2 2
1 1 mv2 + 2. K{ 2 c – c}2 2 2 Solving the equation & putting the values we have =
15 ( 5 1)2 ( 2 1)2 v = 2
1/ 2
m/s = 3.189 ms–1 .
PART - II 3.
W agent + W G = K = 0 W agent = – W G, But W G is independent of the path joining initial and final position. W G is independent of time taken.
5.
W f + W G = K –mgd – mgh = 0 – gd + gh =
1 m v02 2
1 (v02) 2
(0.6) (10) d + 10(1.1) = 18 7.
d=
7 = 1.1666 1.17 6
W S + W f = K – U + W f = – Ki – Uf – mgx = – Ki 1 1 K x 2 + mgx = mu2 2 2
100 x 2 + 2(0.1) (50) (10) x = 50 × 4 x2 + x – 2 = 0 x=1m
RESONANCE
SOLN_Work, Power & Energy - 72
8.
v= s
ds
s
0
s
ds s , dt
2 s = t
t
dt
0
....(1) s = t/2 W = workdone by all the forces = K = 10.
2 t 2 1 1 1 mv2 = m 2s = m 2 2 2 2 4
K.E. + P.E. = constant
fu;r = C (say)
K – mg (tu sin –
1 gt2) = C 2
K = mg [tu sin –
1 gt2] + C 2
[= parabolic]
C 0 so answer is (B) 12.
dU = positive constant dx
For x < a, F = negative constant and for x > a, F = 0 so, ans. (C) 14.
E=
1 p2 , ( E) P = 2m
1 2m
= constant
Rectangular hyperbola (C) 17.
System is block & string. Applying work energy theorem on system (200)10 – 10g(R – R cos60º) =
1 (10)v2 2
2(200 – 10 × 5) = v2 v= 19.
300 = 10 3 . dW = F . ds where ds = dx ˆi + dyˆj and F = – K ( y ˆi + xˆj ) dW = – K ( ydx + xdy = – K d (xy) ( a, a )
W=
20.
( 0, 0 )
( a, a )
dW = – K
( 0, 0 )
(a, a) d ( xy ) = – K [xy] (0, 0)
W = – Ka 2 From given graphs : 3 3 3 t and ay = t 1 vx = t2 + C 8 4 4 At t = 0 : vx = – 3 C=–3
ax =
vx =
3 2 dx = t 3 dt 8
3 2 t –3 8
3 2 dy = t t 4 dt 8
Similarly;
.... (1)
.... (2)
As dw = F. ds = F.( dx ˆi dy ˆj ) W
0
4
dw
3
3
3
4 t ˆi 4 t 1 ˆj . 8 t 0
RESONANCE
2
3 3 ˆi t 2 t 4 ˆj dt 8 SOLN_Work, Power & Energy - 73
W = 10 J Alternate Solution : Area of the graph ;
a
x
and
dt = 6
a
y
= V( x ) f ( 3) V(x)f = 3.
dt = –10 = V( y ) f ( 4) V(y)f = – 6. Thus, u = 5 m/s and v =
45 m/s.
Now work done = KE = 10 J 22.*
W G = K, so
23.*
mgh =
1 1 1 1 mv2 – mu2, mu2 + mgh = mv2 2 2 2 2
v > u and v depends upon u.
dW F = F . ds , if F perpendicular to ds then
ds dW F = 0, ds is displacement of point of application of force, v = . dt (A), (C), (D) are true.
EXERCISE-3 1.
The displacement of A shall be less than displacement L of block B. Hence work done by friction on block A is positive and its magnitude is less than mgL. And the work done by friction on block B is negative and its magnitude is equal to mgL. Therefore workdone by friction on block A plus on block B is negative its magnitude is less than mgL. Work done by F is positive. Since F>mg, magnitude of work done by F shall be more than mgL.
2.
The FBD of block is Angle between velocity of block and normal reaction on block is obtuse work by normal reaction on block is negative. As the block fall by vertical distance h, from work energy Theorem Work done by mg + work done by N = KE of block 1 |work done by N| = mgh – mv2 2 1 mv2 < mgh 2 |work done by N| < mgh (B) Work done by normal reaction on wedge is positive Since loss in PE of block = K.E. of wedge + K.E. of block Work done by normal reaction on wedge = KE of wedge. Work done by N < mgh. (C) Net work done by normal reaction on block and wedge is zero. (D) Net work done by all forces on block is positive, because its kinetic energy has increased. Also KE of block < mgh Net work done on block = final KE of block < mgh.
3.
If the particle is released at the origin, it will try to go in the direction of force. Here
du is positive dx
and hence force is negative, as a result it will move towards – ve x-axis. 4.
When the particle is released at x = 2 + it will reach the point of least possible potential energy (–15 J) where it will have maximum kinetic energy.
1 2 m v max = 25 2
RESONANCE
vmax = 5 m/s
SOLN_Work, Power & Energy - 74
6.
(A) W CL + W f = KE W CL = KE – W f (a) During accelerated motion negative work is done against friction and there is also change is kinetic energy. Hence net work needed is +ve. (b) During uniform motion work is done against friction only and that is +ve. (c) During retarded motion, the load has to be stopped in exactly 50 metres. If only friction is considered then the load stops in 12.5 metres which is less than where it has to stop. Hence the camel has to apply some force so that the load stops in 50m (>12.5 m). Therefore the work done in this case is also +ve.
7.
W CL|accelerated motion = KE – W friction where W CL is work done by camel on load. =
1 2 2 mv 0 k mg.50
=
125 1 1000 5 2 0.1 10 1000 50 = 1000 2 2
similarly,
W CL|retardation = KE – W friction
1 75 2 0 2 mv – [k mg.50] = 1000 2
WCL |accelerate d motion 125 5 = = WCL |retarded motion 75 3
5:3
8.
Maximum power = Fmax × V Maximum force applied by camel is during the accelerated motion. We have V2 – U2 = 2as 25 = 02 + 2 . a . 50 a = 0.25 m/s 2 ; for accelerated motion FC – f = ma FC = mg + ma = 0.1 × 1000 × 10 + 1000 × 2.5 = 1000 + 250 = 1250 N This is the critical point just before the point where it attains maximum velocity of almost 5 m/s. Hence maximum power at this point is = 1250 × 5 = 6250 J/s.
14.
Potential energy depends upon positions of particles
15.
(i)
The net force on the body may have acute angle with its velocity, but one of the constituent force may have obtuse angle with the velocity. Such a force shall perform negative work on the body even though the kinetic energy of the body is increasing. (ii) A net force that is always perpendicular to velocity of the particle does no work but changes the direction of its velocity. (iii) A force which is always constant is also conservative. (iv) From Work - Energy theorem W all forces = KEfinal – KEinitial
EXERCISE-4 PART - I 1.
Power P = F . V = FV
dm = V d( volume F=V dt dt
= density
d( volume = V = V (AV) dt 2 = AV Power P = AV3 or P V3
RESONANCE
SOLN_Work, Power & Energy - 75
Alternate Solution Power output is proportional to number of molecular striking the blades per unit time [which depends on the velocity V of wind] and also proportional to energy to striking molecules or proportional to square of velocity V2 Therefore, power output P V3 2.
F=–
dU dx x
dU = – F . dx
3 U(x) = – ( kx ax ) dx
or ;k
0
2
U(x) = U(x) = 0
kx 2
–
ax 4
and
4
x = 0 and
x=
2k a
2k a From the given function we can see that F = 0 at x = 0 i.e. slope of U-x graph is zero at x = 0. Therefore, the most appropriate option is (d).
U(x) = negative for
3.
4.
x>
Let x be the maximum extension of the spring. From conservation of mechanical energy : decrease in gravitational potential energy = increase in elastic potential energy
Mgx =
or
x=
From F =
1 2 kx 2
2Mg k
dU dx
U( x )
x
x
dU Fdx (kx ) dx 0
U(x) =
0
0
kx 2 2
as U(0) = 0 Therefore, the correct option is (A). 5.
In horizontal plane Kinetic Energy of the block is completely converted into heat due to Friction but in the case of inclined plane some part of this Kinetic Energy is also convert into gravitational Potential Energy. So decrease in the mechanical energy in second situation is smaller than that in the first situation. So statement-1 is correct. Cofficient of Friction does not depends on normal reaction, In case normal reaction changes with inclination but not cofficient of friction so this statement is wrong.
6.
RESONANCE
SOLN_Work, Power & Energy - 76
As springs and supports (m 1 and m 2) are having negligible mass. Whenever springs pull the massless supports, springs will be in natural length. At maximum compression, velocity of B will be zero.
And by energy conservation
1 1 (4K) y2 = Kx 2 2 2 7.
2m1m2
T = m m g = 1 2
y 1 Ans. (C) x 2
2 0.72 0.36 × 10 0.72 0.36
T = 4.8 N m1 m2
g
a = m m g = 3 1 2 Work done by T = (T) (S) = (4.8) × 8.
10 =8J 6
s=
1 g 10 1 2 (1)2 = at = 2 3 6 2
Ans.
Fdt p K.E. =
1 1 ×4×3– × 1.5 × 2 = pf – 0 2 2 81 p2 = ;K.E. = 5.06 J 4 22 2m
pf = 6 – 1.5 =
9 2
Ans.
PART - II 1.
Let initial velocity is u and retardation is a So, (vr%)
u2 = u2 – 2a × (0.03) 4
u2 – 2a × S 4 here S is required distance from equation (i) & (ii) S = 0.01 m = 1 cm
0=
2.
...(i)
..(ii)
W C = – U = – (Ufinal – Uinitial) 1 1 2 2 = – k 15 – k 5 ] 2 2 W C = 8 Joule
3.
K = 5 × 103 N/m x = 5 cm 1 1 W 1 = k × x12 = 5 × 103 × (5 × 10–2)2 = 6.25 J 2 2 k(x1 + x2)2 W2 = 2
× 5 × 103 (5 + 10–2 + 5 × 10–2)2 = 25J 2 Net work done = W 2 – W 1 = 25 –6.25 = 18.75 J = 18.75 N-m =
RESONANCE
SOLN_Work, Power & Energy - 77
4.
M 4 = = 2 kg/m L 2 The mass of 0.6 m of chain = 0.6 × 2 = 1.2 kg Mass per unit length =
0.6 0 = 0.3 m 2 Hence, work done in pulling the chain on the table W = mgh = 1.2 × 10 × 0.3 = 1.2 × 10 × 0.3 = 3. 6 J The centre of mass of hanging part =
7.
m T Instantaneous power F = ma =
0 a T = F = ma
=
= 8.
m m . at = . .t T T T m2 T2
.t
Maximum height attained by the particle
H
u2 52 5 m 2g 2 10 4
W g = -MgH = -0.1 × 10 × (5/4) = -1.25 J 9.
Velocity of ball just after throwing v=
2gh =
2 10 2 =
40 m/s
Let a be the acceleration of ball during throwing, then
10. 11.
40 v2 = = 100 m/s2 2 0 .2 2s
v2 = u2 + 2as = 02 + 2as
F - mg = ma (2) is correct
F = m(g + a) = 0.2(10 + 100) = 22 N
1 mv 2 k 2
a=
1 1 v2 11 K m( v cos 60)2 m mv 2 2 2 4 42 4
Assuming mass of athlete is between 40 kg to 100 kg here we will consider mass of athlete m = 50 kg 100 = 10 m/sec 10 So, K = 1/2 mv2 1/2 × (50 ×102) = 2500 J So Answer is (C)
V = S/t =
12.
K.E. = ct
1 mv2 = ct 2 P2 = ct 2m
P=
2ctm
RESONANCE
SOLN_Work, Power & Energy - 78
TOPIC : CIRCULAR MOTION EXERCISE-1 PART - I SECTION (A) A 1.
Given v = 2i – 2j
(a) when moves in clockwise Ans. : First quadrant (b) When moves in counter clockwise
`
Ans. : Third quadrant A 3.
Given 0 = 0 , = const
1 2 t 2 for first two seconds = 0t +
1 ×(2)2 = 2 2 for next two seconds 1 = 0 +
1 1 (4)2 – (2)2 = 6 2 2 2 / 1 = 3 : 1 Ans. 2 = 4 – 2 =
A 5.
Given = R = 1 cm , V V2 – V1 V =
t = 15 Second
2V
V = R V=
2 1 = cm/sec. 60 30
a=
V 2 = cm/sec2. Ans. t 30 15
V =
2 cm/sec. 30
SECTION (B) B 1.
R = 0.25 m , = 2 rev./sec. = 4 rad/sec. ac = 2R
(at = 0)
= (4)2 × 0.25 = 42 m/s2. Ans.
RESONANCE
SOLN_Circular Motion - 79
B 3.
R = 1.0 cm , V = 2.0 t at t = 1 sec V = 2.0 cm/sec. ac =
v2 = 4 cm/sec2. R
at =
dv = 2.0 cm/sec2. dt
a=
a c2 a 2t =
2 4 2 2 2 = 2 5 cm/sec . Ans.
SECTION (C) C 1.
m = 200 g = 0.2 kg , g = 2 m/s2 Time period = 2
6 cos 1. 2 = 2 = 2 2 5 g
mg 13 0.2 2 N Tension = cos = = 6 12 / 13 C 3.
N=
mv 2 r
given r = 5 m
Ans.
Ans.
, v = 5 5 m/s
for no slipping f mg µmin N = mg µmin = µmin =
C 5.
= 2n =
rg mg = 2 N v
5 10 (5 5 ) 2
=
2 Ans. 5
2 1500 rad/sec 60
d = 60 cm = 0.6 m 2 m = 1 g = 10–3 kg r=
2
2 1500 × 0.6 F = m r = 10 × 60 2
–3
15 2 = 14.8 Ans. 10 This force is exerted by blade of fan and equal force is exerted by particle on blade in same magnitude but opposite in direction. =
SECTION (D) D 1.
R=
v 2 a
=
u2 sin2 g
RESONANCE
Ans.
SOLN_Circular Motion - 80
SECTION (E) E 1.
Tension is maximum in circular motion in vertical plane at lowest position. At lowest position Tmax – mg = m2R 30 – 0.5 ×10 = 0.5 2 × 2 2 =
E 3.
25 0. 5 2
= 5 rad/sec. Ans.
When string become slack apply equation for centripetal force.
mv 2 = mg cos 60º a apply energy conservation
v=
ga 2
1 1 mu2 = mv2 + mga(1 + cos) 2 2 from equation (i) & (ii) u=
....(i)
....(ii)
7ga 2
apply equation for centripetal force at lowest position. T – mg =
mu2 a
put the value of u and we get T = 9mg/2 E 5.
Using energy conservation :
1 mv B2 = mgh 2 Also to complite vertical circle vB =
R=
vB =
2mgh m
vB =
.....(1)
2hg
.....(2)
5gR
2 h = 2 cm 5
Section (F) F 1.
For safe driving vmax =
rg
10 =
rg
for wet road v´ = F 4.
10 rg = = 5 2 m/s Ans. 2 2
v = 48 km/hr = 40/3 m/s. For safe turn without friction v2 h tan = = rg x
F 7.
T = 2
cos geff . = 2
given x = 1m
v2 2 ( 40 / 3)2 h= = = m rg 45 400 10
Ans.
h geff .
geff. = g + a ; T = 2 put geff = 20 g + a = 20 a = 10 m/s2. Ans. Retardation = 10 m/s2 2 Ans. 10 m/s
RESONANCE
SOLN_Circular Motion - 81
PART - II SECTION (A) A 1.
Speed v1 =
2ðr t
v 1 2 1 = r t 1
2ðr t
v2 =
v 2 2 2 = 2r t 2
...(ii)
From eq. (i) and (ii)
A 3.
r=
...(i)
1 t 2 2 t1
t2 1= t 1
20 m, at = constant
n = 2nd revolution v = 80 m/s 80 v = = 4 rad/sec 20 / r = 2 × 2 = 4 from 3rd equation 2 = 02 + 2 (4)2 = 02 + 2 × × (4)
0 = 0, f =
at = r = 2 ×
20 = 40 m/s2
= 2 rad/s2
Ans.
A 5.*
In curved path, may be circular or parabolic. In circular path speed and magnitude of acceleration are constant. In parabolic path acceleration is constant.
A 7.
second =
2 2 = rad/sec. 60 T
2 × 0.06 m/s = 2 mm/s 60 mm/s v v f v i = 2 v = 2 2
v = .r =
Ans. Ans.
SECTION (B) B 1.
Angular velocity of every particle of disc is same aP = 2rp , aQ = 2rQ rP > rQ aP > aQ Ans.
B 3.
ac =
v2 , radius is constant in case (a) and increase in case (b). So that magnitude of acceleration is r
constant in case (a) and decrease in case (b).
SECTION (C) C 1.
r = 144 m, m = 16 kg, Tmax = 16 N T=
v=
mv 2 r
Tr = M
16 144 = 12 m/s 16
RESONANCE
Ans.
SOLN_Circular Motion - 82
C 3.
Uniformly rotating turn table means angular velocity is constant. New radius is half of the original value. r´ = r/2 and = constant v´ = r= r/2 = v/2 = 5 cm/s Ans. a´ = 2 r = r/2 = a/2 = 5 cm/s2 Ans.
C 5.
T1 – T2 =
M 2 L 2 2
T 1 > T2
Ans.
SECTION (D) D 1.
At t = 0
a = g cos ,
R=
v2 u2 = a g cos
SECTION (E) E 1.
Let the car looses the contact at angle with vertical
mv 2 mv 2 N = mg cos – R R During descending on overbridge is incerese. So cos is decrease therefore normal reaction is decrease. mg cos – N =
E 3.
mv 2 ....(1) r from energy conservation.
T – mg cos =
(from centripetal force)
1 1 mu2 = mv2 + mgr (1 – cos ) (here u is speed at lowest point) 2 2 from (1) and (2) T= E 5.*
mu2 + 3mg cos – 2mg r
for = 30º & 60º
T 1 > T2
For normal reaction at points A and B.
mv 2 mv 2 N = mg – r r NA > NB and normal reaction at C is NC = mg, so NC > NA > NB
mg – N = E-7._
either | T | = 0 a= So
V2 0 r
or
mg 2 +
|a|=0
or
= 90º
for whole motion there is velocity.
T = 0,
T + mg =
mV 2
T=0 T=
for
mV 2 – mg
1 1 mV2 = mu2 2 2
V2 = u2 – 4 gl E 9_
Ans.
T . a = | T | | a | cos = 0
T=
mu 2 – 5 mg
T=0
or
T<0
u 5 g
T . V = | T | | V | cos
= 90º every time. So
T . V = 0 for every value of u.
RESONANCE
SOLN_Circular Motion - 83
Section (F) F 1.
Here required centripetal force provide by friction force. Due to lack of sufficient centripetal force car thrown out of the road in taking a turn.
F 3.
When train A moves form east to west
m( v R)2 m( v R)2 N1 = mg – R R N1 = F 1 When train B moves from west to east mg – N1 =
mg – N2 = N2 = F 2 F 5_
m( v R)2 R F1 > F2 Ans.
N2 = mg –
m( v R)2 R
g R
mg = m2 R , =
EXERCISE-2 PART - I 1.
Change in velocity when particle complete the half revolution : v = vf – vi = 2v Time taken to complete the half revolution t =
average acceleration =
3.
ac = a cos 30º = 25 v2 R
v 2v 10 2v 2 2 52 = = = = m/s2 t R / v R 5
3 m/s2 2
Ans.
Ans.
v2 = aCR = 25
3 v = 125 4
m/s
Ans.
at = a sin 30º =
25 m/s2 2
ac =
R v
3 × 2.5 2
1/ 2
5.
Ans.
mv 2 R
(i)
The normal reaction by wall on the block is N =
(ii)
The friction force on the block by the wall is f = µN =
(iii)
The tangential acceleration of the block =
(iv)
dv µv 2 =– dt R
µmv 2 R
f µv 2 = m R v
integrating we get
RESONANCE
or
dv µv 2 v =– ds R
v n v = – µ 2 0
v0
or
dv v =–
2R
0
µ ds R
v = v0 e–2µ
SOLN_Circular Motion - 84
7.
Centripetal acceleration m2 r = T1 cos + T2 cos .... (1) apply Newton law in vertical direction T1 sin = mg + T2 sin .....(2) given m = 4 kg, T1 = 20 kgf = 200 N, r = 3m 3 4 , sin = 5 5
cos =
Put in equation (2) T2 = 150 N Put in equation (1) we get 210 35 = 43 2
2 =
n= 9.
1 = 2 2
Ans.
35 rad/s 2
=
35 rev/sec. 2
n=
30
35 rev/min. Ans. 2
Time take by ring to fall on ground. T=
2h g
from centripetal force m2x = ma = mv
2x = v
dv dx
x = . T =
dv dx
0
v
2 x d x vdv
2
0
2h g
vy = L
L2 v 2 2 2
y = T =
distance of one ring from center is =
vx = L
2h g
y 2 ( x )2
distance between the point on the ground where the rings will fall after leaving the rods. = 2 y 2 ( x )2 12.
(i)
where
x = y =
2h g
CP = CO = Radius of circle (R) COP = CPO = 60º OCP is also 60º
Therefore, OCP is an equilateral triangle. Hence, OP = R Natural length of spring is 3R/4. Extension in the spring x=R–
3R R = 4 4
mg R mg = Spring force, F = kx = R 4 4
The free body diagram of the ring will be a shown.
mg 4 and N = Normal reaction Here,
C
F = kx =
RESONANCE
O
P F mg
SOLN_Circular Motion - 85
(ii) Tangential acceleration ar = The ring will move towards the x-axis just after the release. So, net force along x-axis : mg Fx = F sin 60º + mg sin 60º = 4
Fx =
3 + mg 2
3 2
5 3 mg 8
Therefore, tangential acceleration of the ring. aT = ax = aT =
Fx 5 3 = g m 8
5 3 g 8
Normal Reaction N : Net force along y-axis on the ring just after the release will be zero. Fy = 0 N + F cos 60º = mg cos 60º N = mg cos 60º – F cos 60º =
N= 14.
mg mg mg 1 mg = – – 8 2 4 2 2
3mg 8
(a) at equator T + m2 R = mg. %
4 2 6400 1000 2R T 100 = 0.65 % Ans. = = 2 g T ( 24 60 60 ) 9 .8
mg ....(1) 2 T + m2R = mg ....(2) from (1) and (2) (b) T =
2R = g/2
T= 16.
=
g 2R
2R 2 = 2 g = 2hr
Ans.
Block B rotate in vertical plane. Tension is maximum in string at lowest position. When block B at lowest position and block A does not slide that means block A not slide at any position of B. At lowest position T – mg =
mv 2
T = mg +
mv 2
....(1)
From energy conservation mg(1 – cos ) =
1 mv2 2
...(2)
from equation (1) and (2) T = mg + 2mg (1 – cos ) = 3mg – 2mg cos for no slipping. T = mg = 3mg – 2mg cos min = 3 – 2 cos Ans.
RESONANCE
SOLN_Circular Motion - 86
18.
Constant speed = 18 km/hr = 5m/sec. m = 100 kg, r = 100 m (a)
at B mg – NB =
100 5 2 mv 2 = = 25 100 r
NB = 975 N
mv 2 ND = 1025 N r at B & D friction force act is zero. at D ND – mg =
(b)
Ans.
Ans.
1 at C f = mg sin 45 = 100 × 10 (c)
2
( v = constant) = 707 N
Ans.
for BC part mg cos 45 – NBC =
mv 2 R
NBC = 682 N
mv 2 R
NCD = 732 N
for CD part NCD – mg cos 45 = (d)
f N position where its maximum and N is minimum which is in part BC at C position. f N
mg sin 45 º mv mg cos 45 º r
2
707 = 1.037 682
Ans.
PART - II 1.
QP = 2 – 5 = – 3 rad/s RP = 3 – 5 = – 2 rad/s Time when Q particle reaches at P = t1 =
3.
/2 1 = sec. 3 6
t2 =
5 / 2 5 = sec. 3 6
t3 =
9 / 2 3 = sec. 3 2
Time where R particle reaches at P.
t1 =
1 = sec. 2 2
t2 =
3 3 = sec. 2 2
Common time to reaches at P is
3 sec. Ans. 2
at loose contact N = 0 mv 2 R from energy conservation
mg cos =
mgR(1 – cos ) =
....(1)
1 mv2 2
....(2)
from (1) & (2) cos =
2 3
tangential acceleration
RESONANCE
sin =
5 3
= g sin =
5g 3
Ans.
SOLN_Circular Motion - 87
6.
For M to be stationary T = Mg Also for mass m, T cos = mg
.... (1) .... (2)
mv 2 .... (3) sin
T sin =
Tcos
m
dividing (3) by (2)
v2 tan = v= g sin
T
g . sin cos
M
2 R = v
g . sin cos
From (1) and (2) cos =
9.
12.
m M
then time period = 2
F = kx, T1 = ka = m2 2a = Time period =
mg
Mg
2 sin
Time period =
Tsin
m gM
k 2m
2 2m = 2 =T k
T2 = 2ka = m23a
Time period = 2
3m = T´ 2k
=
2k 3m
3 T´ = 2 T
Ans.
T – mg cos =
mv 2
(i) at angle at = g sin from centripetal acceleration
...(1)
From energy conservation : 0 + mg cos = from (1) & (2) a= (ii)
1 mv2 v = 2g cos 2 T = 3mg cos aC = 2g cos
a 2t a c2 = g
....(2)
1 3 cos 2
Vertical component of sphere velocity is maximum when acceleration in vertical is zero that means net force in vertical direction is zero. Net force in vertical at angle mg cos and tension also from equation T = 3mg cos ....(4) from (3) & (4)
T cos = mg
3 mg cos =
T=
mg cos
...(3)
1
cos =
3
T = mg (iii)
Ans. 3 Total acceleration is directed along horizontal that means avertical = 0
1 cos =
RESONANCE
3
Ans. SOLN_Circular Motion - 88
14.
For vertical circular motion, in lower half circle tension never be zero anywhere. Tension is maximum at lowest point of oscillation. Tension decrease both side in same amount. Therefore correct option is (D).
16.
Maximum retardation a = g For apply brakes sharply minimum distance require to stop. 2
0 = v – 2gs
v2 s= 2g
For taking turn minimum radius is g =
v2 , r
r=
v2 , g
here r is twice of s
so apply brakes sharply is safe for driver. 19.
= 2
d 2d = = 2 × 0.4 = 0.8 rad/s dt dt
vAC = r = 0.8 ×
1 = 0.4 m/s 2
1 = 0.32 m/s2 2 a = aC = 0.32 m/s2 (at = 0) aC = 2r = (0.8)2
21.
Energy conservation from initial and final position
1 = 1 mv2 + 1 mv2 mgr + mg r 1 2 2 2
v=
2gr
1
gr
Ans.
2
Normal reaction at bottom position A 2
N – mg = 23.
mv 2 r
N=
mv 2 r
gr m 2gr 2 + mg = r
mg + mg = 3 mg –
2
= 2.29 mg
The acceleration vector shall change the component of velocity u|| along the acceleration vector. r=
v2 an
Radius of curvature rmin means v is minimum and an is maximum. This is at point P when component of velocity parallel to acceleration vector becomes zero, that is u|| = 0. u|| = 0
R=
u 2 42 = = 8 meter.. a 2 T cos 60o
2
mv T 3 = ( 3 / 2) 2
25.
T = mg 2 Hence T = 2 mg , So (B) holds From (1) & (2) V2 = 3 g/2 3 9.8 1.6 2
V=
V = 2.8
........(1) 60o
.......(2)
60o
/ 2
T 3/2
V o
T sin 60
mg
3 m/s . So (C) hold
RESONANCE
SOLN_Circular Motion - 89
ac = V2/r = t=
27.
( 3 g / 2) ( 3 / 2)
3 × g = 9.8 3 m/s2
=
(D) holds 2 3 /2 2 r = (3 g / 2) v
Speed of cage =
t = 4/7
(A) holds.
gr = const.
Normal reaction at (weight reading) NA – mg =
mv 2 r
NA = 2mg = 2w Ans. Weight reading at G & C = mg = w Ans. weight reading at E mg – NE =
mv 2 r
NE = 0 29.
Ans.
Tangential acceleration = at = gsin Normal acceleration = an = g cos at = an g sin = g cos = 45° vy = vx uy – gt = ux 20 – (10)t = 10 t = 1 sec. During downward motion at = an vy = – vx 20 – 10 t = – 10 t = 3 sec.
EXERCISE-3 1.
From graph (a) = k angular acceleration =
where k is positive constant
d = k × k = k2 d
angular acceleration is non uniform and directly proportional to . (A) q, s From graph (b) 2
d =k d
or
From graph (c) angular acceleration = From graph (d) angular acceleration =
2 = k .
Differentiating both sides with respect to .
d k = d 2
Hence angular acceleration is uniform. (B) p
= kt d =k dt = kt2
d = 2kt dt
Hence angular acceleration is uniform (C) p
Hence angular acceleration is non uniform and directly proportional to t.
(D) q,r
RESONANCE
SOLN_Circular Motion - 90
2.
v = 2t2 Tangential acceleration at = 4t Centripetal acceleration ac =
Angular speed =
v2 R
4t 4 R
v 4t = , R R
tan =
4tR R at 3 = 4t 4 t ac
Sol. 3 to 5. The angular velocity and linear velocity are mutually perpendicular v = 3x + 24 = 0 or x=–8 5 v 1 = = meter 10 2 The acceleration of particle undergoing uniform circular motion is a v = ( 8 ˆi 6 ˆj ) (3 ˆi 4 ˆj ) = 50 kˆ
The radius of circle r =
6.
mu02 r Now, along vertical mg =
r=
1 2 gt 2
u0 =
t=
gr
2r g
Along horizontal ; OP = 2u0t = 2 2 r 7.
As at B it leaves the hemisphere, N=0
A
mV 2 mg cos = r
N
u0/3 B
r
c mg
h mV 2 mg = r r mv2 = mgh .............(1) By energy conservation between A and B
os
v
h
O
2
1 u0 1 mgr + m = mgh + mv2 2 3 2 Put u0 and mv2 8.
9.
h=
19r 27
v2 = g cos r at = g sin anet = g Alternate Solution : when block leave only the force left is mg. anet = g. As ac =
geff g a
–a
Tension would be minimum when it (tension) is along geff mg
tan = 3
4
mg
=
4 3
RESONANCE
g
= 53º .
geff
SOLN_Circular Motion - 91
10.
Vmin =
geff =
5 g 4
=
= 6 mgeff
5g . 2
(geff =
5 15 g) = mg 4 2
11.
Tmax
12.
For conical pendulum of length , mass m moving along horizontal circle as shown T cos = mg .... (1) T sin = m2 sin .... (2) From equation 1 and equation 2,
cos =
g
2 cos is the vertical distance of sphere below O point of suspension. Hence if of both pendulums are same, they shall move in same horizontal plane. Hence statement-2 is correct explanation of statement-1. 13.
The normal reaction is not least at topmost point, hence statement 1 is false.
14.
Let the minimum and maximum tensions be Tmin and Tmax and the minimum and maximum speed be u and v. Tmax = Tmin =
mu2 + mg R mv 2 – mg R
u2 v 2 T = m R R + 2 mg. From conservation of energy
u2 v 2 = 4g is indepenent of u. R R and T = 6 mg. Statement-2 is correct explanation of statement-1. 15.
Statement-2 is wrong. R =
v2 , where a is acceleration component perpendicular to velocity.. a
and as particle goes up, v2 decreases and a increases so radius of curvature R decreases hence statement -1 is true 16.
(i) False. (ii) True. (iii) True.
It has tangential as well as radial acceleration. The angle is less than 180°. The angle between velocity and radial acceleration is 90°. It has no acceleration in verticall direction
(iv) False.
=
(v) False.
aT =
RESONANCE
initial final is valid only for constant angular acceleration. 2 dv a c2 dt
2
> ac
SOLN_Circular Motion - 92
17.
(i)
Given that tangential acceleration = at = 3 m/s2 Centripetal acceleration = ac = a=
a c2 a 2t =
2 4 2 3 2 = 5 m/s
`
Now
(ii)
= average acceleration =
v 2 R 2R2 = = t R / R
= 2R
Instantaneous acceleration
(iii)
20 2 v2 = = 4 m/s2 100 r
a 2 = a Tension before cutting T sin = mg
Ans.
mg sin Tension after cutting. T2 = mg sin
T1 =
T2 2 T1 = sin (iv)
Ans.
h
tan = (v2/rg) =
]
2
b h2 1/ 2
Ans : (v)
ghr 2 2 b h
Acceleration at lowest position v2 R From energy conservation
aL =
mgR (1 – cos ) =
mv 2 2
v2 = 2g(1 – cos) R aL = 2g (1 – cos) acceleration at highest position. aH = g sin according to problem aL = aH 2g(1 – cos ) = g sin 2 (1 – cos ) = sin 2(1 – 1 + 2 sin2 /2) = 2 sin /2 cos /2
tan
1 = 2 2
tan =
2 tan 2 1 tan 2
RESONANCE
2 2
21
1
1 4
=
4 3
= 53º Ans.
SOLN_Circular Motion - 93
EXERCISE-4 PART - I 1.
2.
Net acceleration a of the bob in position B has two components. (i) an = radial acceleration (towards BA) (ii) ar = tangential acceleration (perpendicular to BA) Therefore, direction of a is correctly shown in option (C).
A //////////////////////////
an a
d h = R (1 – cos) 2 velocity of ball at angle is
B
(a)
at
d v2 = 2gh = 2 R (1 – cos)g 2
.......(1)
Let N be the total normal reaction (away from centre) at angle . Then mg cos – N =
mv 2 d R 2
h
v mg
Substituting value of v2 from equation (1) we get mg cos – N = 2mg (1 – cos) N = mg (3 cos – 2) Ans. (b) The ball will lose contact with the inner sphere when N=0
or
3cos – 2 = 0
or
2 = cos–1 3
After this it makes contact with outer sphere and normal reaction starts acting towards the centre. Thus for 2 cos–1 : 3 NB = 0
and
NA = mg (3 cos – 2)
and
for
2 cos–1 3
NA = 0 and NB = mg (2 – 3cos) The corresponding graphs are as follows NB
NA
5mg
mg 2mg
-1
3.
2/3
+1
cos -1
2/3
+1
cos
By energy conservation, 1 1 mu2 = mv2 + mg(1 – cos) 2 2
V2 = U2 – 2g (L – L cos)
5gL = 5gL – 2gL (1 – cos) 4 5 = 20 – 8 + 8 cos cos = –
7 8
3 << 4
RESONANCE
Ans. (D)
SOLN_Circular Motion - 94
4.
T sin = m Lsin2 324 = 0.5 × 0.5 × 2 2 =
324 0 .5 0 .5
=
324 0.5 0.5
18 = 36 rad/sec. 0 .5 Since distance of particle P from point O is initially decreasing then increasing so, its angular velocity will initially increase then decrease. So, angle swept by P is more than angle swept by disc. So it will fall in unshaded portion. Since distance of particle Q from O is continuously increasing so its is continuously decreasing. So angle swept by Q is less than angle swept by disc. So it will fall in unshaded portion.
=
5.
6.
vr = |2 v sin )| = |2v sin t)|
PART - II 1.
For a particle moving in a circle with constant angular speed, velocity vector is always tangent to the circle and the acceleration vector always points towards the centre of circle or is always point towards the centre of circle or is always along radius of the circle. Since, tangential vector is perpendicular to radial vector, therefore, velocity vector will be perpendicular to the acceleration vector. But in no case acceleration vector is tangent to the circle
2.
When a force of constant magnitude acts on velocity of particle perpendicularly, then there is no change in the kinetic energy of particle. Hence, kinetic energy remains constant.
3.
S = t3 + 5 Linear speed of the particle =
dS = 3 t2 dt
at t = 2 s
Linear acceleration
v = (3 × 22) m/s = 12 m/s a1 =
d =6t dt
at t = 2 s,
a1 = 12 m/s2
The centripetal acceleration a2 =
122 2 = m/s2 = 7.2 m/s2 20 R
anet =
a12 a 22 = 12 2 7.2 2 = 14 m/s2
V2 V2 cos ˆi – sin ˆj R R
4.
aC = –
5.
They have same . centripetal acceleration = 2r
a1 2r1 r1 a 2 = 2r2 r2
RESONANCE
SOLN_Circular Motion - 95
TOPIC : CENTRE OF MASS EXERCISE-1 SECTION (A) A 1.
xcm =
1 0 2 1 3 1cos 60 7 1 2 3 12
ycm =
1 0 2 0 3 1sin 60 3 3 3 6 12 4 2 7 3 12 4
r=
A 3.
2
=
49 3 76 2 19 19 m 144 16 144 12 6
3 4 A = M, A = M 4 1 3
xcm
4 3a a M – M / 3 4a –a x1m1 x 2m2 3 2 = m m2 = = + 3 2 1 M 8–3
5 xcm = a 3 x 2 = a 6
5 a 6
Similarly ; ycm = A 5.
M1 = (2R)2 × So Xcm = Xcm =
A 8.
Ans. M2 = (R)2 ×
x1 = 0, x2 = R
m1x1 m2 x 2 m1 m2
4R2 0 R2 R
=
4R2 R2
R towards smaller disc 5
length of the shaded region = 2y = 2kx2 dm = 2y dx × dm = 2kx2 × dx a
a
o
o
2 ka 3 3
2 M = dm 2 kx dx =
a4 2k 4 a3 2k 3
3a 0 0 Xcm = a Xcm = a 4 dm dm 0 0 By symmetry the y-coordinate of the shown plate is zero. a
x dm
a
3 2 kx dx
SECTION (B) 10 7 30 x 40
B 2.
1=
B 3.
2u2 sin cos Xcm = g
x1 =
40 = 20 m 2
RESONANCE
get x = – 1 cm 2 20 20
=
xcm =
1 2
10
m1x1 m 2 x 2 m1 m 2
1 2
= 40 m.
40 =
m 20 m x 2 get x2 = 60 m 2m
SOLN_Centre of Mass - 96
B 5.
So, ms × R = (40 + 60) × x 1 2u2 sin cos = 100 x g
B-7
get x = 0.1 m
Mh Mm since no external force is acting initially Ycom =
yCM =
COM should be at rest.
m1y1 m 2 y 2 m1 m 2
Let baloon descend by a distance x. 0=
m( x ) M( x – h) mM
x=
Mh (Distance decend by ballon) mM
h–x=
Mh = (m + M) x
mh (Distance raised by man) mM
SECTION (C) C 1.
238 × 0 = 4 × 1.17 × 107 + 234 × v2 V2 = –2 × 105 m/ses
C 3.
(a) P1 = 2.4 × 10–26 kg–m/sec. P2 = 7.0 × 10–27 kg–m/sec P1 + P2 + P3 = 0 P3 = – (24 ×10–27 + 7.0 × 10–27) P3 = 31 × 10–27 V3 =
31 10 –27 1.67 10 – 27
= 18.6 m/sec.
(b) Pe = 2.4 × 10–26 ˆi P an = 7.0 × 10–27 ˆj P p = –( P e + P an) = – (24 × 10–27 ˆi + 7.0 × 10–27 ˆj )
PP = C 4.
( 24 )2 (7.0) 2 ×10–27
P1 =20 × 20 ˆi
P3 = 40 × 20 kˆ
Pi = P1 + P2 + P3 = Pf get C 6.
v=
| Pp |
Vp = m = 15.0 m/sec. p
P2 = 30 × 20 ˆj
400i +600 ˆj + 800 kˆ = 30 (10i + 20 kˆ ) + 40v
100i 600 ˆj 200kˆ = 2.5 ˆi + 15 ˆj + 5 kˆ Ans. 40
mc = 20 kg
mT = 180 kg
Pi = 200 × 36 ×
5 = 2000 kg m/sec 18
just before jump Vbg = VbT + VTg = (10 + VT) So MTVT + mc Vc = Pf = Pi 180VT + 20 × (10 + VT) = 2000 1800 10 = 9 m/sec. Time taken to cover 10 m t = = 1 sec. 10 200 distance covered by trolly = 9 × 1 = 9 m.
VT =
RESONANCE
SOLN_Centre of Mass - 97
C 8.
Net ext force = 0
dp = 0 , p = constt dt
F=
COM remain at rest 1 1 mv2 + MV2 2 2
mg R = mv = MV v=
MV mv or V = m M 2 2
mgR =
m v 1 1 × mv2 + M 2 2 M2
M mM m 2 2 v 2gR M m = v2 M
2mgR =
2gR =v m 1 M
SECTION (D) D 1.
Energy Conservation Total change in length of spring = 2x {
1 1 2 2 = kx comp } kx ext 2 2
Time is same no external force
centre of mass is at rest
hence m1x1 = m2x2
x = m & x1 + x2 = 2d 2 1
x1
m2
or, m1x1 = m2x2 & x1 + x2 = 2d x 1=
m2 x 2 m2 x 2 m1 m1 + x2 = 2d
2 x2 m 1 = 2d
m2 m1 = 2d x2 m
x2 = m m & x1 = m m 1 2 1 2
D-3.
m
1
2dm1
1
2dm 2
By momentum conservation mAVA = mBVA K.EA =
(i)
VB =
m A VA mB
P2 m2 V 2 m V 2 A A A A 2m A 2m A 2
Similiarly K.EB =
...(i)
mB VB2
...(iii)
2
dividing (ii) by (iii) we get. m A VA2 K.E A K.EB = mB V 2 B
put VB =
m A VA mB
K.E A
mB
we get K.E = m . B a
hence proved.
SECTION (E) E 2.
Pi = 200 × 10–3 ( 3i – ˆj ) Pf = 200 × 10–3 ( 3i ˆj ) |Pi| = |Pf| P = |Pf| – |Pi| = 0 |P| = |Pf – Pi| = |(200 × 10–3 3 i – 200 × 10–3 ˆj ) – (200 × 10–3 3 i + 200 × 10–3 ˆj )| |P| = |2 × 200 × 10–3 ˆj | = 0.4 kg m/sec.
RESONANCE
SOLN_Centre of Mass - 98
E 4.
v=
2gh =
2 10 4 =
(a) J = P = 2mv = 2 ×
80
1 × 80 2
J = 4 5 N-s (b)
N dt = dP
N × 2 × 10–3 = 4 5
N = 2 5 ×103N.
SECTION (F) F 1.
from momentum conservation mu + 0 = (m + m) v
v=
1 1 u P.E. = mu2 – 2m 2 2 2
from energy conservation
1 mu2 = K 2
P.E. =
u 2 2
K 2
F 3. After first collision After collision of B from wall vB = – v + 2 × 0 = –v so F 5.
Particle B is a rest mv + 0 = mv1 + 2mv2 v = v1 + 2v2
.....(i)
v 2 – v1 =1 v 0
v2 – v1 = v Adding (i) + (ii) 3v2 = 2v 2 v 3 Now, (iii) + (iv)
v2 =
2r
.....(ii)
v1 = v2 – v =
2r
t = v v = 2v 2 1
v 3 3
–v 3
t=
2r Ans. v
SECTION (G) G 1.
m0 = 20 kg ; m = 180 kg. – Fth = (m +M)g = 2×103 N – Fth = vr
dm dt
So
dm 2000 = = 1.25 kg/s. Ans. dt 1.6 103
m
0 v = vr n m – gt.
RESONANCE
SOLN_Centre of Mass - 99
m
(i)
t1 = dm / dt =
180 = 90 s. 2 200
v1 = 1600 n 20 – 10 × 90 M
(ii)
v1 = 2.784 km/s. Ans.
180
t2 = dm / dt = = 9 s. 20 200
v2 = 1600 n 20 – 10 × 9
v2 = 3.59 km/s.
PART - II SECTION (A) A-2.
A1 = R2
A2 =
R2 16
x1 = 0
x2 =
3R 4
R2 3R 16 4 – R 2 20 R R2 – 16
0–
xcen =
A-4.
A1 = 2r × r = 2r2
A2 =
r 2 2
x1 =
r 2
x2 =
4r 3
2 r r 2 r r 3 1 – – 2r 3 2 2 3 = 2 4 – 3 [ 4 – ] r r2 2r 2 – 2 2
2r 2 xcm =
A-9.
1 7 × 0.14 + ×h = 0 8 8
ycm = 0
7h 0.14 =– h = –0.02 below x-axis. 8 8
SECTION (B) B-2.
vcm =
m1v 1 m 2 v 2 m1 m 2
vcm =
m(2ˆi ) m(2ˆj ) 2m
acm =
m(i j) m(0 ) . 2m
vcm has same direction as of acm straight line. B-3.
a=
(n – 1) (nm – m) g = g (n 1) nm m
a1 = a2 = a acm =
acm =
nma1 – ma 2 (n – 1) a = (nm m) (n 1) (n – 1) 2 (n 1) 2
g.
RESONANCE
SOLN_Centre of Mass - 100
B 9.
a cm
m1 a 1 m 2 a 2 m0ma a = = = m1 m 2 (m m) 2
given m1 = m2 = m
a1 = 0
a2 = a
SECTION (C) C-3.
mv ˆi + mv ˆj + 2m v 3 = 0
v v ( v ˆi vˆj ) = – ( ˆi + ˆj ) = – . v3 = – 2 2 2 kf =
kf =
1 1 1 v2 mv2 + mv2 + 2m . 2 2 2 2
3mv 2 . 2
C-4.
500 × 10 = 550 × v
C 6.
Vcom = V cos
v=
500 100 m/s . = 55 11
– m 0 mv 2 2m v2 = 2V cos V cos =
C 8.
M v = m. 0 + (M – m) v´
v´ =
Mv (M m)
SECTION (D) D 1.
Pi = mv1 + mv2
Pf = (m + M) v
mv1 Mv 2 (m M)
Pi = Pf v =
mv12 + Mv22 = (M + m)
By energy consarvation
1 1 1 1 mv12 + Mv22 = (M + m) v2 + kx2 2 2 2 2
(mv1 Mv 2 )2 (M m )2
kx 2
mM . (M m)k
solving x = (v1 –v2)
SECTION (E) E 1.
v1 = k2 =
2gh =
2 10 10 = 10 2
1 1 k v22 = v12 4 1 4
v1 = 5 2 2 |P| = |–mv2 – (mv1)| = m |–v2 – v1| v2 =
|P| = 50 × 10–3 ×
15 10 –1 3 × 10 2 = 2 2
J = P = 1.05N-s.
RESONANCE
SOLN_Centre of Mass - 101
E 3.
From momentum conservation mu = 2mv
v=
u 2
from energy conservation 1 u × 2m × 2 2
2
= 2 mgh
h=
u2 8g
SECTION (F) F 5.
0.05 × vp + m × 0 = 5.05 v
vf 0.05 v = = 10–2 5 i
F-6.
1 m( v f )2 2 = (10–2)2 = 10–4. 1 2 m( v i ) 2
m1 2gh + 0 = (m1 + m2) v
v=
m1 2gh (m1 m 2 )
v2 – u2 + 2g ×
h h gh = 6 + 2g × = 9 4 2
v=
gh 2
Also,
m 2gh gh = 2m1 + m1 + m2 m 2 1 m2
m1 m 1. 2 F-8.
4 3 1 r e= 3 2 mA v + 0 = mAv1 + mBv2 ev = v2 – v1
MA = ×
Adding (i) + (ii) = 9v2 = v + v1 = v2 – F-9.
F-11.
MB = ×
4 (2r)3 = 8MA 3
.........(i) .........(ii)
v 3v = 2 2
v1 v v v v /3 v = – =– . v = = 2. 6 2 3 v/6 2 2
V2 = Z0 Vel. of Sep = Vel of approach ( elastic) 20 + 5 = V – 5 V = 30 m/s Ans. vb = –(v0 + 2v) m1 > > m2 vb = –(20 + 10) = –30 m/sec.
2d t = v (time for succeesive collision) 0 N × t = dP = mv0 – (–mv0)
2d N × v = 2mv0 0 N=
mv 02 d
RESONANCE
SOLN_Centre of Mass - 102
F-14.
If mass = m first ball will stop v = 0 so K.E. = 0 (min) (K.E. can't be negative )
SECTION (G) G 1.
F=
dm dt
210 = 300 ×
dm dt
dm = 0.7 kg/s. dt
EXERCISE-2 PART - I 1.
2H g =
2 80 = 16 = 4 s. 10
2h g
=
2 (80 60 ) = 10
t=
t = T – t = 4 – 2 = 2 s
V=
2 d = = 1 m/s t 2
Mv = mv
v=
60 1 = 30 m/s Ans. 2
T=
40 = 2 s. 10
R = vt = 30 × 2 = 60 m Ans.
4. By momentum conservation Mu = mV (i) V=
M u m
By energy conservation mgh =
1 1 Mu2 + mV2 2 2
=
1 1 M Mu2 + m 2 2 m
mgh =
1 1 M2 2 Mu2 + u 2 2 m
=
2 1 2 Mm M u m 2
2
u2
2m 2gh = u2 (Mm + M2) 2 m2gh
Mm M2
u=m
= u 2.
2 gh
...(ii)
Mm M2
By momentum conservation mV = (M + m) V1 V1 =
mV Mm
...(iii)
By energy conservation 1 1 mV2 = (m + M) V12 + mgh1 2 2 1 1 mV mV2 = (m + M) 2 2 Mm
RESONANCE
2
+ mgh1
1 1 m2 V 2 mV2 – = mgh1 2 2 (M m)
SOLN_Centre of Mass - 103
2 1 2 m – m V M m = mgh1 2
2gh
M
Put V = – m × u and u = m
Mm M2
put value of V from eqn (iv) to (iii)
1 MmV 2 = mgh1 2 (M m)
V=M
h' =
2gh Mm M2
...(iii)
....(iv)
M2h (M m)2
6.
By mechancnical energy conservation 1 1 ( 2m)Vb2 mVr2 = 2 mgl 2 2
2 Vb2 + Vr2 = 4gl using momentum conservation mVr = 2 mVb Vr = 2 Vb 2Vb2 + 4Vb2 = 4 gl
....(1) ....(2)
6Vb2 = 4gl
Vb =
2 gl 3
2 gl 3
(a) Vr = 2 Vb = 2
when string be comes vertical velocity of block wrt to string. Vbr = Vb – (– Vr) = 3Vb = 3 (b) T – 2 mg = 8.
2m( Vbr )2
2 gl 3
T = 2 mg +
9 2gl ( 2)m = 14 mg 3
m = 20 × 10–3 kg ; M = 5 kg u = 400 d = 0.2 m V = 200 =? PBullet = PBlock m (u – v) = 20×10–3 (400 – 200) = 4 kg. m/s. 2
KEBlock =
4 P2 = = 1.6 J = Mgd 2M 25
1 .6
1 .6
= Mgd = = 0.16 Ans. 5 10 0 . 2 10.
string will taut when A waves a distance of (.7 – .25) m at that Pt VA =
2gh and Now B starts on using with same velocity as A.
let us suppose it is u.
T dt
= mB u + 0
....(1)
– T dt = mA (u –VA)
....(2) m A VA
6
from (1) and (2) u = m m = 5 A B
RESONANCE
6
Tdt 3 5
= 3.6 m/s
SOLN_Centre of Mass - 104
12.
R1 = V cos T 1 R2 = V cos T 2 R3 = V cos T 3 R = R1 + R2 + R3 = u cos [T 1 +T 2 + T 3]
2 u sin 2 eu sin 2 e 2u sin = V cos g g g (1 e e 2 ) V 2 sin 2 R= g 14.
( V )2 2a (a) e = 1 so after collision VA = 0 and VB = 5 m/sec V2 = U2 + 2as
0 = (V)2 – 2as
s=
(5 ) 2 s = 6.25 m 2 0.2 10 (b) when e = 0 applying momentum conservation m × 5 + 0 = (m + m) × V V = 2.5m/sec So mg = ma s =
so V2 = u2 + 2as 16.
s=
Px = 5×2 = 10
; P = Px ˆi + Py ˆj
Py = 10 3
= 10 ˆi + 10 3 ˆj = (5 + 10) V
(2.5)2 2 2 0.2 10
a=
g 2
s = 3.12 m.
10 ˆ i 3ˆj 15
=
2 ˆ i 3ˆj V = 3
V =
4 m/s. 3
Ans.
1 1 1 2 2 H = E = Ei – Ef = 2 m1v1 2 m2u2 – (m1+m2) V2 2
= 18.
1 1 × 5 × 22 + ×10 × 2 2
(a) V= In x dir
2
3
–
(b)
20.
= 3 mV
v=
V 2 – 2gl(1 – Cos )
=
2
V =
Ans.
2gl
v 2m ×
1 40 35 × (10+5) (4/3)2 = 25 – = 2 3 3
3 2
V
=
2
2gl
v = 3 gl
Ans.
2gl – 2gl 2glCos
For = 60°
V =
Vx =
(at heighest point)
gl Cos 60°
3
gl
Vx =
gl 2
Applying momentum conservation in horizontal direction mV0 = Mu M = 2m u=
mV0 V0 M 2
Eqn of e along normal V cos u sin e= V0 sin V
1
e = V cot + 2 0
RESONANCE
V V cos 0 sin 2 = V0 sin
...(i) SOLN_Centre of Mass - 105
Along incline surface of wedge friction is negligible so change in momentum mV0 cos = mV sin V V0 = cot
...(ii)
Put value of (ii) in (i) e = cot2 +
1 2
given tan = 2 =
1 1 3 4 2 4
Ans.
(b) h = (ut) tan By (2)nd eq. of motion or
– u tan = V – 2
– h = Vt – 1 gt 2
– (ut) tan = Vt –
2
V0
2V0 tan 2e V0 tan t= (e) = g g
1
t = g V0 tan (cot2 + ) 2 2
1 gt2 2
1 gt = V + u tan 2
t = g (V0 cot + tan ) 2
22.
1 gt2 2
substituting values :
3 10 2 4 = 3sec 10
(a) At highest point V = 50 cos ...(i) After striking bullet get embedded with bob so by momentum conservation. MV = 4Mu from (i)
u=
u=
V 4
....(ii)
50 cos 4
By energy conservation after collission 1 (4m) u2 = 4mg (1 + cos 60º) 2
50 50 cos2 = 100 16
cos 2 =
(b) Max height R
y=
1 50 cos 2 4
16 25
1 u2 sin 2 502 2 sin 37 º cos 37 º g 2 10
(c) x = 2 2
=
=
cos =
( 50 sin 37 º )2 2g
2
10 10 1 1 3 2 4 5
= 37º
1 3 3 50 50 = 5 × 9 = 45 20 5 5
= 120 m Ans
(a) = 37º (b) x = 120 m and y = 45 m
PART - II 1.
COM can lie anywhere, within or at the radius r.
3.
Since no external force is acting on the system hence VCM remain constant.
5.
when cylinder reaches pt B. then block get shifted by x but since than there is no ext force therefore com remain at its position [(R–r) – x]m = Mx x=
8.
m (R – r ) Mm
Pi = 0 Pf = MV – mV1
...(i) ....(ii)
MV – mV1 = 0
using
u=
M V.. m
V12 = u2 + 2ax. a = g.
RESONANCE
MV m
2
= 0 + 2g x.
x=
M2 V 2 2m2g
SOLN_Centre of Mass - 106
10.
Taking the origin at the centre of the plank. m 1x 1 + m 2 x 2 + m 3x 3 = 0 ( x CM = 0) (Assuming the centres of the two men are exactly at the axis shown.) 60(0) + 40(60) + 40 (–x) = 0 , x is the displacement of the block. x = 60 cm i.e. A & B meet at the right end of the plank.
12.
yCM = 0 yCM =
m 3m y1 + y2 4 4 y2 = –5 cm
y1 = + 15
14.
I. Since velocity of both R and S is positive they will move in same direction. II. At mid point velocities of R and S are same. III. Change in velocity of R is small as compare to change in velocity of S. But change in momentum is same for both in magnitude. Hence mass of R should be greater than S. Hence all three are correct.
16.
If we treat the train as a half ring of mass 'M' then its COM will be at a distance the circle. Velocity of centre of mass is : VCM = RCM .
18.
2R V . R
=
2R .
VCM =
The linear momentum of the train =
=
( =
2R from the centre of
V ) R
2MV 2V MVCM = As the linear momentum of any system = MVCM
I = f × t
and
F=
2MV
Ans.
m( 2gh 2 2gh1 ) t
100 10 –3 ( 2 9.8 0.625 2 9.8 2.5 ) 0.01 F = 105 N
F=
20.
Using momentum conservation p1 p 2 p 3 p 4 0 p1 – p 2 – p 3 – p 4
p1 p 22 p 32 p 24
p 2 p 32 p 24 p12 2 = E0 + E0 + E0 2m 2m Total energy = 3E0 + E0 + E0 + E0 = 6E0
K. E1 =
v sin 22.
e=
2gh cos
apply conservation of momentum m 2gh sin = m vcos e
2gh cos × m = mv cos
tan = cot. e e = tan2
......(i) ......(ii)
on solving
RESONANCE
SOLN_Centre of Mass - 107
25.
m2vcos = 3vy
vy v cos
vy
Also e =
27.
v cos
=
2 . 3
R/2 ; = 30º R Both have equal mass it means along LOI particle transfer it velocity to disc which is vcos. sin =
so 29.
2 3
=
VD = Vcos = Vcos 30º =
3V 2
v r v mc
vr = v m – v c = v – u = 0. since vr = 0 so Ft = Fnet = m 30.
dv dt
vrdm = 0. dt
F + 0 = (m0 – t)
dv dt
F = (m0–t)
dv . dt
Neglecting gravity,
m
0 v = un m ; t
u = ejection velocity w.r.t. balloon.
m0 = initial mass
mt = mass at any time t.
m0 = 2n2. = 2n m0 / 2 34.
mv = nvm
v =
v n
time for first collisen is t1 = 2nd collisions
t2 =
L (2nd block) V
2 = 2t1 V
(3rd block)
so t = t1 + 2t1 + 3t1 + at1 ...........(n–1) t1. t = t1 [1 + 2 + 3] .......................(n–1)] =
36.
(n – 1) (n – 1 1) n(n – 1) = 2 2
f m 2 v1 = 0 + 2ad a=
vb12 =
so t =
L n (n – 1). 2V
for elastic collission e = 1
2F .d m
vb1 =
2Fd m
after collisin vb2 = 0.
RESONANCE
SOLN_Centre of Mass - 108
39. Pi = mv (i) Pf = (m + m) v at maximum conservation Pi = Pf v' = v/2 By energy compression
1 1 1 mv2 + 0 = (2m) (v)2 + kx2 2 2 2 at maximum compression k =
kx2 =
mv 2 2
x=
m v. 2k
1 v2 × 2m × k = mv'2 = mv2/4. 2 4
EXERCISE-3 1.
(A) If velocity of block A is zero, from conservation of momentum, speed of block B is 2u. Then K.E. of 1 2
block B = m(2u)2 = 2mu2 is greater than net mechanical energy of system. Since this is not possible, velocity of A can never be zero. (B) Since initial velocity of B is zero, it shall be zero for many other instants of time. (C) Since momentum of system is non-zero, K.E. of system cannot be zero. Also KE of system is minimum at maximum extension of spring. (D) The potential energy of spring shall be zero whenever it comes to natural length. Also P.E. of spring is maximum at maximum extension of spring. 2.
(A) (B)
(C) (D) 3.
Initial velocity of centre of mass of given system is zero and net external force is in vertical direction. Since there is shift of mass downward, the centre of mass has only downward shift. Obviously there is shift of centre of mass of given system downwards. Also the pulley exerts a force on string which has a horizontal component towards right. Hence centre of mass of system has a rightward shift. Both block and monkey moves up, hence centre of mass of given system shifts vertically upwards. Net external force on given system is zero. Hence centre of mass of given system remains at rest.
(a) The acceleration of the centre of mass is
F 2m The displacement of the centre of mass at time t will be a COM =
x =
1 Ft 2 a COM t 2 = 2 4m
Ans.
4&5 Suppose the displacement of the first block is x 1 and that of the second is x 2. Then, x =
mx1 mx 2 2m
or,
Ft 2 x1 x 2 4m 2
or,
x1 + x2 =
Ft 2 2m
...(i)
Further, the extension of the spring is x 1 – x 2. Therefore, x1 – x2 = x0 From Eqs. (i) and (ii),
RESONANCE
1 x1 = 2
Ft 2 x0 2m
and
...(ii)
1 x2 = 2
Ft 2 x0 2m
SOLN_Centre of Mass - 109
6.
During collision, forces act along line of impact. As collision is elastic and both the balls have same mass, velocities are exchanged along the line of impact. Therefore ball B moves with velocity VB||, that is equal to u cos 30°. Ball A moves perpendicular to the line of impact with velocity VA = u cos60°. Along the line of impact, ball A does not have any velocity after the collision. Therefore velocity of ball A in vector form after the collision y
VA 30°
VA||
u
x
30°
R 60°
VB||
= VA cos60°i + VA cos 30°j = (u cos 60°) cos60°i + (u cos 60°) cos 30°j
1 1 1 3 .j = 4. . . i + 4. . 2 2 2 2 7.
= (i 3 j) m/s
x
Using impulse-momentum equation for ball B N dt p f p i and as pi 0
N dt VB||
N dt p f
B
= (mu cos 30°) cos 30 i – (mu cos30°) cos 60° j = m. 4 . 8.
3 3 3 1 .i – m. 4 . . . . j = (3 m i – 2 2 2 2
m 3 m j) kg s
Suppose V2 is velocity of ball B along the line of impact and V1 is velocity of ball A along the line of impact, after the collision, as shown. Then
1 (Velocity of approach) = Velocity of separation 2
1 3 . u 2 2 = V2 – V1
.... (1)
A
V1
Conserving momentum along the line of impact m.
3 u = m. V2 + mV1 2
V2
B .... (2)
Solving and using u = 4 m/s V2 =
3 3 m/s 2
9 3 3 3 3 3 3 V2 cos 30 i cos 60 j = 4 i 4 2 2
j m/s
EXERCISE-4 PART - I 1.
vCOM =
=
m1v 1 m 2 v 2 m1 m 2
10 14 4 0 10 4
RESONANCE
= 10 m/s.
SOLN_Centre of Mass - 110
2.
Angular speed of particle about centre of the circle,
v2 v2 , = t = t R R
=
v p = (– v2 sin ˆi + v2 cos ˆj )
v v v p = v 2 sin 2 t ˆi v 2 cos 2 t ˆj
or
and
R
R
v m = v1 ˆj
linear momentum of particle w.r.t. man as a function of time is L pm = ( Vp – Vm ) v2 v2 = m v 2 sin t ˆi v 2 cos t v 1 ˆj R R
3.
(i)
X1 = V0 t – A (1– cost) Xcm =
(ii)
a1 =
m1x1 m 2 x 2 = V0 t m1 m 2 d2 x1 dt 2
m1 X2 = 0 t + m A (1–cos t) 2
Ans.
= – 2 A cos t
The separation X2 – X1 between the two blocks will be equal to 0 when a1 = 0 or cos t = 0 x2 – x1 =
m1 A (1–cos t) + A (1– cos t) m2
Thus the relation between 0 and A is, 8.
0 =
m1 m 1 A 2 m
(cos t = 0)
1 0 = m 1 A 2
According to Newton’s Law v 2 v1 e = u u 1 2 For elastic collision cofficient of restitution e = 1 so Statement - 1 is correct v 2 v 1 = u1 u 2
9.
Linear momentum is conserved in both elastic & non elastic collision but it’s not the explanation of statement -1 so it is not the correct explanation of the statement A. P2 p ˆi P1 p ˆi as there is no external force so momentum will remain conserved P1 P' 2 P1 P2 P1 P2 0 Now from option (A) P1 P2 = (a1 a 2 ) ˆi (b1 b 2 )ˆj c 1kˆ (B) P1 P2 = (c 1 c 2 ) kˆ (C) P1 P2 = (a1 a 2 ) ˆi (b1 b 2 )ˆj (D) P1 P2 = (a1 a 2 ) ˆi 2 b1 ˆj and it is given that a1 b1 c 1 , a2, b2, c 2, 0
in case of A and D it is not possible to get P1 P2 = 0 Hence Ans. (A) and (D)
RESONANCE
SOLN_Centre of Mass - 111
10.
11.
At point B there is perfectly inelastic collision so component of velocity to incline plane becomes zero and component parallel to second surface is retained velocity immediately after it strikes second incline V=
2gh cos 30
V=
45 m/s
=
2 10 3 ×
3 = 2
2 10 9 4
At point ‘C’
VC2 VB2 2gh VC2 = 45 + 2 × 10 × 3 VC = 12.
105 m/s
The block coming down from incline AB makes an angle 30° with incline BC. If the block collides with incline BC elastically, the angle of block after collision with the incline shall be 30°. Hence just after collision with incline BC the velocity of block shall be horizontal. So immediately after the block strikes second inclined, its vertical component of velocity will be zero.
13.
ycm =
m1y1 m2 y 2 m3 y 3 m4 y 4 m5 y 5 m1 m2 m3 m4 m5
ycm =
6m(0) m(a ) m(0 ) m( a) m(– a ) m m m m 6m
=
a . 10
14.
Since masses of particles are equal and collisons are elastic, so particles will exchange velocities after each collision. The first collision will be at a point P and second at point Q again and before third collision the particles will reach at A.
15.
from momentum conservation : 9m = (2m) V1 – (m)V2 9 = 2V1 – V2 ..... (1) e=
V1 V2 1 ......(2) 9
from eqn(1) and eqn(2) V1 = 6 m/sec. for second collision between second block and third block : (2m) 6 + m(0) = (2m + m) VC VC = 4 m/sec.
16*.
Since collision is elastic, so e = 1 Velocity of approach = velocity of separation So, u=v+2 .............(i) By momentum conservation : 1 × u = 5v – 1 × 2 u = 5v – 2 v + 2 = 5v – 2 So, v = 1 m/s and u = 3 m/s
RESONANCE
SOLN_Centre of Mass - 112
Momentum of system = 1 × 3 = 3 kgm/s Momentum of 5kg after collision = 5 × 1 = 5 kgm/s
1 So, kinetic energy of centre of mass = (m1 + m2) 2
Total kinetic energy =
17.
2h g
R= u
2
2
m1u 1 3 = 1 (1 + 5) = 0.75 J m m 2 6 2 1
1 × 1 × 32 = 4.5 J. 2
20 = V1
25 25 and 100 = V2 10 10
V1 = 20 m/s , V2 = 100 m/sec. Applying momentum conservation just before and just after the collision (0.01) (V) = (0.2)(20) + (0.01)(100) V = 500 m/s
18.
= 0.1
1 1 mu 2 = mg × 0.06 + kx2 2 2
1 × 0.18 u2 = 0.1 × 0.18 × 10 × 0.06 2 0.4 =
N 10
N = 4 Ans.
PART - II 14.
If initial momentum of particles is zero, then they loss all their energy in inelastic collision but here initial momentum is not zero. Principle of conservation of momentum holds good for all collision.
RESONANCE
SOLN_Centre of Mass - 113
TOPIC : RIGID BODY DYNAMICS EXERCISE-1 PART - I SECTION (A) A 1.
i = 0 = it +
t = 5 sec
= 50 (2) rad.
1 t2 2
(50) (2) = 0 +
1 (5)2 2
(50) (2) = 0 +
25 2
(50 )(2)(2) = 4 (2) = 4 rev/ se2 25 f = i + t f = 0 + 4(5) = 20 rev/ sec
=
SECTION (B)
B 1.
For first solid spheres IAB = Iam + Md2 7 2 2 MR2 + MR2 = MR 5 5 Similar way for second sphere
IAB =
IAB =
7 MR2 5
14 2 I = 2 IAB = MR 5
B 4.
mR 2 I0 = 2 0 = cm + md2 4R mR 2 = cm + m 3 2 cm
2
4R mR 2 = –m 3 2
2
2 MR 2 4R M ICM = 2 3
RESONANCE
SOLN_RIGID BODY DYNAMICS - 114
SECTION (C) C 2.
F1 = 2 ˆi – 5 ˆj – 6 kˆ
at point
at point
F2 = – ˆi + 2 ˆj – kˆ r0(–1,0,5)
r = (1 ˆi + ˆj + 0 kˆ ) – (– ˆi + 0 ˆj + kˆ ) 1
`
= 2ˆi ˆj – kˆ × 2ˆi – 5 ˆj – 6kˆ
r = 2ˆi ˆj – kˆ 1
= r F 1 1 1
= ( – 10 kˆ + 12 ˆj – 2 kˆ – 6 ˆi – 2 ˆj – 5 ˆi ) 1
= (–11 1 ˆi + 10 ˆj – 12 kˆ ) 1
2
Total
T
C 4.
T
= r F = ˆi ˆj kˆ × – ˆi 2ˆj – kˆ 2 2
T
1
2
= +
= – 14 ˆi 10 ˆj – 9kˆ
= 2ˆi ˆj – kˆ × 2ˆi – 5 ˆj – 6kˆ + ˆi ˆj kˆ × – ˆi 2ˆj – kˆ
(a) 0 = mg R/2 = mg
= mg 0
v 2 sin 2 2g
mv 2 sin 2 v 2 sin 2 = 2 2g
= (mv2 sincos) 0
(b)
= mgR 0 = (2mv2 sincos)
RESONANCE
SOLN_RIGID BODY DYNAMICS - 115
SECTION (D) D 2.
The F.B.D. of rod is as shown For rod to be in translational equilibrium N1 = P ....(1) N2 = W = mg ....(2) For rod to be in rotational equilibrium, net torque on rod about any axis is zero. Net torque on rod about B is zero
i.e.,
cos – N2 cos + P sin = 0 2 from equation (2) and (3) solving we get mg
P= D 3.
.......(3)
mg cot 2
For translational equilibrium Fx = 0 Fy = 0 N1 = f N2 = 75g + 24g = 99g = 990 N Rotational equilibrium = 0 (about any point) B = 0 N1 × 6 = 24g (5cos 37º) + 75g (8cos 37º)
4 4 ) + 75g (8 × ) 5 5 N1 × 6 = (96g + 480 g) N1 = 96g = 960 N f = N1 N2 = N1
N1 × 6 = 24g (5 ×
N1 96g 32 = N = 99g = 33 2 Ans.
990 N, 960 N ,
32 33
SECTION (E) E 1.
(a) Torque about hinge (m1g – m2g) = . 2
=
(m1 m 2 )g( / 2) 2
2
m1 m 2 2 2
2(m1 m 2 )g = (m m ) 1 2 =
2(6 3)10 10 = rad/sec2. 2(6 3) 3
RESONANCE
SOLN_RIGID BODY DYNAMICS - 116
(b)
If mass of rod is 3 Kg Torque about hinge (m1g – m2g)
' =
' =
= '' 2
(m1 m 2 )g 2 2 2 m 2 m1 m 2 3 12 2 2
2(m1 m 2 )g m m1 m 2 3 3
=
2(6 3)10 2 3 = 3 rad/s 2 6 3 3
For m1 block m1g – T1 = m1a m1 T1 = m1g 2
623 = 42 N 2 For m2 block T2 – m2g = m2a T1 = 60 –
T2 = m2g + m2 E 3.
323 = 30 + 2 2
T2 = 39 N
Let be the angular acceleration of the pulley system. For 6 kg block 6 g – T1 = 6 (2) .........(i) for 3 kg block T2 – 3g = 3 .........(ii) for pulley system 2T1 – T2 = = 3 .........(iii) From equation (i) and (ii) putting the values of T1 and T2. 2[6g – 12] – [3g + 3] = 3 12 g – 24 – 3g – 3 = 3 30 – 9g
=
90 = 3 rad/s2 Ans. 30
SECTION (F)
F 2.
initial positial
final position
Using Energy conservation Ki + Ui = Kf + Uf
1 = 2 2 2 = (1 + 2) 0 + 3mg
[ =
m 2 + m2] 3
RESONANCE
SOLN_RIGID BODY DYNAMICS - 117
1 3mg = 2 2 3g = F-4.
m 2 m 2 2 + 0 3
4 2 3
=
2 2 3g= 3 2
9g 4
Initial and final positions are shown below
5R 5mgR = Decrease in potential energy of mass ‘m’ = mg 2 4 2 R mgR Decrease in potential energy of disc = mg 2 = 4 2 Therefore, total decrease in potential energy of system
=
1 2 2 = moment of inertia of system ( disc + mass ) about axis PQ. = moment of inertia of disc + moment of inertia of mass
Gain in kinetic energy of system W here
5mgR mgR + = 3 mgR 2 2
=
2 2 mR 2 R 5R 15mR 2 m = + m = 4 4 8 4 From conservation of mechanical energy Decrease in potential energy = Gain in kinetic energy
2 16 g 1 15mR 2 3 mgR = = 8 5R 2 Therefore, linear speed of particle at its lowest point 5R 5R 16 g = v = or v = 5 gR 5R 4 4
SECTION (G) G 2.
– 3x 5 y 4 4
3x + 4y = 5 P = mv = 2 × 8 = 16 (kg – m/s) L = (5/4) × mv cos 37º L = 5/4 × 2 × 8 ×
4 = 16 kg m2/s 5
RESONANCE
SOLN_RIGID BODY DYNAMICS - 118
G 3. initial position
Final position
No external torque so
L = cont.
Li = Lf (i0 = f0) mr 2 mr 2 2 2 4 4 0 = ( + mr + mr ) 2 mr 2 2mr 2 0
G 5.
Angular momentum conservation about O = mvR MR 2 = mvR 2 MR = 2mv
v M
o
m
R
MR v = 2m With respect to bord man's rotation v + R velocity so in one rotation when velocity v + R angle taken by man (2). 2R t V R 2 Angular velocity bord is so at the same time angle covered by disc = . t R . R V
2R 4m MR M 2m R 2m
SECTION (H) H1
VA
= (Vcm – /2) = 50 – 5 × 5 = 25 m/s
VB
= Vcm 2
= 50 + 25 = 75 m/s H4
(a) vA sin = v0 cos vA =
v0 4v 0 = tan 3
4v 3v 0 4 0 v 0 sin v A cos 9v 0 16 v 0 5v 0 3 = (b) = = 15 3 5
(c) vx = vy =
v Ax vBx v = 0 2 2
1 v Ay v By = 2
2v 0 3
RESONANCE
SOLN_RIGID BODY DYNAMICS - 119
SECTION (I) I-3.
For linear motion : mg – T = ma For angular motion :
............(i)
mR 2 T.R. = 2
mR 2 For no sliping : a = R From equation (i), (ii) & (iii) T=
a= I-4.
............(ii) ............(iii)
2 g. 3
Let R & r be the radii of hemispherical bowl & disc respectively From energy conservation,
1 1 mv2 + 2 2 2 For pure rolling, v = r mg(R – r) =
1 1 2 v 1 mg(R – r) = mv2 + mr 22 2 r 3 mv2 4 From FBD of bottom : mg(R – r) =
N – mg =
2
...........(i)
mv 2 (R r )
...........(ii)
From equ. (i) & (ii), N= I-5.
7 mg. 3
Let v1 & v2 be minimum speed of ring of bottom & top of cylindrical part At top of path mv 22 (R – r ) for minimum speed N = 0 v22 = g (R – r) .......... (i) From energy conservation between bottom & top point of cylindrical part
N + mg =
1 1 1 1 mv12 + 12 = 2 mg (R – r) + mv22 + 22 2 2 2 2 For pure rolling 1 =
v1 v2 , 2 = r r
v 12 v 22 1 1 1 1 mv12 + (mr2) 2 = 2 mg (R – r) + mv22 + (mr2) 2 2 2 2 2 r r mv12 = 2 mg (R – r) + mv22 .......... (ii) from equation (i) & (ii) mv12 = 2 mg (R – r) + mg (R – r)
v1 =
3g(R – r )
RESONANCE
SOLN_RIGID BODY DYNAMICS - 120
I-6.
For linear motion, F = ma For angular motion,
..........(i)
2 2 F.R. = mR 5
=
5F 2mR
..........(ii)
1 2 at 2
= 0t +
8mR 1 5F 2 t t2 = 5F 2 2mR Distance covered by sphere during one full rotation
2 = 0 +
S = ut +
S=
1 2 at 2
=0 +
1 F 8mR 2 m 5F
4R 5
SECTION (J) J 2.
(a) Pi = m2v Pf = (m1 + m2) Vcm m2v = (m1+m2) Vcm
m2v Vcm = m m 2 1 (b) v1 = (u – Vcm) m1u m 2u V1 = v – m m = m m 1 2 2 1 – m1u (c) V1 = – Vcm = m m 2 1
L m1(0) m 2 m 2L 2 (d) Xcm = = 2(m1 m 2 ) (m1 m 2 )
L1 =
m 2L L – 2(m m ) 2 1 2
L1 =
1 2
m1L m1 m 2
momentum of particle m 2m12u m1L m 2u 1 1 m ( u V ) L m u Pi = 2 = cm 2 2 2(m1 m 2 ) m1 m2 2(m1 m 2 ) 2
Momentum for rod = m1Vcm L cm
m1L m2 u 2 (m1 m 2 ) 2
(e) For particle : 2
1 =
m2L2
m 2m1 L2 = 4(m1 m 2 )2
= 1 + 2 =
RESONANCE
m L2 m2 L 2 = 1 m1 12 2 ( m m ) 1 2
2
m1(m1 4m 2 )L2 12(m1 m 2 ) SOLN_RIGID BODY DYNAMICS - 121
(f) Velocity of centre of mass
m2v = m m 2 1 Using angular momentum conservation m2v × Lcm = cm m1L = m 2u 2(m m ) = cm . 1 2 m1L m1(m1 4m 2 )L2 = m 2u 2(m m ) = × 12(m1 m 2 ) 1 2
6m2 v = (m 4m )L . 1 2
SECTION (K) K 1.
Force balance N= mg cos f = mg sin Torque balance (about centre of mass) Nx = f ×
a mg sin amg sin a tan a = and x = = 2 2mg cos 2 2
Torque of normal force Nx = mg sin
a 2
PART - II SECTION (A) A 1.
0 = 3000 rad/min 3000 rad/sec = (50 rad/sec) 60 t = 10 sec f = 0 f = 0 + t = 50 – (10) = 5 rad/sec2
0 =
= o t +
1 t2 2
1 (–10) (10)2 2 = 500 – 250 = 250 rad
= (50) (10) +
A 3.*
Sphere is rotating about a diameter so , a = R but, R is zero for particles on the diameter.
SECTION (B) B 3.
B > A B > A so, If the axes are parallel
B 6.
Moment of inertia of the elliptical disc should be less than that of a circular disc having radius equal to the major axis of the elliptical disc. Hence (D)
RESONANCE
SOLN_RIGID BODY DYNAMICS - 122
B 7.
0 = 1 + 2 2
2
m / 2
m / 2
2 3
0 =
2 3
+
m 2 12
=
SECTION (C) C 1.
F2 = –2i – 3j – 4k F1 = 2i + 3j + 4k Net force Fnet F1 F2 0 the body is in translational equilibrium.
r2 = i
r1 = 3i + 3J + 4k
= (3 ˆi 3ˆj 4kˆ ) × (2ˆi 3ˆj 4kˆ )
1 = r1 × F1
1 = 9kˆ – 12ˆj – 6ˆj 12ˆi 8ˆj – 12ˆi
1 = – 4ˆj 3kˆ 2 = r2 × F2 = ( ˆi ) × ( – 2ˆi – 3ˆj – 4kˆ ) 1 2 –4ˆi 3kˆ – 3kˆ 4ˆj 0
= –3 kˆ + 4 ˆj
body in rotational equilibrium C 3.
ˆj ˆ ˆ F = 2 i + 3 – k at point (2,–3,1) torque about point (0, 0, 2)
r = 2ˆi – 3ˆj kˆ – 2 kˆ
= r × F = (2ˆi – 3ˆj – kˆ ) (2ˆi 3ˆj – kˆ )
= (6 ˆi 12kˆ )
= (6 5 )
SECTION (D) D 2.
N1 = N2 , N1 + N2 = mg , A = o
3 N2 – 4 N1 – Hence =
3 mg = o 2
1 Ans. 3
Aliter Using force balance f1 = –N1 f2 = N2
N1 + f2 = mg (1) N2 = f 1 N2 = N1 (2)
Using aq (1) N1 + N2 = mg N1 + N1 = mg
mg N1 + 2 1 torque about point B B = 0 f1 × 4 + mg (5/2 cos 53º) = 3N1
RESONANCE
For rotational equilibrium
SOLN_RIGID BODY DYNAMICS - 123
4N1 +
3mg = 3N1 2
3mg = (3 – 4) 2
mg 1 2
3mg = (3 – 4) N1 2
3 – 4 3 = 2 1 2
3 + 32 = 6 – 8 32 + 8 – 3 = 0 32 + 9 – – 3 = 0 3( + 3) –1 ( + 3) ( = 1/3)
w1
w D4
x
–x
weight of object = w
w ( – x) = w1x ...........(i) If weight is kept in another pan then : w2( – x) = wx ...........(ii) By (i) & (ii)
w w1 = w2 w w=
w2 = w1 w2
w 1w 2 .
SECTION (E) E-3. 2 N = m 2
E-4.
Initial velocity of each point onthe rod is zero so angular velocity of rod is zero. Torque about O = 20g (0.8) =
m 2 3
20g (0.8) =
20 (1.6 )2 3
3g = = angular acceleration 3 .2
=
15 g 16
SECTION ( F ) F 2.
By energy conservation 2
1 7 m 2 2 mg = . 4 2 48 0 =
7 ml 2 48
RESONANCE
[ (about O)
=
24g 7
m 2 m ] = 12 4
Ans.
SOLN_RIGID BODY DYNAMICS - 124
SECTION ( G ) G 3.
x = v0 cos 45º × t = = mgx =
L=
mgv 0 t
mgv 0 2
v0t
=
2
2 dL dt
v0 / g
t dt =
0
mv 30 2 2g
G 5.
external torque ext = 0 11 = 22 when he stretches his arms so 1 < 2 then (1 > 2) so, (L = constant)
G 7.*
External force will act at hinge so linear momentum of system will not remain const. but torque of external
force is zero about hinge so L = const., collision is elastic so K.E = const.
SECTION (H) H 3*.
for pure rolling V = R VA = 2V VB = 2 V (VC = 0)
SECTION (I) -3.
mg sin – f = ma mg sin – f a= .......(i) m a is same for each body.
f.R =
For solid sphere k2 =
f .R mk 2
2 R2 is minimum there fore is maximum hence, k.E. for solid sphere will be max 5
at bottom. -5.
mg sin – f = ma
mg sin – f m a is equal for each body so all the object will reach at same time. a=
-7.
There is no relative motion between sphere and plank so friction force is zero then no any change in motion of sphere and plank.
SECTION (J) J-2.*
at the moment when ring is placed friction will act between them due to relative motion. Friction is internal force between them so angular momentum of system is conserved. I11 = I22 mR 2 2 mR 2 0 = 2 mR 2
RESONANCE
=
0 3
SOLN_RIGID BODY DYNAMICS - 125
J-3.
Conservation of angular momentum about C.O.M. of m and loop of mass m gives 2 2 R R mVR m R 2 m m = 2 2 2
V = 3 R
=
V 3R
J-4.
velocity of COM after collision is V friction will act such that = o at some intant after some time (V = R)
SECTION (K) K-2.
For no slipping µmg cos mg sin For toppling
.........(1)
h a mg cos. 2 2 for minimum µ (by dividing) mg sin
µ.
.........(2)
2 2 = a h
µmin =
a . h
[ Ans.: a/h ] Sol.(2) If f > mg sin mg cos > mg sin ( > tan ) block will topple before sliding torque about point A A =0
a mg sin h 2 = mg cos 2
tan = a h > a h
If > tan (block will slide)
N
a a/2
K-3. b/2
mg The block will not topple if mg acts from within the base area of the block. So,
a b cos 2 2
RESONANCE
cos
b a
SOLN_RIGID BODY DYNAMICS - 126
EXERCISE-2 PART - I 3.
linear density =
m
m
dm = dx
AB =
m
4.
0
dm · x
2
m
2
dx · ( x cos 45)
=
o
m x2 dx = 2 2
x3 3
m2 6 0
dm = (2xdx) R
=
dm· x
2
=
(2xdx).x
2
o
R
= 2 x 3 dx
0
R
R
3 = 2 x ·( x )dx = 2
0
o
R
x 3 dx x 4 dx
o
R 4 R 5 = 2 4 5 5.
h y R r
r=
R y h
dm = (r2dy) dAB =
1 (dm) r2 2 h
AB =
1
2
2 r dy r
2
y 0
=
R 4 2 h4
h5 5
m 3 4 m = = R h . 1 2 ........... mR2 10 10 R h 1 2 R h 3 3
RESONANCE
SOLN_RIGID BODY DYNAMICS - 127
m 2 R
7.
=
For small ring friction force dfr = K(2rdr)g Torque of the friction = (–rdfr) = – 2krgr2dr R
= –2krg
2
r dr = –
2 krgR3 3
0
For rotation about z-axis ( = ) –
2 ( R 2 )( ) 2 krgR3 = R 3 2
–4kg 3R
=
From equation of motion = 0 + t –4kg
0 = 0 + 3R t 8.
3R
m
0 t = 4kg
m
= (a = R)
m1 = x = x m 1g – T = m 1a
............ (i)
2
MR
T R = 2 + (m – m1) (R2) MR + (m – m1) R 2 Ma m 1g – – (m – m1) a = m1a 2
T=
Ma
............(ii) Ma + (m – m1) a 2 Ma m 1g – – ma + m1a = m1a 2
T=
m1g = 2 ma 2m1a = (M 2m)
10.
mx 2 g = (M 2m)R
2mgx
(M 2m)R
Using energy conservation mgh =
1 1 1 kx2 + 2 + mv2 2 2 2
String does not slip So(V = r) mg x =
1 1 v2 1 kx2 + 2 + mv2 2 2 r 2
x = 0.1m m = 11 kg 11 × 10 × 0.1 =
= 0.1 kg – m2 r = 0.1 m
K = 100 N/m
V2 1 1 1 × 100 × (0.1)2 + × 0.1 × 11 × V2 2 + (0.1) 2 2 2
22 = 1 + 10 V2 + 11 V2 21 V2 = 21 V = 1 m/s
RESONANCE
SOLN_RIGID BODY DYNAMICS - 128
11.
(a)
Energy conservation loss in P.E. = gain in rotational K.E.
(b)
1 m2 2 (1 – cos ) = 2 3 2
mg
2 =
=
= mg
sin = 0 2
mg
m2 sin = 2 3
=
3g sin 3g = sin 2 4
3g (1 – cos )
3g (1 cos )
fy = may
mg – N2 = mat sin
N2 = mg – mat sin = mg – m
3g sin2 4
2 N2 = mg 1 3 sin
4
fx = max
N1 = ma1 cos =
Ans.
Normal reaction = where N1 =
12.
m3 g sin . cos 4
N12 N22
3mg sin cos 4
N2 = mg 1
3 sin2 4
(a) About the axis of rotation of rod, the angular momentum of the system is conserved velocity of the flying bullet is V
2
3
M mv = m2
=
3mv mv (m <<< M) = M M m 3
RESONANCE
................. (i)
SOLN_RIGID BODY DYNAMICS - 129
conservation of mechanical energy of the system (rod + bullet) 1 2
2 M 2 m 2 = (M+m)g (1– cos ) 2 3
——(ii)
From (i) and (ii) V=
M m
2g 3
sin 2
(b) P = m () M 2 – mv From v and w g 1 mv = M 6 sin 2 2
P =
M 2
2 mvx = 3 mx ´
3mvx
´ =
M2
final momentum
pf = mx ´ +
y´ 0
M M dy = ´ 2
3x
3 x mv 2
p = pf – pi = mv 2 – 1 = 0 5.
=
x
2 3
a = R (Pure rolling) v = u + at (v = at) For pure rolling = (v = R) (a) After 2 sec VA = V + R = 2V = 2at VB = Vi + R (–j) = ( 2 V) = V0 = V – R = 0
2 at
(b) a = R aA = 2a ˆi + 2R (– ˆj ) 2R 2 aA = 2a ˆi + (– ˆj ) R
4a 2 t 2 (2a )2 R
aA =
4a 2
aA = 2a 1
2
16a 4 t 4 R2 4a2 t 4 R2
aB = (a – 2R) ˆi + (R) (– ˆj ) 4a2t 2 ˆ aB = a – i + a (– ˆj )
R
aC = 2R aC =
a2 t 2 v2 = R R
RESONANCE
SOLN_RIGID BODY DYNAMICS - 130
16.
V = R = (For pure rolling) (linear acceleration = 0) rolls with out slipping so acc. only centripetal acc. v2
aA = R V = R VA = (V – V cos ) ˆi + V sin ˆj VA = ( v – v cos è)2 ( v sin è ) j VA = (2V sin /2) = t t ds = 2 V sin \2 = 2 V sin 2 dt s
2 /
ds =
0
18.
O
t 2v sin dt = 8v = (8R) 2
Kinetic energy can become zero only for the case shown in figure ; Torque equation : (mg).R =
MR 2 . 2
Therefore , t =
=
0R 0 = 2g
2g R ............(1)
For translational motion
v0 t = g ............(2) v0 0R = 2g g
From (1) & (2)
21.
0 = (i) (a)
2(10) 2v 0 = = 100 rad/sec. 0. 2 R
mg – 4f = ma fR = =
a R
Ans.
......... (i) M
M
fR2 = a fR2 =
MR 2 a 2
Ma f= 2 4Ma ma = (2M + m)a mg = 2 M = 2kg, m = 5 kg
a=
RESONANCE
M
M
f
f
5g () 9
SOLN_RIGID BODY DYNAMICS - 131
(b)
If
M=0 f=0 mg = ma a = g()
(c)
m=0 mg = (2M + m) a 0=a a=0
(a)
(m + 4M)g – 4f = (m + 4M)a Torque about centre of disk ( = a / R)
(ii)
f.R=
a MR 2 . R 2
Ma f= 2 (m + 4M) g – 2Ma = (m + 4M) a (m + 4M) g = (m + 6M) a (5 + 8) g = (5 + 12) a 13 g () a= 17
(b)
If
M=0 mg = ma a = g()
22.
If
M=0 4Mg = 6Ma a=
2g () 3
(a) mg sin – f = ma –(i) Torque about com fR = I fR =
2 mR2 . 5
f=
2 m (R) 5
f=
2 2 m (R) = ma 5 5
For pure rolling a = R
mg sin –
2 ma = ma 5
mg sin =
2 7ma ma + ma = 5 5
5g sin a= 7 5g sin 2mg sin 2 = m 7 5 7 f = N
f=
2mg sin 2 f = 7mg cos = tan 7 N (b) torque about com =
f.R =
2 mR2 . 5
NR =
2 mR2. 5
RESONANCE
SOLN_RIGID BODY DYNAMICS - 132
1 2 tan (mg cos ) R = mR2 7 5
5g sin 14R mg sin – f = ma =
mg sin –
1 tan . mg cos = ma 7
mg sin –
1 mg sin = ma 5
6 a = g sin 7 1 2 1 2 K.E mv 2 2
v2 = u2 + 2as v2 = 0 + 2
6 g sin 7
12 v2 = g sin 7
s = ut + =0+
1 2 at 2 1 6 × g sin t2 2 7
7 t = 3g sin
½
1 12 12 2 5g sin K.E. = m g sin mR 2 7 25 14R
K.E. =
2
7 3g sin
5 6 mgsin + mgsin 84 7
11 mgsin. 12 Given mass of disc m = 2Kg FBD of any one disc is K.E. =
23. (i)
and
radius R = 0.1 m
z
Truck
2
a = 9m/s
×
y
RESONANCE
x
SOLN_RIGID BODY DYNAMICS - 133
Frictional force on the should be in forward direction.
a0 f
P
Let a0 be the acceleration of COM of disc the angular acceleration about its COM. Then – f Q 2
a = 9m/s
f f = m 2
a0 =
......(i)
f. R 2f 2f = = = = 10 f 1 I mR 2 0.1 mR 2 2 Since there is no slipping between disc and truk therfore. Acceleration of point P = Acceleration of point Q a0 + R = a
=
f + (0.1)(10 f) 2
or
or
3 f=a 2
.......(2)
f =
2a 2 9 .0 = N 3 3
f = 6N Since this force is acting in positive x-direction. Therefore, in vector form f =( 6 ˆi ) N
(ii)
= r ×f Here f = ( 6 ˆi ) N ( for both the discs ˆ r = r = 0.1 ˆj – 0. 1 k and P
1
Ans. 3 (i)
20cm = 0.2 m
2
1
z
ˆ rQ = r2 = 0.1 ˆj – 0. 1 k and
O y
x
Therefore, frictional torque on disk 1 about O (centre of mass ) – = r × f = ( –0.1 ˆj – 0.1 kˆ ) × ( 6 ˆi ) N-m
Q
P f
f
= ( 0.6 kˆ – 0.6 ˆj ) or
ˆ ˆ ˆ ˆ r1 = 0.6 ( k – j ) N-m 0.6 ( k – j )
and
| r |=
Similarly,
ˆ ˆ ˆ r1 = r2 × f = ( 0.1 j –0.1 k ) × ( 6 i ) N-m
1
(0.6)2 (0.6)2 = 0.85 N-m
ˆ ˆ ˆ ˆ r1 = 0.6 ( – j – k ) 0.6 ( k – j ) and
| r2 | = | r1 | = 0.85 N-m
RESONANCE
Ans. 3 (ii)
SOLN_RIGID BODY DYNAMICS - 134
(a) The cylinder rotates about the point of contact. Hence, the machanical energy of the cylinder will be conserved i.e.,
co s
R
R
24.
R
(PE + KE ) 1 = ( PE + KE ) 2
1 1 I 2 + mv 2 2 2 but = v / R ( No slipping at point of contact. )
mgr + 0 = mgr cos +
V’
1 I= mv 2 2
and Therefore,
mgR = mgR cos +
1 1 mR 2 2 2
or
3 2 v = gR ( 1 – cos ) 4
or
v2 =
v2 1 2 R 2 + 2 mv s co g m
4 gR ( 1 – cos ) 3
V
N=0 mg
2
4 v or = gR ( 1 – cos ) ...........(1) 3 R At the time of leaving contact, normal reaction N = 0 and = c hence,
mg cos =
mv 2 R
v2 = g cos R From Eqs. (1) and (2)
or
...........(2)
4 g ( 1 – cosc ) = g cos c 3
or
7 cos c = 1 4
(b)
v=
or
4 gR (1 cos ) 3
cosc = 4 / 7
or
c = cos – 1 ( 4 / 7 )
[From Eq. (1)]
At the time of losing contact cos = cos c = 4 / 7
v=
4 4 gR 1 3 7
v=
RESONANCE
4 gR 7
SOLN_RIGID BODY DYNAMICS - 135
Therefore, speed of COM of cylinder just before losing contact is Therefore, rotational kinetic energy
or
KR =
4 gR 7
1 I 2 2
4 1 1 mR v 2 1 1 = mv 2 = m gR 2 2 7 R 2 4 4
KR =
mgR 7 Now, once the cylinder losses its contact, N = 0, i.e., the frictional force , which is responsible for its rotation, also vanishes. Hence, its rotational kinetic energy now becomes constant, while its translational kinetic energy increases. Applying conservation decrease in gravitational PE = Gain in rotational KE + translational KE Translational KE (K T) = Decrease in gravitational PE – K R or
KR =
or
KT = (mgR) –
mgR 6 = mgR 7 7
From Eqs. (3) and (4)
KT KR
28.
(i) (a) CM
6 mgR 7 = mgR 7
KT KR = 6
or
C2 2 C2 M = 5 MC = 12 4 12 12 4
A = CM + Mx 2 A =
5 MC 2 5 M C2 + 16 12 4
A
mg ×
=
20 MC 2 48
6g C = A = 5 C 2
(b) acm = x =
6g 5C 5C 4
6g = 5C x
C 6g 6g ax = – acm cos =– 5 C x . 4 . x = – 20 = – 0.3 g 6g C/ 2 ag = – acm sin = 5 C x . x = – 0.6 g
a 0.3 g ˆi 0.6 g ˆj (ii) (a) Mg – T = M acm
.... (1)
CM
5 MC2 C 5 MC T= .... (2) = 2 24 48 As aA = 0 (we know : acc. along the string is zero) acm – x cos(90 – ) = 0 T×
RESONANCE
SOLN_RIGID BODY DYNAMICS - 136
acm = x sin = ax. acm = T=
C 2
.... (3)
5 MC 2 a cm 24 C
Mg = M acm + =
C 2x
=
.... (4)
5 M a cm 12
17 M a cm , 12
acm =
2 a cm C
(a) =
5 M a cm 12
=
12 g 17
24 g 17C
(iii) (a) cm =
mg C 2 2
= I cm
5 MC2 mg C = 4 48 (b)
FA =
12 g = 5C
Mg 2
Mg – FA = m acm 31.
acm =
g = 0.5g 2
Coefficient of restitution
m m
V2 sin 2 e== V1
sin ) 2 angular momentum about point A (V1 = V2 +
Li = mV1
..... (i)
sin 2
Lƒ = LCM + LA
= ICM mV2 sin 2
Li = Lƒ
m 2 sin = – mV2 sin 2 2 12 Put equation (i) in (ii) equation mV1
..... (ii)
mV1
m 2 sin sin = – m V1 sin 2 2 12 2
mV1
sin 2
=
m 2 m 2 – mV1 sin + sin2 2 12 4
2 2 mV1 sin m m sin2 V sin = + sin2 1 12 4 12 4
RESONANCE
V1 (12 sin ) 3 sin 2 1
SOLN_RIGID BODY DYNAMICS - 137
PART - II 1.
The given structure can be broken into 4 parts For AB =
= CM + m × d2 =
AB
4 2 ml 3
For
m2 3 For composite frame : (by symmetry)
BO
=
4m 2 m 2 10 2 = 3 3 = ml 2.] 3
= 2[AB + OB] 3.
m 2 5m 2 ; 12 4
M of the system w.r.t an axis to plane & passing through one corner 2 2 ML M 3 L ML2 ML2 = + + 12 2 3 3
ML2 3ML2 2ML2 = + 12 4 3
=
3 2ML2 3ML2 10 ML2 18ML2 + = = = ML2 2 3 3 12 12
Now
[ Ans.: 4.
3 ML2 = 3k 2M 2 2
k=
2
]
= 1 + 2 + 3
3 mr2 2
1 = 2 = 3 =
mr 2 2 7 mr2 2
= 1 + 2 + 3 =
Moment of inertia = 3mk2 3mk2 =
5.
7 mr2 2
where k is radius of gyration.
k=
Taking mass of plate m =
7 r 6
M 6
Then M of two plates through which the axis is passing =
m a2 m a2 ×2= 6 3
M. of 4 plates having symmetrical position from the axis 2 m a2 m a2 a m = 4 × 12 =4× 3 2
Total M =
4 m a2 m a2 5 m a2 + = 3 3 3
RESONANCE
using
M =m 6
= M =
5 Ma 2 18 SOLN_RIGID BODY DYNAMICS - 138
6.
Taking cylindrical element of radius r and thickness dr M
dm =
(R 22 R12 ) R2
AB =
d e =
dm r 2 =
(R
R1
× (2r dr)
2M 2 2
R12 )
.r 3 dr = 1 M (R 2 R 2 ) 2 1 2
Using parallel axis theorem IXY =
1 M M (R 22 R12 ) + MR22 = (3R 22 R12 ) 2 2
10. For rotational equilibrium = N2 × 6 4 N1 : N2 = 4 : 3
N1 ×
11.
Balancing torque about the centre of the rod :
– N2 . =0 4 4 N 1 = N 2. N1 .
12.
Fnet ( 400 100 )ˆi (200 200 )ˆj = 300 ˆi 400 ˆj | F | = 500 N 300 = 37° Angle made by Fnet with the vertical is = tan–1 400 also = 500 R therefore point of application of the resultant force is at a distance R from the centre. Hence (C).
14.
Immediately after string connected to end B is cut, the rod has tendency to rotate about point A. Torque on rod AB about axis passing through A and normal to plane of paper is 3g m 2 = mg = 2 2 3 Aliter : Applying Newton’s law on center of mass mg – T =ma .....(i) Writing = I about center of mass
T
m 2 = 2 12
2 From (i) , (ii) and (iii) Also a =
=
....(ii) ....(iii)
3g 2
RESONANCE
SOLN_RIGID BODY DYNAMICS - 139
16.
For rigid body separation between two point remains same. v1 cos60° = v2 cos30°
v1 = 2
3 v2 2
v 2 sin 30 v 1 sin 60 d
disc =
= 18.
v1 =
v 2 3 3v 2 2d
=
3 v2 v2 3v1 2 2 d
=
2v 2 v2 = 2d d
disc =
v2 d
When ball at maximum height block and ball has equal velocity So Using momentum conservation Pi = mv Pf = 2mv0 (v0 final velocity) Pi = Pt mv = 2 mv0 V
V0 = 2 Using energy conservation 1 2 1 1 1 + mv2 = 2 + 2mv02 + mgh 2 2 2 2
( = mR2) v = R 1 1 mv2 = 2mv02 + 2mgh 2 2
v2 – 2 20.
2 h` v 4g
v2 = 2gh 4
As torque = change in angular momentum F t = mv (Linear) 2 F t = m 2 12 Dividing : (1) and (2)
and
2= Using
12v
=
..... (1)
(Angular)
..... (2)
6v
S = ut :
Displacement of COM is
Dividing
6v t = t = 2
2x = 6
x=
and
12
x = vt
Coordinate of A will be , 0 12 2
Hence (D). 24.
=
dL =4 dt
from figure
r 2 2 m
Hence = rF 4 = 2 2 .F
RESONANCE
F
=
2 N
Ans.
SOLN_RIGID BODY DYNAMICS - 140
28.
By conservation of angular momentum about hinge O. L = 2 Md2 d d m mv = 12 2 2
mvd 3 md2 2 4 30.
mvd md2 md2 2 2 4
2v 3d
FBD for sphere & block
fr mg a1 = = m m a1 g ˆi arel a1 a 2 2g ˆi arel = 2g. 31.
fr mg a2 = = m m a 2 g ˆi
m
a1 fr
fr m
a2
Using Energy conservation, (at maximum distance V = 0 V0 = 0) 1 Kx2 = (mg x sin ) 2 2mg sin K
x= 33.
Since the two bodies have same mass and collide head-on elastically, the linear momentum gets interchanged. Hence just after the collision 'B' will move with velocity 'v0' and 'A' becomes stationary but continues to rotate v0 at the same initial angular velocity R
.
Hence, after collision. (K.E.)B =
1 mv 02 2
2 (K.E.)B 3 12 1 2 v 2 = mR . 0 Hence (D). ( K.E.) A 2 23 2 R Note : Sphere 'B' will not rotate, because there is no torque on 'B' during the collision as the collision is headon.
and
35.
(K.E.)A =
Decrease in PE = Increase in rotational K.E
2mg.
1 – mg. = . 2 2 2 2
=
1 2
2 2m m. 2 4 4
3m 2 2 mg 1 3m 2 = . .= 2 2 8 4 =
4g 3
[ Ans.: (a) V = g / 3 ,
RESONANCE
and
v = r =
2
4g = 3
g 3
= 4g / 3 ]
SOLN_RIGID BODY DYNAMICS - 141
36.
Just before collision Between two Balls potential energy lost by Ball A = kinetic energy gained by Ball A.
mg
h 1 1 2 2 = cm mv cm 2 2 2 2
1 2 1 2 v cm 2 mv cm = mR 2 5 R 2
=
1 1 mv 2cm + mv 2cm 5 2
1 mgh 5 mgh = mv 2cm = mv 2cm 5 7 7 After collision only translational kinetic energy is transfered to ball B
So just after collision rotational kinetic energy of Ball A = 39.
Torque about COM f.R = ·
(a = R)
mR 2 mR 2 f.R = a = 2 ·R 2
41.
1 mgh mv 2cm = 5 7
Here, u = V0, 0 =
ma f 2
V0 2R
At pure rolling ; Ff V = V0 – t m
&
V0 Ff V t = 2R m.R R
V0 – V = V +
Ff .R = ) mR 2
V0 2
V0 V V= 0 Ans. 2 4 As the disc is in combined rotation and translation, each point has a tangential velocity and a linear velocity in the forward direction. From figure vnet (for lowest point) = v – R= v – v = 0.
42.
(In pure rolling V = R) ( =
2V =
and Acceleration =
v2 v2 +0= R R
(Since linear speed is constant) 43.
Since there is no slipping at any interface, the velocities of bottom and upper most point of lower and upper cylinder are shown in figure. Angular velocity of upper cylinder =
2V V 3V = 2R 2R
Angular velocity of lower cylinder =
V 0 V = 2R 2R
The ratio is
3 1
RESONANCE
SOLN_RIGID BODY DYNAMICS - 142
44.
For maximum a, normal reaction will shift to left most position. for rotational equilibrium P = = 0 [in frame of truck] ma
b = mg 2 2
a=
gb
45.
Torque about point A
TA = Fr d 2 + F2 (d)
3d 1A = Fr 4 + F1 (d)
(F1 + F2)
3d d + F2d = (F1 + F2) 4 + F1d 2
3 3 F1 F2 + F2 = 4 F1 4 F2 F1 2 F2 3 3 F1 – F1 – F1 = 4 F1 F2 – 2 2 4 F –F1 –F – F1 = 2 – 2 2 4 4 5F1 3F2 = 4 4
5F1 = 3F2 F1 3 F2 = 5 .
47.
By angular momentum conservation ; L = mv
R + mvR = 2mR2 2
3 mvR = 2mR2 2 3v 4R Also at the time of contact ; =
mgcos – N =
mv 2 R
N = mg cos –
mv 2 R
RESONANCE
SOLN_RIGID BODY DYNAMICS - 143
when it ascends decreases so cos increases and v decreases. mgcos is increasing and
mv 2 is decreasing R
we can say N increases as wheel ascends. 48.
Torque about point A ( mg) R =
2 mR 2 . 5
A
5 g = 2 R
v0
w0 mg
v = u + at 0 = v0 – gt
v t = 0 g 0 =
= 0 + t
0 = 0 –
v0 5g . g 2P
5v 0 2R
EXERCISE-3 1.
Since all forces on disc pass through point of contact with horizontal surface, the angular momentum of disc about point on ground in contact with disc is conserved. Also the angular momentum of disc in all cases is conserved about any point on the line passing through point of contact and parallel to velocity of centre of mass. The K.E. of disc is decreased in all cases due to work done by friction. From calculation of velocity of lowest point on disc, the direction of friction in case A, B and D is towards left and in case C is towards right. The direction of frictional force cannot change in any given case.
2.
(A) Speed of point P changes with time (B) Acceleration of point P is equal to 2x ( = angular speed of disc and x = OP). The acceleration is directed from P towards O. (C) The angle between acceleration of P (constant in magnitude) and velocity of P changes with time. Therefore, tangential acceleration of P changes with time. (D) The acceleration of lowest point is directed towards centre of disc and remains constant with time
3.
Let the angular speed of disc when the balls reach the end be . From conservation of angular momentum
4.
0 1 1 m 2 m 2 mR2 0= mR2 + R + R or = 3 2 2 2 2 The angular speed of the disc just after the balls leave the disc is 0 3 Let the speed of each ball just after they leave the disc be v. From conservation of energy
=
1 1 1 1 1 m 2 1 m 2 2 mR 2 2 + v + v mR 02 = 2 2 2 2 2 2 2 2 solving we get v= NOTE : v = 5.
2R0 3
(R )2 v r2
;
Workdone by all forces equal Kf – Ki =
RESONANCE
vr = radial velocity of the ball 1m mR 2 20 v2 = 2 2 9
SOLN_RIGID BODY DYNAMICS - 144
6 to 8 The free body diagram of plank and disc is Applying Newton's second law F – f = Ma1 .... (1) f = Ma2 .... (2) 1 FR = MR2 .... (3) 2 from equation 2 and 3 R a2 = 2 From constraint a1 = a2 + R a1 = 3a2 .... (4)
3F F and = 4M 2MR If sphere moves by x the plank moves by L + x. The from equation (4) Solving we get a1 =
L 2 The moment of inertia about both given axis shall be same if they are parallel. Hence statement-1 is false. L + x = 3x
9.
or
x=
10.
As x increases, the required component of reaction first decreases to zero and then increases (with direction reversed). Hence statement-1 is false.
11.
For a disc rolling without slipping on a horizontal rough surface with uniform angular velocity, the acceleration of lowest point of disc is directed vertically upwards and is not zero( Due to translation part of rolling, acceleration of lowest point is zero. Due to rotational part of rolling, the tangential acceleration of lowest point is zero and centripetal acceleration is non-zero and upwards). Hence statement 1 is false.
12.
The acceleration of any point on the disc rolling with uniform angular velocity in the frame of ground is equal to centripetal aceleration of that point in the frame of centre of mass of disc. Hence Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
13.
Let v be the speed of centre of mass, be its angular speed and R be the radius of disc respectively. v = R, because the disc rolls without slippping. The velocity of any point P on disc distant x from centre O due to translation is v T = (vT = v) in horizontal direction and its velocity due to rotation is v R (vR = x) as shown. Since the magnitude of v R is less than or equal to magnitude of v T , the horizontal component of v R cannot cancel v T to make velocity of any point P vertically upwards. Hence statement-1 is true. In statement-2 R may be more than v. Hence condition of statement-1 may not be satisfied. Otherwise statement-2 is true.
14.
The applied horizontal force F has tendency to rotate the cube in anticlockwise sense about centre of cube. Hence statement-2 is false.
15. (i)
=
I
= F × r : Torque is same in both the cases but moment of inertia depends on distribution of mass from the axis.Distribution of mass in both the cases is different.Therefore,moment of inertia will be different or the angular acceleration will be different. =
I
RESONANCE
= F × r : SOLN_RIGID BODY DYNAMICS - 145
(ii)
S1 : The line of action of each action reaction pair is same. Therefore magnitude of couple of each such pair is zero. Hence net torque of all internal forces is zero. S2 : The direction of angular acceleration and angular velocity shall be opposite if the body is slowing down. S3 : If net torque on a rigid body is zero, its angular velocity will be constant. The constant may necessarily not zero. S4 : Since the centre of mass is fixed, that is, at rest; hence net force on rigid body is zero. Therefore torque on this rigid body about any point is same.
(iii)
I 1 1 = I 2 2
(iv)
I1 2= I .1 2
MR 2 2 = MR 2 M R 2 2 4 . 2
4 = 5
KR In case of ring : K = 1 T In case of pure rolling or
KR = KT =
K 2
1 K (0.3) v1 2 = 2 2
.........(1)
2 1 (0.4) v22 = K 3 2
.........(2)
KR 1 In case of disc : K = 2 T 2 K 3 From Eqs. (1) and (2) ,
or
KT =
v1 v2 = 1 i . e ., (v)
v1 = v 2
True Angular momentum will be conserved if the net torque is zero . Now for the sphere to move down: mg sin > mg cos Let x be the perpendicular distance of the point (as shown in figure) about which torque remains zero. for = 0 ; x > R as shown
Note: As mgsin > mgcos, the point should be inside the sphere. 16. (i)
Let the mass of disc without hole = m The mass of cut out hole of radius
R m is 2 4
4 m =M or m = M. 3 4 Moment of inertia of given body about axis passing through O = MI of complete disc – MI of cut out hole.
m–
1 M R 2 M R 2 1 4M 2 13 R = – 2 3 2 3 2 = MR2 2 3 24
RESONANCE
SOLN_RIGID BODY DYNAMICS - 146
(ii)
mvr = K ( a constant ) v =
k mr 2
T=
m K mv 2 K2 1 = = . r mr r m r3
= Ar Hence , (iii)
2 where, A k m
–3
n=–3
under the given conditions only posibility is that friction is upwards and it accelerates downwards as shown below : The equations of motion are : a=
g f mg sin – f mg sin 30 º– f = = – m m 2 m
fR 2f fR = = = mR 2 I I mR 2 For rolling (no slipping) a = R =
(iv)
......(2)
or
3f = g/2 or m Talking moments about of point O :
.......(1)
g/2 – f/m = 2f/m f = mg/6
Moment of N (normal reaction) and f (force of friction) are zero. In critical case normal reaction will pass through O. To tip about the edge, moment of F should be greater than moment of mg. or, a a F > (mg) 4 2 F > 2mg
RESONANCE
SOLN_RIGID BODY DYNAMICS - 147
EXERCISE-4 PART - I 1.
mg sin component is always down the plane whether it is rolling up or rolling down. Therefore, for no slipping, sense of angular acceleration should also be same in both the cases. Therefore, force of friction f always act upwards.
2.
Since, there is no external torque, angular momentum will remain conserved. The moment of inertia will first decrease till the tortoise moves from A to C and then increase as it moves from C and D. Therefore will initially increase and then decrease. Let R be the radius of platform m the mass of disc and M is the mass of platform. Moment of inertia when the tortoise is at A
MR 2 2 and moment of inertia when the tortoise is at B 1 = mR2 +
O
MR 2 2 = mr2 + 2
ra B C
D
2 2 2 vt here r2 = a2 + [ R a vt ] From conservation of angular momentum 0 1 = (t)2 substituting the values we can see that variation of (t) is nonlinear.
3.
(a) The distance of centre of mass (COM) of the system about point A will be : r=
3 Therefore the magnitude of horizontal force exerted by the hinge on the body is F = centripetal force or F = (3m) r2 2 F = (3’m) 3
or
or F = 3 m2 (b) Angular acceleration of system about point A is
Ans.
A = I A 3 (F) 2 = 2 2m A
3F 4m Now acceleration of COM along x-axis is
y ,
=
3 X = r= 3 4 m or
ax =
RESONANCE
x
3/2
COM
F B
C
F 4m
SOLN_RIGID BODY DYNAMICS - 148
Now let Fx be the force applied by the hinge along x-axis. Then : Fx + F = (3m) ax or
F Fx + F = (3m) 4m
3 F F or Fx = – 4 4 Further if Fy be the force applied by the hinge along y-axis. Then : Fy = centripetal force or
Fx + F =
or
Fy =
Ans.
3 m2
Ans.
4.
In uniform circular motion the only force acting on the particle is centripetal (towards center). Torque of this force about the center is zero. Hence angular momentum about center remain conserved.
5.
Let ‘’ be the angular velocity of the rod. Angular impulse = Change in angular momentum about centre of mass of the system J·
6.
L = C 2
M
L (MV) = (2) 2
ML2 4 ·
=
V L
M
J=MV
From conservation of angular momentum ( = constant), angular velocity will remain half. As,
1 2 2 The rotational kinetic energy will become half. Hence, the correct option is (B). K=
7.
In case of pure rolling bottommost point is the instantaneous centre of zero velocity. C
Q
P O
Velocity of any point on the disc, v = r, where r is the distance of point from O. rQ > rC > rP vQ > vC > vP Therefore, the correct option is (A). 8.
0 = 1 – 2 where 1 = (M.. of full disc about O) 2 (M.. of small removed disc about O) since mass area
R2 mass of cut disc 1 9 mass of total = R 2 = 9 mass of cut disc = m
0 =
(9m)R 2 2
R 2 2 3 2R –m 2 3 =
RESONANCE
9mR 2 – mR2 2
(by theorem of parallel axis.)
1 4 9mR 2 mR 2 8mR 2 – = = 4mR2. 18 9 = 2 2 2 SOLN_RIGID BODY DYNAMICS - 149
9.
Only direction of L (angular momentum) is constant because the direction of rotation is unchanged.
10.
From equilibrium, friction = mg N=F about centre of mass =0 mg a = torque due to normal Normal will produce torque since F passes through centre its torque is zero.
11.
Initial angular momentum about fixed point = mvL
ML2 2 final angular momentum = = 3 mL where is moment of inertia of the system about the fixed point and is angular velocity about the fixed point. angular momentum before collision = angular momentum after collision M mLv = L2 m 3
12.
=
mv 3mv = M (M 3m) L L m 3
2/5 MR2 = 1/2 Mr2 + Mr2 2/5 MR2 = 3/2 Mr2 r2 =
4 2 R 15
2R r= 13.*
15
necessary torque for rolling as mg sin – f = ma but a = r as = fr =
= fr,
mg sin – f = mr = fr/
mg sin – f = mrfr/ = 5f/2
(frictional force provides this torque)
2 2mr 5
r
f
q
7f 2 thus friction increases the torque in hence the angular velocity and decreases the linear velocity. If decreases friction will decrease. mg sin =
14*.
As total mechanical energy at points A,B and C will be constant A = B = C mghA + KA = KB = mgh + KC KB > KA (mghA + KA = KB) and KB > KC (mghC + KC = KB) Also
hA – hC =
KC K A mg
when
mghA + KA = mghC + KC
if hA > hC KC > KA (if LHS is positive then RHS have to be positive) 15.
(As collision is elastic) dP 2mV 2mV dt 1
F=
b b torque about hinge = 2mV × × 100 2 4 V = 10 m/s
RESONANCE
= 2mV
3b b × 100 = Mg 4 2
SOLN_RIGID BODY DYNAMICS - 150
16.
1 1 (2)2 kx12 2 2 1 1 2 2 kx 22 2 2
17.
x1 = x2
2
Apply conservation of angular momentum ( × 2) + (2 × ) = ( + 2) ’
’ =
4 3
For Disc A t = × (2 – ’) 18.
Initial Kinetic Energy k1 =
=
2 3t
1 1 × × (2)2 + ×2 × 2 2 2
1 1 × × ’2 + × 2 ’2 2 2 Loss of Kinetic Energy= k1 – k2 Final Kinetic Energy k2 =
2 3 From the conservation of energy loss in KE of body = Gain in potential energy =
19.
1 mv2 + 2
1 2
v r
2
= mg
3 4
v g
2
on solving mr 2 2 The body is a disc
=
20.
If torque external = 0, then angular momentum = constant =
21.
The acceleration of centre of mass of either cylinder a=
g sin 1
K2
where K is radius of gyration.
R2 So acceleration of centre of hollow cylinder is less than that of solid cylinder. Hence time taken by hollow cylinder will be more. So statement-1 is wrong. Ans. (D)
22.
(A) Since there in no resultant external force, linear momentum of the system remains constant. (B) Kinetic energy of the system may change. (C) Angular momentum of the system may change as in case of couple, net force is zero but torque is not zero. Hence angular momentum of the system is not constant. (D) Potential energy may also change.
23*.
VA V( ˆi ) R( ˆi ) ; VB V ˆi ; VC V ˆi Rˆi VC VA 2Rˆi 2 VB VC 2 [ V( ˆi ) V( ˆi ) R( ˆi )] = –2R( ˆi ) Hence VC VA = 2( VB VC )
RESONANCE
SOLN_RIGID BODY DYNAMICS - 151
| VC VA | = | 2( VB VC ) | VC VB = R( ˆi ) VB VA = R( ˆi ) VC VB VB VA VC VA 2R( ˆi ) ; VC VB VB VA VC VA 2( VB )
so
Hence
Hence 24.
4VB = 4V( ˆi ) = 4R ( ˆi )
Angle of repose 0 = tan–1 = tan–1 3 = 60º tan =
5 2 = . 15 / 2 3
< 45º.
Block will topple before it starts slide down.
25.
...(i)
f '2 F 2 2
a R F – f' = 2ma = 1.2 ...(ii) From (i) & (ii) (1.2 + f ')2 + f '2 = 22 2f '2 + 2.4f ' + 1.44 = 4 f '2 + 1.2f ' + 0.72 – 2 = 0 f '2 + 1.2f ' – 1.28 = 0 FR – f'R = 2mR2
f' =
1.2 1.44 4 (1.28 ) 2
= 0.6 ± 0.36 1.28 = –0.6 ±
1.64
= 0.68
From eq. (2) F = 1.88 0.68 P = P = 3.61 4 Ans. 1.88 10 Note : In Hindi friction force is aksed, so the answer is P = 6.8. (for Hindi)
=
Note : But if only normal reaction applied by the rod is considered to be 2 N Law 2 – f = 2 [0.3] f = 2 – 0.6 f - 1.4 Nx ...(i) a = R 0.3 = [0.5]
3 rad/s 5 c = c fR – 2R = mR2 f – 2 = mR
=
1.4 – 2 = 0.8 = 2µ
RESONANCE
....(ii)
2 3 1.4 – 0.6 = 2µ 2 5
= 0.4 =
P 10
P=4
Ans.
SOLN_RIGID BODY DYNAMICS - 152
Note : In Hindi friction force is aksed, so the answer is P = 8. (for Hindi) 26.
2 2 2 2 2 = MR 2 + MR Mx 2 5 5 2 2 2 2 = MR 2 + MR 2 + (Mx2) 2 5 5 2 2 = 4 MR + 2mx2 5
=
8 MR 2 + 2mx2 5
2 8 0.5 5 2 (0.5) ( 4 2)10 4 = 5 2
5 = 8 × 10–4 5 = 9 × 10–4 = N × 10–4 So, N = 9 Ans.
27. 28.
Friction force on the ring. L = [m(vt)2] L = mv2t2 So
dL = 2mv2t dt t straight line passing through (0, 0) =
30.
I0 =
3 ( 4m) (2R )2 – mR2 2 2 = mR2 [8 –
3 13 ] = mR2 2 2
RESONANCE
SOLN_RIGID BODY DYNAMICS - 153
IP =
mR 2 2 2 3 11 37 mR 2 = mR 2 (4m) (2R)2 – 2 m[( 2R) R ] = 24 mR2 – 2 2 2
37 IP 37 2 3 13 = IO 13 2 Ans. 3 31.
Angular Velocity of rigid body about any axes which are parallel to each other is same . So angular velocity is .
32.
Since z- coordinate of any particle is not changing with time so axis must be parellel to z axis.
33.
IP > IQ g sin aP =
IP mR 2
g sin aQ =
IQ mR 2
aP < aQ V = u + at t
1 a
t P > tQ V2 = u2 + 2as v a VP < VQ
1 mV2 TR KEP < TR KEQ 2 V = R V P < Q Translational K.E. =
34.
V0 = 3R ˆi VP (3R – =
R R cos 60º) ˆi + sin 60 ˆj 2 2
11R ˆ 3 R ˆ i i 4 4
PART - II 1.
1 M 2 AC = 2 6
2 = M , 12
RESONANCE
EF =
M 2 , AC = EF . 12
SOLN_RIGID BODY DYNAMICS - 154
2.
mg sin – ƒ = maCM ƒ.R = aCM = R On solving (i),(ii) & (iii) aCM =
g sin 1 MR 2
..........(i) ..........(ii) ..........(iii)
.
3.
Central force is directed towards a point, therefore torque of the central force is zero.
4.
a IA = Icm + m 2
5.
=
ma 2 ma 2 + 6 2
=
2 ma2 3
2
1 2 = mgh 2
2 1 m 2 = mgh 2 3
h=
2 2 6g
6.
1 2 Angular Momentum = m ( v 0 sin gt )( V0 cos t ) ( V0 cos ) V0 sin t 2 gt =–
1 mg V0 t2 cos 0 kˆ 2
7.
From angular momentum conservation about vertical axis passing through centre. When insect is coming from circumference to center. Moment of inertia first decrease then increase. So angular velocity inecrease then decrease.
RESONANCE
SOLN_RIGID BODY DYNAMICS - 155
8.
mg – T = ma
mR 2 2
TR =
3ma = mg 2 9.
T=
mR ma = 2 2
a=
2g 3
To reverse the direction
mg –
ma = ma 2
Ans.
d 0 (work done is zero)
= (20 t – 5t2) 2 = 40t – 10t2
40 t 10 t 2 4t t 2 10
=
t
=
dt = 2t2 –
0
t3 3
is zero at
t3 =0 3 t3 = 6t2 t = 6 sec. 2t2 –
6
=
dt =
0
( 2t 2
t3 )dt 3
6
2t 3 t 4 = 216 12 3 0 No of revolution 10.
2 3
1 = 36 rad. 2
36 Less than 6 2
L0 = Pr L0 = mv cos H = mg
2 2 3 v sin 30º . = 2g 2
RESONANCE
3mv 3 . 16g
SOLN_RIGID BODY DYNAMICS - 156
TOPIC : UNIT AND DIMENSIONS EXERCISE-1 PART - I 3**.
ML2 T 3
U = AT4
[] =
(ii)
T = b
[b] = [] [T] = LK
(iii)
F = 6rv
[] =
(iv)
=
(v)
Energy =
(vi)
[B 2 ] [U] [B 2 ] [ V ] = = [0] = [V] [2 0 ] [U]
Also ,
F = qVB
[ A ][T 4 ]
=
L2K 4
= MT–3 K–4
[F] MLT 2 = = ML–1 T–1 [r ][ v ] L. LT 1
ML2 T 3 P = = MLº T–3 A L2
[0] = 8.
[U]
(i)
[E]
1 2 Mi 2
[M] =
B=
(F ) 2 [ V ] [q 2 v 2 ][U]
= MLT T–2A–2
[i 2 ]
= ML2T–2 A–2
F qv Ans.
We have the equation Gm1m 2
[G][M] 2
=F
r2 [G] = M–1L3T –2
[L] 2
= MLT T –2
.......... (i)
[h] [c ] hc = Energy = ML2T –2 [ ] [] = L [h] = ML2T –1 .............. (ii)
[c] = LT T –1
[c] = LT –1 ................ (iii) taking the product of (i) & (ii) [G] [h] = L5T –3 [c]3 = L3T –3
[G][h]
[c ] 3
= L2
G1 / 2h1 / 2 c 3 / 2 = L again from (iii) [T] =
[L] = 1 / 2 1/ 2 3 / 2 1 = G1 / 2h1 / 2 c 5 / 2 [c ] G h c
From (ii) [h] = ML2T –1 [h] = [h] =
MGhc 3 G
1 / 2 1 / 2 5 / 2
h
c
MG1/2h1/2c –3+5/2
RESONANCE
or
G-1/2 h1/2 c 1/2 = M SOLN_UNIT & DIMENSIONS - 157
10.
Let, w = KMarbGc where K is a dimensionless constant. Writing the dimension of both the sides and equating then we have, T–1 = MaLb(M–1L3T–2)c = Ma–c Lb+3c T–2c Equating the exponents
1 , 2
– 2c = – 1
or c =
b + 3c = 0
or – 3 c = b = –
a–c=0 . c=a =+
3 2
1 2
Thus the required equation is = K
Gm r3
PART - II 5.
All the terms in the equation must have the dimension of force [A sin C t] = MLT–2 [A] [M0L0T0] = MLT–2 [A] = MLT–2 –2 Similarly, [B] = MLT [A] = MºLºTº [B]
Again
[Ct] = MºLºTº [Dx] = MLTº
[C] = T–1 [D] = L–1
[C] = MºL1T–1 . [D]
a V2
[a] = [P] [V2]
6.
[P] =
10.**
V = R V has the dimensions of
[ work ] ML2 T 2 [V] = [ch arg e] = = ML2 T–3 –1 T [v] = ML2 T–3 –2 []
[R] = 11.
[v] = [k] [a b gc] LT–1 = Mb La – 3b + c T–2c so,
LT–1 = La Mb L–3b Lc T–2c a = ½, b = 0, c = ½
v2 = kg
13.
G = 6.67 × 10–11 N m2 (kg)–2 = 6.67 × 10–11 × 105 dyne × 1002 cm2 / (103)2 g2 = 6.67 × 10–8 dyne-cm2-g–2
14.
dx 2ax x 2
x a n sin 1 1 . a
L.H.S. is the dimensionless as denominator 2ax – x 2 must have the dimension of [x]2
RESONANCE
SOLN_UNIT & DIMENSIONS - 158
(we can add or substract only if quantities have same dimension) 2 2ax x = [x]
Also, dx has the dimension of [x] x dx
is having dimension L
2ax x 2
Equating the dimension of L.H.S. & R.H.S. we have x a
[an] = M0L1T 0
{ sin–1 1 must be dimensionless}
n=1 16.
ma
[a] = ....(i) 0 0 0 ma = M L T
17. 18.**
ma = [] = [] = L
[g] = LT–2 and numerical value
1 unit
[h] ML2 T 1 [] = [ 4 ] = = L2 T2 MT – 3K 4 .K 4 So, unit of will be m2s2. ( weber ) ( )2 (Farad )2 Tesla
=
Tm 2 . 2F2 = m2s2 T
EXERCISE-2 PART - I 1.**
Reynold’s number and coefficient of friction are dimensionless quantities. Curie is the number of atoms decaying per unit time and frequency is the number of oscillations per unit time. Latent heat and gravitational potential both have the same dimension corresponding to energy per unit mass.
2.**
X = 3YZ2 [X] = [Y] [Z]2
[ X] [Y] =
[Z]2
=
M 1L2 Q 2 T 2 M2 Q 2 T 2
= M–3L–2Q4T 4
3.*
Torque and work have same dimensions = ML2 T–2 Light year and wavelenth have dimension of length = L
4.
Micron, light year & angstrom are units of length and radian is unit of angle.
5.*
(A)
L=
(B)
di e=–L dt
(C)
U=
(D)
U=
i
or
Henry =
Weber Ampere
L=
1 2 Li 2
L=
1 2 Li = i 2 Rt 2
RESONANCE
e di / dt
Volt sec ond Ampere
or
Henry =
2
or
Henry =
[L] = [Rt].
or
Henry = ohm-second
2U i
Joule ( Ampere )2
SOLN_UNIT & DIMENSIONS - 159
6.*
1 q1q2 we have F = 4 r2 0 [ 0] =
[q1 ] [q2 ] [F] [r 2 ]
=
2T 2 MLT 2 L2
= M–1L–3T 42
1 Also C (speed of light)
=
0µ0
1 [µ0]1/2 = [c ] [ ] 0 [0] = MLT–2 –2 7.**
(None of the four choices)
1 E2 is the expression of energy density (Energy per unit volume) 2 0
ML2 T 2 1 2 2 0 E L3
[ML–1 T–2] From this question, students can take a lesson that even in IIT-JEE, questions may be wrong or there may be no correct answer in the given choices. So don’t waste time if you are confident.
8.**
X = 0L
v t
v [X] = [ 0 ] [L ] t [v] = [Electric field] [Length]
[Force] MLT 2L = [ch arg e] [Length] = = MQ–1L2T –2 Q [ 0] = M–1 L–3T 42 (as in question no. 6) = M–1L–3Q2T2 [X] = M–1L–3Q2T 2 L
MQ 1L2 T 2 T
= QT –1 = [x] is that of current Z k = [M0 L0 T0]
9.
k [] = Z
k Further, [P] = [] = = P ZP Dimensions of k are that of energy. Hence,
ML2 T 2 0 2 0 [] = 1 2 = [M L T ] LML T Therefore, the correct option is (A). 10.**
[Dipole moment] = LIT, [E] = ML3 /T 3 [E] = ML/T 3 .
11.**
(A)
GMeMs Re
2
= Force
[GMeMs] = [Force] [Re2] = MLT–2 L2 = ML3T–2
RESONANCE
SOLN_UNIT & DIMENSIONS - 160
Hence S unit of GMeMs, will be (kilogram) (meter3)(sec–2) ie same as (volt) (coulomb) (metre)
3RT = [VR.M.S.]2 = L2T–2 M0 Hence S unit will be (metre)2 (second)–2 ie same as (farad) (volt)2 (kg)–1 3RT = VR.M.S. M
(B)
[F 2 ]
(C)
[ q2 B 2 ]
(D)
=
[q2 v 2B 2 ] [q2B 2 ]
= [V2] = L2T–2
Hence S unit (metre)2 (second)–2
i.e. same as (farad) (volt)2 (kg)–1
GMe [Force] [R e ] MLT 2L = = = L2T–2 [Mass] Re M Hence S unit will be (meter)–2 (second)–2
i.e. same as (farad) (volt)2 (kg)–1
PART - II 1. 2.
The dimensions of torque and work are [ML2 T2] h = Planck’s constant = J–s = [ML2T–1] P =momentum = kg m/s = [MLT–1]
3.
As we know that formula of velocity is
1 v=
4.
0 o
1 v2= = [LT T–1]2 0 o
1 2 –2 0 o = [L T ]
From Newton’s formula
F = A v / z x Dimensions of
5.
[MLT –2 ] dim ensions of force = 2 –1 = [ ML–1 T–1] [L ] [L ] dim ensions of area dim ensions of velocity – gradient
I = mr2 [I] [ML2] and
6.
=
= moment of force = r F = [L] [MLTT–2 ]
Energy stored in inductor
U
1 2 LI 2
L
2U
I2 Since Henry is unit of inductance L (4) is correct.
7.
From
[L ]
L2 T –2 Q2 / T 2
ML2 Q2
F = qvB [MLT–2] = [C] [LT–1] [B] [B] = [MC–1T–1]
RESONANCE
SOLN_UNIT & DIMENSIONS - 161
