KINEM ATICS
AB = Example 3. A particle goes from A to B with a speed of 40 km/h and B to C with a speed of 60 km/h. If AB 6BC the average avera ge speed in km/h between betwee n A and C is _______ AB = 40t 1 ...(1) Sol. BC = 60t 2
...(2)
Avera verage ge speed speed = Vav =
total total distance distance trave travelle lled d
time time take taken n AB + BC
t1 + t 2 From eqn. (1) and (2) 40 t1 + 60 t 2 Vav = t1 + t 2 According to question AB = 6BC 40t1 = 6 × 60t 2
A
...(3)
From eqn (1) and (2)
t1 = 9t2 From eqn (3)
Vav
Vav =
40 9 t 2 60t 2
from eqn (3)
9t 2 t 2 420t 2 10t 2
Vav = 42 km km / h
B
C
Example 6. At a distance L = 400m from the traffic tr affic light light brakes are applied to a locomotive loco motive moving moving at a velocity v = 54 km/hr. km/hr. Determi Det ermine ne the position of the t he locomotive relative to the traff t raffic ic light light 1 min after the application of the brakes br akes if its acceleration is –0.3 m/sec 2. 5 15 m / s Sol. u 54 18 a = –0.3 m/s2 v = u + at 0 = 15 – 0.3 t 0 15 t0 = = 50 sec 0.3 After 50 second, locomotiv locomot ivee comes co mes in rest permanently. permanently. 2 2 v = u + 2as
O2 = 152 – 2 × 0.3 S 0
S0 =
225 0.6
=
2250 6
= 375 m
the distance of the t he locomotive from tr traffic affic light light = 400 – 375 =
25 metr etre
2 Example 7. A car moves in the x–y plane with acceleration accelerat ion (3 ˆi + 4 ˆj) m / s .
(a) Assuming Assuming that the t he car is at rest r est at the origin at t = 0, derive expression for the velocity as function of time. (b) Find the equation o off path of car and find the position vector vecto r as function of time t ime.. Sol. Here, ux = 0, uy = 0, uz = 0 ax = 3 m/s2 , ay = 4 m/s2 (a) vx = ux + axt or and
vx = 3t vy = uy + ayt
or
vy = 4t
v = v x ˆi + v y ˆj
v = (3tˆi + 4 tˆj)
(b)
x = uxt + 1
x=
and and
y = uyt +
or
y=
y = 2´
2
4
1 2
3 2
t2
a y t2
( 4) t 2 = 2 t 2 2 3
x=
4 3
x
x 3 Hence, the path pat h is straight line. The position of car is 3 r = x ˆi + y ˆj = t 2 ˆi + 2 t 2 ˆj 2
y=
1
2
axt2
´ 3t 2 =
or
2
1
x=
3 2ö t ÷ 2 ø
Example 9. A ball is thrown upwards from the ground with an initial speed of u. The ball is at a height of 80 m at two times, the time interval being 6s. Find u. Take g = 10 m/s 2. Sol. Here, u = u m/s, a = g = – 10 m/s 2 and s = 80 m. Substituting the value in s = ut + 80 = ut – 5t
2
1 2
2
at , we have
5t2 – ut + 80 = 0
or or
t=
and
u 2 - 1600
u+
10
s = 80 m
2
u-
t=
+ve
u - 1600 10
–ve
u
Now, it is given that
u 2 - 1600
u+
10 or
u 2 - 1600 5
or
-
u-
=6
u 2 - 1600 10
or
= 6
u 2 - 1600 = 30
u2 – 1600 = 900
u2 = 2500 or u = ± 50 m/s Ignoring the negative sign, we have u = 50 m/s
A
Example 10. A disc arranged in a vertical plane has two groves of same length directed along the vertical chord AB and CD as shown in the fig. The same particles slide D down along AB and CD. The ratio o f the time t AB/t CD is : (A) 1 : 2
(B) 1 : 2 SAB =
Sol.
SCD =
But
1 2 1
gt
g cos 60º t 2CD
2 S = S AB CD SAB SCD
=
2 1 2
or
(D)
1= 2
B
gt
A 60º 6 0 g c o s D
gcos60ºt
2 CD
t AB
C 6 0 º º
2 AB
t 2AB t 2CD
2 :1
2 AB
1
(C) 2 : 1
60º
t CD
=1 :
2
g
B
g
C
Example 11. A stone is dropped from a height h. Simultaneously another stone is thrown up from the ground with such a velocity that it can reach a height of 4h. Find the time when two stones cross each other. Sol. For second stone, v 2 = v 02 - 2g (4h) 0 2 = v 02 - 8gh
v0
8gh
but they meth at height H in time t0. Displacement of 1st stone is 1 h – H gt 20 ...(1) 2 and that of second stone is 1 H = v0 t 0 - gt 02 ...(2) 2 After solving eqn (1) and (2) 1 1 h – v 0 t 0 g t 02 g t 02 or 2 2 h = v0t0
v=0
(1) (2) h
v0
4h
t0 =
h v0
h
=
t0 =
8gh
h 8g
Example 12. A rocket is launched at an angle 53º to t he horizontal with an initial speed of 100 ms –1. It moves along its initial line of motion with an acceleration of 30 ms –2 for 3 seconds. At this time its engine fails & the rocket proceeds like a free body. Find : (i) the maximum altitude reached by the ro cket (ii) total time of flight. Sol. S0 = ut + ½ at 2 = 100 × 3 + ½ × 30 × 9 = 300 + 135 = 435 m In OAB h sin53º = S0 4 h S0 sin 53º 435 87 4 348 m 5 v0 = u + at = 100 + 30 × 3 = 190 m/s After engin switch off 3 v0x v 0 sin 37º 190 114 m / s 5 4 v0y v0 cos 37º 190 152 m / s 5 2 ay = – 10 m/s , a x = 0 (i) At maximum altitudes vy = 0
(velocity at the time of switch off)
v 2y = v 20y + 2a y h 0 2
0 = v 0y + 2a y h 0 h0 h0
S0
2 0y
v – 2a y
h0 =
A u O
2 10 11552
h 0 = 1155.2 m 10 The maximum altitude reached by the rocket is
= h0 + h = (1155.2 + 348) m
1503.2 m
(ii)Total time of flight. 1 2
a yt2
- h = v0 y t 0 +
1 2
a y t 02
1155.2 152t 0
1 2
2
10 t 0
v0
Engin fail x h
53º
152 152
y = v0yt +
37º
37º
B
- 1155.2 = 152t 0 - 5t 20 5 t 20 - 152 t 0 - 1155.2 = 0 t 02 - 30.4 t 0 - 231.04 = 0
t 0 = 35.54 sec.
Example 13. A particle is moving with a velocity of v = (3 + 6t + 9t 2) cm/s. Find out : (a) the acceleration of the particle at t = 3 s. (b) the displacement of the particle in the interval t = 5 s to t = 8 s. Sol. (a) Acceleration of particle a=
dv dt
2 = ( 6 + 18t ) cm / s
At t = 3 s, a = (6 + 18 × 3) cm/s2 a = 60 cm/s2 (b) Given, v = (3 + 6t + 9t2) cm/s ds
or
= ( 3+ 6t + 9t
2
)
dt ds = (3 + 6t + 9t 2)dt
or
8
8
ò (3+ 6t + 9t ) dt
ò
2 3 s = éë 3t + 3t + 3t ù û5 or
5
ds =
2
5
8
s = 1287 cm
Example 14 : The motion of a particle along a straight line is described by the function x = (2t – 3)2 where x is in metres and t is in seconds. (a) Find the position, velocity and acceleration at t = 2 s. (b) Find the velocity of the particle at origin. Sol. (a) Position, x = (2t – 3) 2 Velocity,
v
dx dt
4 2t 3 m / s
and acceleration, a =
dv dt
= 8m/s
2
At t = 2 s, x = (2 × 2 – 3) 2 = 1.0 m v = 4(2 × 2 – 3) = 4 m/s and a = 8 m/s 2 (b) At origin, x = 0 or (2t – 3) = 0 v=4×0=0
v 20
Example 15. The fig. shows the v–t graph of a particle moving in straight line. Find the time when particle returns to the starting point. Sol. When the particle comes at initial position, total displacement is zero. Since, the area of v – t graph gives displacement. In this case area of v – t graph should be zero. 1 1 S 20 25 t 25 v 0 2 2 v0 t 25 or 0 250 – 2 500 or t – 2 5 ...(1) v0 Also tan or
4
20
25 20
t 25
from figure
10 20
25 t
v 20
v0
t–25 v 0 4 t – 25 From eqn (1) and (2) 500 t–25 4 t – 25
t – 25
2
20
...(2)
25
t
20
t(s) 40
v0
125 t – 25
125
t – 25 125
v0
10
5 5
t = (25 + 5 5 ) sec.
t = 36.2 sec
Example 16. From the velocity-time plot shown in figure, find the distance travelled by the particle during the first 40 seconds. Also find the average velocity during this period. Sol. The distance fravelled by the particle during the first 20 second. 1 S1 v t 2 1 S1 5 20 S1 = 50 m 2 The distance travelled by the partile during next 20 second is 1 S2 5 20 2 S2 = 50 m Since distance is a sclar quantity therefore total distance = S1 + S2 = 50 + 50
V 5m/s
–5m/s
S = 100 m
Example 17. A ball is dropped from a height of 80m on a floor. At each collision the ball losses half of its speed. Plot the speed-time graph and velocity-time graph of its motion till two collisions with the floor, [Take g = 10 ms –2].
Sol. The time in first collision
h = ut +
or
80 = 0 +
2 1
gt 2 (during downward motion) ´ 10 ´ t 2
(
2 2´ 80
u = 0)
= 4s 10 Final speed just before first collision v = 0 + 10 × 4 = 40 m/s or
t=
1
æ 40 ö 20 ÷m / s. It now loses half of initial speed after the collision i.e., when it first bounces its initial speed is ç = è 2ø So, the time is loosing half of its speed. 0 = 20 – 10 × t´ (during upward motion) 20 t´= = 2s (final speed = 0) 10 In 2 s, it attains height 1 2 h´= 20´ 2 - ´ 10 ´ (2) 2 h´ = 40 – 20 = 20 m Now, it is dropped again from 20m with zero initial speed. Time taken in reaching the ground 1 20 = 0 + ´ 10(t) 2 2 t=2s Also final speed v´2 = 0 + 2 × 10 × 20 (from v2 = u2 + 2gh) v´ = 20 m/s Thus, with the above data, we can draw the speed-time graph. Speed (m/s) 40 20 4
6
8
Time(s)
Since, velocity is a vector quantity so from the above graph, we can now draw the velocity-time graph. [Take downward motion positive and upward motion negative in case of v–t graph] Velocity (m/s) 40 20 4
6
Time(s)
8
–20
Example 18. The velocity-time graph of an object moving along a straight line is as shown in the fig. Calculate the distance covered by the object : A
20 ms –1
B
v O
0
A´ 2
B´ 5
C 10
(a) between t = 0 to t = 5 s and (b) between t = 0 to t = 10 s. Sol. (a) Let x be the distance covered in the time interval between t = 0 to t = 5 s. Then, x = area of the trapezium 1 1 OABB´ æ AB + OB´ ö 3+ 5 =ç ´ 20 = 80 m ÷´ AA´= è ø 2 2 (b) Let x2 be the distance covered in the time interval between t = 0 to t = 10 s. Then, x 2 = area of the trapezium OABC. AB + OC 3 + 10 = ´ AA´= ´ 20 = 130 m 2 2
Example 21. A, B & C are three objects each moving wit constant velocity. A’s speed is 10m/sec in a
direction PQ . The velocity of B relative to A is 6 m/sec at an angle of cos –1(15/24) to PQ. The velocity
of C relative to B is 12 m/sec in a direction QP , then find the magnitude of the velocity of C. Sol. v A = 10 ˆi
v B 6 cos ˆi 6sin ˆj
Here And
or
cos
15 24
v BA = v B - vA
6 cos i 6 sin j vB 10 ˆi v B cos ˆi 6sin ˆj 10 ˆi
15 ˆ 351 ˆ i6 j 10 ˆi 24 24 351 ˆ 15 ˆ 4 10 i 4 j
55 ˆ 351 ˆ i j 4 4
v CB = vC - v B 55 ˆ 351 ˆ v C = vCB + v B = - 12 iˆ + i+ j 4 4 7 351 ˆ = ˆi + j 4 4
7 351 vC 4 4 2
24
v CB = - 12 ˆi
=
351
6
But
, sin
=
49 + 351 = 4
2
400 20 = 4 4
Ans.
5 m/s
Example 23. A man with some passengers in his boat, starts perpendicular to flow of river 200m wide and flowing with 2m/s. boat speed in still water is 4m/s. When he reaches half the width of river the passengers asked him they want to reach the just opposite end from where they have started. (a) Find the direction due to which he must row to reach the required end. (b) How many times more tot al time, it would take to that if he would have denied the passengers. Sol. (a) Event (1) — From A to B, Time taken by boat to recent from A to B is 100 100 t= = = 25sec. vy 4 Also, AD = vx t = 2t = 50 m Event (2) — From B to C, Actual velocity of boat should be along BC. This actual velocity is found by resultant of v rel and vr .
v rel = v b - vr x–comput of actual velocity is
y C
vx = vr – vrel sin = 2 – 4sin and
y – comput of actual velocity is vy = vrel cos
100 v rel cos
Also,
100
4cos
25
cos EC = –vx t 0 = –(2 – 4 sin
But
4 sin
vrel
vrel
The time taken to go from B to C is BE t0 = vy
2
EC = AD = 50 m
) t0 25
cos
E v b
A
B 200m D
vr
EC 4sin 2 50 4 sin 2
25 cos 25
cos 50 cos = 100 sin – 50 cos = 2 sin – 1 1 + cos = 2 sin 2 cos 2 a
tan
4sin
2
2
cos
2
1
=
2
2
1 tan 1 2 2
(b)
1 2tan 1 2
If boat crosses the river with initial condition 200 200 t1 = = vy vrel t1 =
200
= 50sec. 4 If boat cross the river with final condition, t 2 = t AB + t BC t 2 = t + t 0 25
t 2 25
tan
2
cos
2
cos 1
2 2
5
cos 2 cos 2
2 =
8
2 4
5
1
1
- 1=
3
5 5 25 125 200 t 2 = 25 + = 25 + = 3 3 3 5 t2 t1
200 3 50
4 3
t2 t1
=
4 3
Example 24. To a person going west wards with a speed of 6 km/h rain appears to fall vertically downwards with a speed of 8 km/h. Find the actual direction of rain. Sol. Let
v M = velocity of man = 6 km/h
v = relative velocity of rain w.r.t. man = 8 km/h
Vertical
v R = actual velocity of rain
N
In this case
v = v R + (- vM )
W
v = vR - vM
or
v R =
O
E
A vR
vR = vM + v
vM
v B
C
S
(6) 2 + (8)2 = 10 km / h
S
The velocity of rain ( v R ) is given by the vector OC , the resultant of vectors OA and OB as shown in figure.
If is the angle that v R makes with the vertical, then
BC | v M | 6 0.75 tan OB | v | 8 (east of vertical) = 36º 52´
or
Example 25. Rain is falling vertically with a speed of 20 ms –1 relative to air. A person is running in the rain
with a velocity of 5 ms –1 and a wind is also blowing with a speed of 15 ms –1 (both towards east). Find the angle with the vertical at which the person should hold his umbrella so that he may not get drenched. Sol. v ra = - 20 kˆ z
N
v m = 5 ˆi v a = 15 ˆi
E
v ra = v r - va v r = - 20 kˆ + 15 ˆi
S
x
W
v rm = v r - vm = - 20kˆ +15iˆ - 5iˆ = - 20 kˆ + 10 ˆi 10 1 tan 20 2 1
tan
1 2
vr m
vertically
Example 26. An aircraft flies at 400 km/h in still air. A wind of 200 2 km / h is blowing from the south. The pilot
wishes to travel from A to a point B north east of A. Find the direction he must steer and time of his journey if AB = 1000 km.
Sol. Given that vw = 200 2 km / h . vaw = 400 km/h and v a should be along AB or in north-east direction. Thus,
the direction of v aw should be such as the resultant of v w and v aw is along AB or in north-east direction. N B va 45º A
45º vw = 200 2 km/h C vaw = 400 km/h
E
Let v aw makes an angle with AB as shown in fig. Applying sine law in triangle ABC, we get AC BC sin 45º sin BC sin sin 45º or AC
200 2 1 1 sin 400 2 2 = 30º therefore, the pilot should steer in a direction at an angle of (45º +) or 75º from north towards east.
| va | 400 sin(180º 45º 30º ) sin 45º
Further,
sin105º km (400) sin 45º h km 0.9659 km cos15º | va | (400) (400) sin 45º h 0.707 h km | v a | 546.47 h The time of journey from A to B is AB 1000 t h t = 1.83 h | va | 546.67
| va |
or
Example 27. A glass wind screen whose inclination with the vertical can
be changed, is mounted on a cart as shown in figure. The cart moves uniformly along the horizontal path with a speed o f 6 m/s. At what maximum angle to the vertical can the wind screen be placed so that the rain drops falling vertically downwards with velocity 2 m/s, do not enter the cart ? Sol. v c = 6 ˆi
v=6m/s
vc
D
v r = 2 ˆj
vr
v rc = v r - vc
tan
vc 6 3 v r 2
cos
AE
2sin
vrc
y
AE cos sin ED tan BE cos sin 3 2sin 2 / 2 3
x
2
cos
2sin 2
2
2 cot 2
A cos sin
E
vr
B C
D
or
cot
or
tan
3
2
2
or
1 3
tan 1
2
1 3
1 2 tan 1 3
Example 28. A bird flies in the x–y plane with a velocity v = t 2 iˆ + 3t ˆj . At t = 0, bird is at origin. Calculate position and acceleration of bird as function of time. Sol. We have given
2 v = t iˆ + 3t ˆj
Here,
vx = t2, vy = 3t and vz = 0
Since,
vx = t2
or
dx t2 dt
or
x
0
dx
x
Also,
vy = 3t
or
dy dt
y
0
3
3t
dy
y
t
0
t 2 dt
t3
or
or
t
0
3t dt
3t 2 2
Thus, position of bird is r x ˆi y ˆj t 3 ˆ 3t 2 ˆ r i j 3 2 vx = t2
and
dv x d(t) 2 ax 2t dt dt vy = 3t
or
dvy dt
3
dt dt
ay = 3 unit
Thus, acceleration of bird is a a x ˆi a y ˆj
a
2t ˆi 3 ˆj
Example 29. Prove that the maximum horizontal range is four times the maximum height attained by the projectile; when fired at an inclination so as to have maximum horizontal range. Sol. For = 45º, the horizontal range is maximum and is given by
R max
u2 g
Maximum height attained
H max or
u 2 sin 2 45º
2g
u2
R max
4g
4
R max = 4 Hmax
Example 30. There are two angles of projection for which the horizontal range is the same. Show that the sum of the maximum heights for these two angles is independent of the angle of projection. Sol. There are two angles of projection and 90º – for which the horizontal range R is same. H1
Now,
H2
and
H2 Therefore, H1 H 2
u 2 sin 2
2g
u 2 sin 2 90º 2g
u 2 cos 2 2g
u
2
2
2
u2
sin cos 2g 2g Clearly, the sum of the heights for the two angles of projection is independent of the angle of projection. Example 31. A particle is projected upwards with a velocity of 100 m/sec at an angle of 60º with the vertical. Find the time when the particle will move perpendicular to its initial direction, taking g = 10 m/sec2. Sol. Here
ax = 0 ay = –g ux = 100 sin60º = 50 3 uy = 100 cos60º = 50
u 0 = u x ˆi + u y ˆj = 50 3 ˆi + 50 ˆj
y
v y = u y + a y t = 50 – gt
v x = u x = 50 3 m / s 60º
v = 50 3 ˆi + (50 - gt )ˆj
x
But u and v are perpendicular..
or or
u v0
50
3 ˆi 50 ˆj 50 3 iˆ 50 gt ˆj 0
7500 + 2500 – 500 t = 0 10000 t= 500
t 0 = 20 second
Example 32. Fig. shows a pirate ship 560 m from a fort defending the harbor entrance of an island. A defense cannon, located at sea level, fires balls at initial speed v 0 = 82 m/s.
(a) At what angle 0 from the horizontal must a ball be fired to hit the ship ? (b) How far should the pirate ship be from the cannon if it is to be beyond the maximum range of the cannonballs? y
º 6 3
27º
x R = 560 m
Sol. (a) Because the cannon and the ship are at the same height, the horizontal displacement is the range. The horizontal range is
R
v 20 g
sin 2
...(i)
which gives us
gR 2 v
20 sin 1
0
9.8 560 (82)2
20 sin 1
20 = sin –1(0.816) ...(ii) 1 0 (46.7º ) 23º 2 = 90º – 0 and = 90º – 23º = 67º The commandant of the fort can elevate the cannon to either of these two angles and (if only there were no intervening air!) hit the pirate ship. (b) We have seen that maximum range corresponds to an elevation angle 0 of 45º. Thus, from Eq. (i) with
0
= 45º. 2
(82) 2
sin 20 sin(2 45º ) g 9.8 R = 686 m 690 m As the pirate ship sails away, the two elevation angles at which the ship can be hit draw together, eventually merging at 0 = 45º when the ship is 690 m away. Beyond that distance the ship is safe. R
v0
Example 33. A particle is projected in the X–Y plane. 2 sec after projection the velocity of the part icle makes an angle 45º with the X-axis. 4 sec after projection, it moves horizontally. Find the velocity of projection. Sol. After 4 sec the particle reach at maximum hight. At maximum height it move horizontally. So that, T y =4 2 v 2usin 2usin T8 T g g O x 80 40 u sin ...(1) 2 u cos = v cos 45º v sin45 = 45 sin – 10 × 2 0
1
u sin 20
20
u cos u cos From eqn (1) and (2) u2 sin2 + u 2 cos 2 = 40 2 + 20 2 u2 = 1600 + 400 = 2000 u 20 100 20 5 m / s
10 20
ucos = 20
...(2)
Example 34. A projectile is fired horizontal with a velocity of 98 m/s from the top of a hill 490 m high. Find (a) the time taken by the projectile to reach the ground (b) the distance of the point where the particle hits the ground from foot of the hill and (c) the velocity with which the projectile hits the ground (g = 9.8 m/s2) Sol. Here, ux = 98 m/s, a x = 0, uy = 0 and ay = g (a) At A, sy = 490 m. So, applying 1 sy u y t a y t2 2 1 490 0 (9.8)t 2 2 t = 10 s 1 BA s x u x t a x t 2 (b) 2 or BA = (98) (10) + (0) or BA = 980 m (c) vx = ux = 98 m/s
O
u = 98 m/s x y
B
A
vx
vy
vy = uy + ayt = 0 + (9.8) (10) = 98 m/s v 2x v 2y (98) 2 (98) 2 98 2 m / s
v
and
tan
= 45º
vy vx
98 98
1
Thus, the projectile hits the ground with a velocity 98 2 m /s at an angle of = 45º with horizontal. Example 35. A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute. Find
(a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed. Sol. (a) S1 = The distance moved by the train in first half minute. 1 S1 = u1t 1 + a1t12 2 1 1 S1 = 0 × t1 + × 2 × (30) 2 { t1 = minute} 2 2 1 S1 = × 2 × 900 S1 = 900 m 2 The distance moved by the train after brakes applied. 1 S2 = u2t2 + a2t 22 2 u2 = u1 + a1t1 u2 = 0 + 2 × 30 u2 = 60 m/s v2 = u2 – a2t2 0 = 60 – a 2 × 60
a 2 = 1 m / sec2
v22 = u 22 + 2 a 2S2 0 = 60 × 60 – 2 × 1 × S 2 S2 = 1800 m Total distance(s) = S1 + S2 S = (900 + 1800) m S = 2700 m
S = 2.7 km
(b) the maximum speed attained by the train is u 2 = 60 m / s
(c) the position of the train at half of minimum speed. u22 = u12 + 2a1S’ (30)2 = 0 + 2 × 2 × S’ 900 = S' 4
S' = 225 m
Example 36. A car is moving along a straight line. It is taken from rest to a velocity of 20 ms –1 by a constant acceleration of 5 ms –2. It maintains a constant velocity of 20 ms –2 for 5 seconds and then is brought to rest again by a constant acceleration of –2 ms –2. Draw a velocity-time graph and find the distance covered by the car. Sol. v22 = u2 + 2aS (20)2 = 0 + 2 × 5 × S 400 S= S = 40 m 10 Car maintains a constant velocity of 20 m/s for 5 Second. S’ = v × t S’’ = 20 × 5
S '' = 100 m
The car comes to rest by a constant acceleration of –2 m/s (v’)2 = v2 + 2 × a’S’’ 0 = (20)2 – 2 × 2 S’’ 400 S '' = S '' = 100 m 4 The total distance covered by the car. = S + S’ + S’’ = 40 + 100 + 100 = 240 m Example 37. A ball rolls off the edge of a horizontal table top 4m high. If it strikes the floor at a point 5m horizontally away from the edge of the table, what was its speed at the instant it left the table ? 1
Sol. Using h
2
gt 2 we have, v
h AB t AC
or Further,
1
A
2 gt AC
2 2h AB
g
BC = vt AC
or
24 9.8
4m
0.9 s
C
B
v
BC t AC
5.0 0.9
5.55 m / s
5m
Example 38. A ball is projected at an angle of 30º above with the horizontal from the to p of a tower and strikes the ground in 5 sec at an angle of 45º with the ho rizontal. Find the height of t he tower and the speed with which it was projected. Sol. ux = v0 cos30º uy = v0 sin30º Vx = uX = v0 cos30º
y
vy = uy + ay t
v0
vy = v0 sin30 – gt
vy
But
v0 2
30º x
10 5
H
v0
vy
– 50
2 – tan 45º v x v 0 cos 30 v v 0 cos 30 0 50 2 1
v0
2
v0 =
=
3
50 2
100 3 +1 100
3 –1
3 1 50
3 –1
3 –1
a =0, a = –g x
y
Also,
y = - H = u yt +
1 2
a yt2
H v0 sin 30 5
1
2
10 5 2
H 125 2 – 3 m
Example 39. A ball is thrown horizontally from a cliff such that it strikes ground after 5 sec. The line of sight from the point of projection to the point of hitting makes an angle of 37º with t he horizontal. What is the initial velocity of projection. Sol. vx = v0cos37º vy = v0sin37º
37º
y
ax = gsin37º = 6 m/s
2
O 37º
2
ay = –gcos37º = –8 m/s From O to A, displacement along y-axis is zero.
v0
37º
y = u yt + ½ ayt 2 0 = uy × 5 + ½ × –8 × 25
{ t = 5 sec.}
A
x
5u y = ½ × 8 × 25 100 uy = = 20 m / s 5 3 20 v0 5 100 v0 = m/s 3 Example 40. A ball is projected from top of a tower with a velocity of 5 m/s at an angle of 53º to horizontal. Its speed when it is at a height of 0.45 m from the po int of projection is : (A) 2 m/s (B) 3 m/s (C) 4 m/s (D) data insufficient Sol. According to conservation principle of machenic energy Ui + T i = Uf + Tf 1 1 0 + mu 2 = mgh + mv 2 2 2 2 2 u = 2gh + v v=
u 2 - 2gh
v 52 2 10 0.45 v
25 9
16
v
4 m/s u
Example 41. In the figure shown, the two projectiles are fired simultaneously. 20m/s What should be the initial speed of the left side project ile for the two 60º 45º projectile to hit in mid-air ? 10 m Sol. When two projectiles are projected from same height, then for collision, vertical component of velocities of both projectiles should be same
u sin 60º = 20 sin45º
u=
20sin 45º sin60º
20 2
u
2 3
u = 20
2 3
m/s
Example 42. The speed of a particle when it is at its greatest height is 2 / 5 times of its speed when it is at its half the maximum height. t he angle of projection is _____ a nd the velocity vector angle at half the maximum height is ______. uy = usin
Sol.
ux = ucos vx = ucos
u
u cos
...(1)
v2y = u2y +2ayS v 2y u 2 sin 2 2g
u cos
2
cos
u g
2
1
5
1
u
5
2
1
2
2
2g
sin 2 2
2 2
2
sin 2
1 sin
,
5 5 5sin 2 sin 2 2 2 3 – 4 sin = 0
or or sin and
5
3
2 tan
vy vx
2
u sin u sin g 2g
tan
2
2
sin
5
2 5
sin sin 2 cos 2
tan
tan
2 tan cos
2
ucos 2
sin 2
{from eqn. (1) and (2)}
u cos
2
2
60º
u 2 sin 2 gH
2
sin
sin
5
H
u sin 2
2
5
cos
2
u 2 cos 2 u 2 sin 2 2g
5
u cos
cos
v 2x v 2y
2
2
2
...(2)
2 2
5
u cos
or
H
tan 2
tan 60º 2
3
tan
tan 1
2
3 2
Example 43. A projectile is to be thrown horizontally from the top of a wall of height 1.7 m. Calculate the initial velocity of projection if its hits perpendicularly an incline of angle 37º which starts from the ground at the bottom of the wall. The line of greatest slope o f incline lies in the plane of motion of pro jectile. Sol. ux = v0 uy = 0
ay = g = 10 m/s 2 vx = v1 cos53º = 0.6 v 1 vy = v1 sin53º = 0.8 v 1
vx = ux = v0 or 0.6 v1 = v 0
v1 =
v0
=
10v0
=
0.6 6 vy = uy + ayt
5v0
...(1)
3
0.8v1 = 10 t
0.8
v0 =
5v0 3 6
0.8 In BAC tan37º = or or or or
v0
O
h = 1.7 m
A
x
v1 C
37º
1 2
2
gt
10 t 53º
37º
t=
60 8
t
AB
3
AC
B
...(2)
4
1.7 =
1
2 uxt
gt 2 =
y
grad
x
53º
v1
1.7 - 5t 2 v0 t
3v0t = 6.8 – 20t 2
3v 0
8v 0
2
64 v0
6.8 20 60 3600 2 24 2 64v 0 v0 + = 6.8 60 180 6.8 180 9 v 20 136
72 64 2 180 v 0 6.8
v0 = 3 m / s
Example 44. A hunter is riding an elephant of height 4m moving in straight line with uniform speed of 2 m/ sec. A dear running with a speed V in front at a distance of 4 5m moving perpendicular to the direction of motion of the elephant. If hunter can throw his spear with a speed of 10 m/sec, relative to the elephant, then at what angle to it’s direction of motion must he thrown his spear horizontally for a successful hit. Find also the speed ‘V’ of the dear. 1 2 Sol. h = gt 2 2h 24 8 t g 10 10
Assume horizontal plane at x–y–plane. v rel 10 cos ˆi 10sin ˆj
u 2 ˆi u v rel 2 ˆi 10 cos 2 ˆi 10 sin ˆj The deer is moving along y–axis. So, displacement of deer and displacement of spear along y–axis will be same in time t. v t = u yt v = 10 sin ...(1) Also, along x–axis :
v rel
ux t = 4 5 or or
10cos 2
8 10
10cos 2 4
10 cos = 8 8 4 cos 10 5 From eqn (1)
5
4m
4 5 10 8
2m/s
45 2
x
10
v = 10 sin = 10 sin37º = 10
3 = 5
= 37º
6m/s
y A Example 45. An object A is kept fixed at the point x = 3 m and y = 1.25 m on a P 1.25m plank P raised above the ground. At time t = 0 the plank starts moving along the + x direction with an acceleration 1.5 m/s 2. At the same instant a stone is u x O 3.0m projected from the origin with a velocity u as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45º to the horizontal. All the motions are in x – y plane. Find u and the time after which the stone hits the object. Take g = 10 m/s 2.
Sol.
ux = u cos uy = u sin ax = 0 ay = –g If the stone hitt the object after time t. So that virtical displacement of stone is 1.25 m 1 y = uyt + ayt2 2 1 2 therefore 1.25 (u sin )t gt 2 1 1.25 (u sin )t 10 t 2 2 1.25 = (u sin) t – 5 t 2 (u sin)t = 1.25 + 5t 2 Hotizontal displacement of stone is x = 3 + displacement of object A. Initial velocity of object is zero.
u
u =ucos x 45º
n i s u
1.25 u cos
...(1)
so displacement of object is 1 1 1.25 at 2 1.5 t 2 2 2 2 x = 3 + 0.75 t
0.75 t
2
(u cos)t = 3 + 0.75 t 2 Since velocity vector inclined at 45º with horizontals. u y u sin gt tan(–45º) ux u cos u cos = – (u sin – gt) u cos = gt – u sin (u cos) t + (u sin) t = 10 t 2 Add eqn (1) and (2) (u cos) t + (u sin) t = 4.25 + 5.75 t 2 from eqn (3) and (4) 10 t2 = 4.25 + 5 .27 t 2
...(2)
...(3)
...(4)
4.25 t 2 = 4.25 t2 = 1 From eqn (1) and (2) uy = u sin = 6.25 m/s
t = 1 sec.
uy = 6.25 m/s ux = 3.75 m/s 2
2
u=
ux + u y
u=
(6.25) 2 + (3.75) 2
u = 7.29 m / s
Example 46. A particle is projected with a velocity of 20 m/s at an angle of 30º to an inclined plane of inclination 30º to the horizontal. The particle hits the inclined plane at an angle of 30º, during its journey. Find the (a) time of impact, (b) the height of the point of impact from the horizontal plane passing through the point of projection. Sol. The particle hits the plane at 30º (the angle of inclination of plane). It means particle hits the plane horizontally. (a)
(b)
t
t
T
u
u sin
2
g
20 sin 30º 30º
H H
9.8 2 u sin 2
1.76 s 30º 30º
2g
20
2
sin
2
60º
2 9.8
15.3 m
Example 47. A particle is projected up an inclined plane with initial speed v = 20 m/s at an angle = 30º with plane. Find the component of its velocity perpendicular to plane when it strikes the plane. Sol. Component of velocity perpendicular to plane remains the same (in opposite direction) i.e., u sin = 20 sin 30º = 10 m/s Example 48. A particle is thrown horizontally with relative velocity 10 m/s from an inclined plane, which is also moving with acceleration 10 m/s2 vertically upward. Find the time after which it lands on the plane (g = 10 m/s2). Sol.
ux = 10 cos30º = 5 3 m/s uy = 10 sin30º = 5 m/s arel =20 m/s 2
2
10m/s 30º
y 10sin30º 10 m/s 2
20
A
x
2
10 3 m / s 2 when ball lands on inclined plane, y = 0 1 y = uyt + ay t 2 2 1 0 5 t 10 3 t 2 2 5 3 t =5
t=
1 3
O º 0 3 o s c 1 0
KINEMATICS
30º
ax = a rel sin30º = 10 m/s 2 ay = – a rel cos30º 3
v0= 10m/s
30º
sec.
Example 49. A particle is projected from point P with velocity 5 2 m/s perpendicular to the surface of a hollow right angle cone whose axis is vertical. It collides at Q normally. Find the time of the flight of the particle. Sol.
u0x = 5 2m / s u0y = 0 ax = –g sin45º ay = –g cos45º At point Q, v x = 0 { the particle collied normally at point Q} The time taken by particle to go from P to Q is t 0. vx = ux + ax t 0 = 5 2 – g sin45º t 0 t0
5 2 g sin 45º
5 2 2 10
y
P
Q 45º
y
x
x P
Q
º 5 4
g 45º
45º
1 sec.
t 0 = 1 sec.
Example 50. A ball is projected on smoot inclined plane in direction perpendicular to line of greatest slope with velocity of 8m/s. Find it’s speed after 1 sec. Sol. vy = uy + ayt vy = 0 + 10 sin37º × 1 3 vy = 10 1 5 vy = 6 m/s vx = ux = 8 m/s
v=
v 2x + v 2y
= 82 + 6 2 Ans.
10 m / s
37º
y
x
7 º i n 3 g s
8 m/s
º 3 7
ASSERTION & REASON it of (A) (B) (C) (D) (E) 1.
THE NEXT QUESTIONS REFER TO THE FOLL OWING INSTRUCTIONS A statement of assertion (A) is giv en and a Corresponding statement of reason(R) is given just below t he statements, mark the correct answer as – If both A and R are true and R is the correct explanation of A. If both A and R are true but R is not the correct explan ation of A. If A is true but R is f alse. If both A and R are false. If A is false but R i s true.
As sert io n (A) : Reason (R) :
2.
As sert io n (A) :
Horizontal component of the velo city of angular projectile remains unchange during its flight. At highest poi nt the velocity reduces to zero. (A) (B) (C) (D) (E) Suppose a particle starts moving in a straight line with initial velocity +u and an acceleration –a, t hen velocit y at displacement s comes out to be, between
v
2
v
2
u
2
2as . If we draw a graph
and s, it will be a straight li ne as shown in figure. 2
v
S0
Reason (R) :
v
2
S
versus s graph is a straight line passing through origin with positive intercept and
negative slope. (A)
(B)
(C)
(D)
(E)
3.
As sert io n (A) : Reason (R) :
A body can h ave acc eleration ev en i f its v elocit y is ze ro at a gi ven instant of t ime. A body is mome ntarily at rest when i t rev erses its di rection of motion. (A) (B) (C) (D) (E)
4.
As sert io n (A) :
When a body is projected with an angle 45°, its range is maximum.
Reason (R) :
For maximisation of range sin 2 should be equal to one. (A)
(B)
(C)
(D)
(E)
5.
As sert io n (A) : Reason (R) :
Rocket in flight is not an illustration of projecti le. Rocket takes flight due to combustion of fuel and does not mov e under the gravity eff ect alone. (A) (B) (C) (D) (E)
6.
As sert io n (A) :
The slope of displacement-time graph of a body moving with high velocity is steeper than the slope of displacement-time graph of a body with low velocity. Slope of displacement-time graph = Velocity of the body. (A) (B) (C) (D) (E)
Reason (R) :
7.
As sert io n (A) : Reason (R) :
Displacement of a body may be zero when distance travelled by it is not zero. The displacement is the longest distance between initial and final position. (A) (B) (C) (D) (E)
8.
As sert io n (A) : Reason (R) :
9.
The relative v elocity between any two bodies moving in opposite direction is equal to sum of the veloci ties of two bodies. Sometimes relativ e velocit y between two bodies is equal to difference in veloci ties of the two. (A) (B) (C) (D) (E)
As sert io n (A) : Reason (R) :
The displacement-time graph of a body mov ing with uniform acceleration is a straight line. The displacement is proportional to time f or uniformly accelerated moti on. (A) (B) (C) (D) (E)
10. Assertion (A) : Reason (R) :
A body fall ing freely may do so with constant v elocity. The body falls f reely, when acceleration of a body is equal to acceleration due to gravi ty. (A) (B) (C) (D) (E)
11. Assertion (A) : Reason (R) :
The speedometer of an automobile measure the average speed of the automobile. Average velocity is equal to total displacem ent per total time taken. (A) (B) (C) (D) (E)
Level # 1 1.
A trai n starting f rom rest t rav els the fi rst p art of its journey with con stant accel eration a, second part wi th constant velocity v and third part with constant retardation a, being brought to rest. The average speed for the whole journey is (A)
2.
3
7v . The train travels with constant velocity for .... of the total ti me 8
(B)
4
7 8
(C)
5
(D)
6
9 7
A parti cle mov ing in a stra ight line with uniform acceleration is observ ed to be at a distance a from a fixed point initi ally. It is at distances b, c, d from the same point after n, 2n, 3n second. The acceleration of the particle is (A)
c 2b a n
2
(B)
cba 9n
(C)
2
c 2b a 4n
2
(D)
c ba n2
3.
If a, b and c be the distances travelled by the body during xth, yth and zth second from start, then which of the following relations is true ? (A) a(y – z) + b(z – x ) + c(x – y) = 0 (B) a(x – y) + b(y – z) + c(z – x) = 0 (C) a(z – x) + b(x – y) + c(y – z) = 0 (D) ax + by + cz = 0
4.
A body of mass 3 kg falls from th e multi -storeyed building 100 m high and buries itself 2m deep in the sand. The time of penetration will be (A) 0.09 s (B) 0.9 s (C) 9 s (D) 10 s
5.
Two cars A and B, each hav ing a speed of 30 km/hr are h eading towards each other alo ng a straight path. A bird that can fly at 60 km/hr flies off car A when the distance between the cars is 60 km, heads directly towards car B, on reaching B, the bird direct ly flies back t o A and so forth, then the t otal distance the bird travels till t he cars meet is. (A) infinite (B) 30 km (C) 60 km (D) 120 km
6.
Two cars A and B, each hav ing a speed of 30 km/hr are h eading towards each other alo ng a straight path. A bird that can fly at 60 km/hr flies off car A when the distance between the cars is 60 km, heads directly towards car B, on reaching B, the bird directly f lies back to A and so forth, the total no. of trips which the bird makes till the cars meet is (A) Four (B) Eight (C) Sixteen (D) Infinite
7.
Three particles start from the origin at the same time, one with a v elocity v1 along x-axis t he second along the y-axis with a velocity v2 and the third along x = y line. The vel ocity of the third so that the three may always lie on the same line. (A)
8.
v1 v 2 v1 v 2
(B)
2 v1 v 2 v1 v 2
(C)
3 v1 v 2 v1 v 2
(D) Zero
Choose the correct statement (A) A body start s from re st and mov ing wit h constan t accel eration travels a distance y1 and the 3rd second and y2 in 5th second. The ratio
y1 y2
5 . 9
(B) A ball fall s from the top of a t ower i n 8 secon d in 4 sec ond, i t til l cov er the fi rst quarter of the distance starting from top. (C) The distance traveled by a freely falling stone released. With zero velocity in the last second of its motion to that traveled by it in the last second of its motion to that traveled by it in the last but one second is 7 : 5. The stone strike the ground with velocity 39.2 m/ s. (D) None of these
9.
A steam boat goes across a lake and comes back (i) on a quiet day when the water is still, an d (ii) on a rough day when there is a uniform current so as to help the journey onward and to impede the journey back. If the speed of t he launch, on both days, was same, the tim e required for the comp lete journey on the rough day, as compared to that o n the quiet day, will be. (A) Less (B) Same (C) More (D) Cannot be predicted
10.
A car i s mo ving along a straigh t road with a unif orm acceleration. It pa sses t hrough two points P and Q separated by a distance with velocity 30 km/hr and 4 0 km/hr respectively. The velocity of the car midway between P and Q is (A) 33.3 km/hr
11.
(B) 20 3 km/hr
(C) 25 2 km/hr
(D) 35 km/hr
Starting from rest a particle moves in a straight line with acceleration a = (25 – t2) 1/2 m/s 2 for 0 t 5 s, 3 m/s 2 for t > 5 s. The velocity of particl e at t = 7 s is. 8 (A) 11 m/s (B) 22 m/s (C) 33 m/s
a=
(D) 44 m/s
12.
A body is released from a great height and falls fre ely towards the earth. Exact ly one sec later another body is released. W hat is the distance between the two bodies 2 sec after the release of the second body. (A) 4.9 m (B) 9.8 m (C) 24.5 m (D) 50 m
13.
The distance moved by a freely fall ing body (starting from rest) during 1st, 2nd, 3rd, .... nth s of its motion are proportional to. (A) Even numbers (B) Odd numbers (C) All integral numbers (D) Square of integral numbers
14.
A ball is thro wn vert ical ly upwards with a speed of 10 m/s from the top of a tower 200 m hig h and another is thrown vertically downwards with the same speed simultaneously. the time difference between them in reaching the ground in s(g = 10 m/ s2) is (A) 12 (B) 6 (C) 2 (D) 1
15.
Between two stations a train first acceler ates uniforml y, then mov es with uniform speed and finally, retards uniformly. If the ratios of the time taken f or acceleration, uniform speed and retarded motions are 1 : 8 : 1 and the maximum speed of the train is 60 km/h, the av erage speed of the train over the whole journey is. (A) 25 km/h (B) 54 km/h (C) 40 km/h (D) 50 km/h
16.
A bal l is thrown vertically upwards. It was obse rv ed at a hei ght h twice with a time interval t. The initial velocity of the ball is. (A)
8 gh g ( t ) 2
2
(B)
g t 8 gh 2
2
(C)
1 8 gh g2 ( t )2 2
(D)
8 gh 4g2 ( t) 2
17.
A target is made of two plate s, one of wood and the other of i ron. The thi ckness of the woode n plate is 4 cm and that of iron plate is 2 cm. A bullet fired goes through the wood first and then penetrates 1 cm into iron. a similar bullet f ired with the same velocity from opposite direction goes through iron first and then penetrates 2 cm into wood. If a1 and a2 be the retardation off ered to the bullet by wood and iron plates respective ly then (A) a 1 = 2a 2 (B) a2 = 2a 1 (C) a1 = a2 (D) Data insufficient
18.
The motion of a body fal ling from rest in a resisting medium is described by the equation a and b are constants. The velocit y at any time t is a a b (A) v t = (1 – e –bt) (B) vt = e –bt (C) v t = (1 + e –bt) b b a
19.
(D) vt =
dv = a – bv where dt
b bt e a
A self-propelled vehic le of mass M whose engine deli vers constant power P has an accelerat ion a =
P
Mv increase the velocity of t he vehicle from v1 to v2, the distance travelled by it (assuming no friction) is (A) s =
3P 2 (v 2 v12 ) M
(B) s =
M 2 (v 2 v 12 ) 3P
(C) s =
M 3 (v 2 v 13 ) 3P
(D) s =
3P 3 ( v 2 v 13 ) M
. To o
20.
For an airplane to take-off it accelerates according to the graph shown and takes 12 s to take-of f f rom the rest position. The distance travel led by the airplane is. (A) 21 m (B) 210 m (C) 2100 m (D) 120 m
A
5
B
s / m
2
6
t (in s)
12
21. A river is flowing from west t o east at a speed of 5 met ers per minute. A man on the south bank of the riv er, capable of swimming at 10 meters per minute in still water, wants to swim across the river in the shortest time. He should swim in a d irection. (A) due north (B) 30° east or north (C) 30° west of north (D) 60° east of north 22. A boat whi ch has a speed of 5 km/ hr in still water crosses a riv er of wi dth 1 km along t he shortest possible path in 15 minutes. The velocity of the river water in km/hr is A (A) 1
(B) 3
(C) 4
(D)
41
23. In 1, 0, s, a particle goes from point A to point B, mov ing in a semicircle of radius 1.0 m (see Figure). The magnitude of the average velocity (A) 3.14 m/s (B) 2.0 m/s (C) 1.0 m/s (D) Zero
1.0 m
B
24. A ball is dropped ver tically from a height d above the ground. It hits the ground and
bounces up vertically to a height d 2 . Neglecting subsequent motion and air resistance, its velocit y v varies with the height h above the ground as v
v d
(A)
h
(B)
v
d
v d
h
d
h
(C)
(D)
25. A body st arts from rest at time t = 0, t he acce leration tim e graph i s shown in the figure. The maximum v elocity attained by the body will be (A) 110 m/s (B) 55 m/s (C) 650 m/s (D) 550 m/s
h
Acceleration 2 (m/s ) 10
Time 11 (Sec.)
26. The veloci ty is displacement graph of a particle mov ing along a straight line is shown v v0
x0 The most suitable acceleration-displacement graph will be a
a
a
x (A)
x
a x
(B)
(C) x
x (D)
Multipl e Choi ce Question 27. A parti cle is mov ing eastwards with a velocity of 5 m/s. In 10s the v elocity changes to 5 m/s northwards. The average acceleration in this time is (A) Zero (C) 1
(B) 1 1
2 2 m s towards north-east
(D)
2
2 2 m s towards north-west 2
m s towards north-west
28. A parti cle of mass m mov es on the x-axi s as follows : it starts f rom rest at t = 0 from the point x = 0, and comes to rest at t = 1 at the point x = 1. NO other inform ation is available about its motion at int ermediate times (0 < t < 1). If denotes the instantaneous acceleration of t he particle, then: (A) cannot remain positive for all t i n the interval 0 t 1 . (B) cannot exceed 2 at any point in it s path. (C) must be 4 at some point or points in its path. (D) must change sign during the motion, but no other assertion can be made with the information giv en. 29. The coordinates of a particle mov ing in a plane are given by x(t ) = a cos(pt) and y(t) = b sin(pt) where a, b(
(B) tan ( + )
(C) sin 2 – cos2 = sin2 – cos2
(D) cot = cos sec
Fill in the blanks 1.
A particle mov es in a circle of radius R. In hal f the period of rev olution its disp lacement is _________ and distance covered i s ___________.
2.
Four persons K, L, M, N are initial ly at the four corners of a square of sided. Each person now move s with a uniform speed v in such a way that K always moves directly towards L, L directly towards M, M directly towards N, and N directly towards K. The four persons will meet at a time __________.
3.
Spotlight S rotates in a horizontal plane with constant angular veloc ity of 0.1 radian /second. The spot of light P moves along the wall at a distance of 3 m. The velocity of the spot P when 45 (see figure) is __________ ms/. S
4.
The trajectory of a projectil e in a vert ical plane is y = ax – bx2, where a, b are constants, and x and y are respectively t he horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is ________ and the angle of pro jection from the hori zontal is _____ ___.
3m
P
True / False 5.
Two Two balls of d ifferent ma sses are are thrown vertic ally upwards with the same speed. They pass through through the point of projection in their downward motion with the same speed (Neglect air resistance).
6.
A projecti proj ecti le fi red fr om the ground gr ound f ollo ws a parabol para bolic ic path. path . The speed of t he project proj ectile ile is i s mini mum at th e top of its path.
7.
Two Two identical trains are mov ing on rails along the equator on the earth in opposite directions with the same speed. They will exert the same pressure pressure on the rail s.
8.
An electr ele ctr ic line li ne of force fo rces s in i n t he x -y plane pla ne is gi v en by the equ ati on x2 + y 2 = 1. A particle with unit positive charge, initiall y at rest at the point x = 1, y = 0 in the x-y plane, will mov e along the circular line of force.
Table Match 9.
Match List I and List II and select the correct answer using the codes given below in the lists: Co l u m n -I Co l u m n -I I
I. Deceleration decreasing
A. time
II . Deceleration increasing
B. time
velocity
III. Accel Accel erati on decreasing decreasi ng
C. time velocity
IV. Uniform acceleration
D. time velocity
E. time (A) I—D, (A) I—D, II—E, III—C, IV—A (C) I—C, (C) I—C, II—E, III—B, IV—A
(B) I—E, (B) I—E, II—B, III—C, IV—D (D) I—D, (D) I—D, II—B, III—A, IV—C
Pass Passage age Type Type Quest Quest io ns THE NEXT QUESTIONS REFER TO THE FOLL OWING PASSAGE When an airplane f lies, its total veloci ty with respect to the ground is –
vtotal
v plane v wind
Where v plane , denotes the plane’s velocity through motionless air, and vwind denotes the wind’s velocit y..
Crucially, Crucially, all t he quantities in this equation equation are vectors. The magnitude of a v elocity vector is often called the “speed”. Consider an airplane whose speed through motionless air is 100 m/s. To To reach its destination , the plane must fly east. The “heading” of a plane is the direction in which the nose of the plane points. So, it is the direction in which the engines propel the plane. 1.
If the plane has an eastward heading, and a 20 m/s wind blows towards the southwest, then the plane’s speed is – (A) 80 (A) 80 m/s (B ) more than 80 m/s but less than 100 m/s (C) 100 (C) 100 m/s (D) more (D) more than 100 m/s
2.
The pilot mai ntains an eastward heading while a 20 m/s wind wind blows northward. The plane’s veloci ty is deflect ed from due east by what angle? 1
(A) sin
3.
1
1
(B) cos
5
1
5
1
(C) tan
1
(D) None (D) None of these
5
Let denote the answer to previous question. The plane in question 2 has what speed with respect to the ground? (A) 100 m s sin
4.
(C)
(D)
sin
100 m s cos
Because the 20 m/s northward wind persists, the pilot adjusts the heading so that the plane’s total v elocity is eastward. By what angle does the new heading differ from due east? (A) sin
5.
(B) 100 m s cos
100 m s
1
1
(B) cos
5
1
1 5
(C) tan
1
1 5
(D) None (D) None of these
Let denote the answer to previous question. What is the total speed, with respect to the ground, of the plane in previous question? (A) 100 m s sin
(B) 100 m s cos
(C)
100 m s
(D)
sin
100 m s cos
THE NEXT QUESTIONS REFER TO THE FOLL OWING PASSAGE During a car crash, the more r apidly a person decelerates, t he more likel y she is to be injured. A large deceleration is dangerous, even if it lasts for a short time. Airba gs are designed to decrease decre ase the magni tude of the dec elera tion. Before Be fore the th e airbag inflat in flat es, the driv er continues forward at constant speed. But once the airbag inf lates, the driv er decelerates gradually, instead of getting thrown into t he windshield or steering wheel. 20 v (m/s)
Airbag Airb ag 1
.01 .02 .03 .04 .05 t(s)
20 v (m/s)
Airbag Airb ag 2
.01 .02 .03 .04 .05 t(s)
Airbag Airba g3
20 v (m/s)
.01 .0 1 .02 .0 .03 3 .0 .04 4 .0 .05 5 t(s) Three diff erent models of airbags were tested using identical crash test dummies. Sensors measured the velocity of the crash test dummy as a function of time, when the car crashed at 20 m/s into a brick wall. The shown shown velocit y vs, tim e graphs resulted. resulted. Time t = 0 is the moment t he car crashes. 6.
All t hree airbags air bags are ar e the same sa me si ze and shape. sha pe. W hich hi ch one inf i nflat lat es mos t qui ckl y? (A) Airbag (A) Airbag 1 (B) Airbag (B) Airbag 2 (C) Airbag (C) Airbag 3 (D) W (D) W e cannot determine the answer answer from the given information.
7.
Let
a
max
denote the largest instantaneous acceleration that the crash test test dummy ex periences during the
crash. The best airbag is the one for which (A) Airbag (A) Airbag 1 (C) Airbag (C) Airbag 3 8.
max
is as small as possible. Which airbag is best?
(B) Airbag (B) Airbag 2 (D) W (D) W e cannot determine the answer answer from the given information.
For airbag 2, which of the f ollowing graphs best represents theposition of the crash test dummy as a function of time? Let x = 0 m be the dummy’s position at time t = 0s. A
B
n o i t i s o P
(A) A (A) A
C
n o i t i s o P
.01
9.
a
time(s)
n o i t i s o P
time(s)
.01
(B) (B ) B
D
(C) C (C) C
n o i t i s o P
.01
time(s)
.01 time(s)
(D) D (D) D
For airbag 2, approximate ly how much distance does the dummy dummy cov er between the moment the car crashes and the moment the dummy first makes contact with the airbag? (A) 0.2 (A) 0.2 m (B) (B ) 0.4 m (C) 0.6 (C) 0.6 m (D) 1.0 (D) 1.0 m THE NEXT QUESTIONS REFER TO THE FOLL OWING PASSAGE Recently, college teams from all ov er the country sent tennis players to participate in a series of experiments conducted by the Physical Education Department of a major university. A variety of coaching met hods was used to improv e the players’ serv es, described below. Experiment Experiment 1 Two Two groups of 50 tennis players worked on the speed of their basic serves for two weeks. one group consisted solely of right-handed players; the other consisted consisted solely of left-handed players. Half of each group watched watched vide os of a right-handed tennis coach. Each player was told to pattern his or her serve on that of the coach in the video. T he players received no v erbal or physical guidance. the average speed of each player’s serve was measured at the begi nning and end of the two-week perio d, and changes were recorded in Table Table 1.
Table 1 Players' handedness Right Right Left Left
Coach's handedness Right Left Right Left
Averag Av erag e ch ange an ge i n speed (mph) 5 2 -1 8
Experiment Experiment 2 For two weeks, a second group group of 100 right- handed tennis players watched the same videos of the righthanded tennis coach. The coach also physically physically guided 50 of the those players through the moti ons of the serve. Again, no verbal i nstruction was given during the experiment. The av erage speed and accuracy of each player’s serves were recorded at the beginning and end of this two-week two-week period. The r esults are recorded in Table abl e - 2.
Table 2 Guided No Yes
Aver age Chan ge in speed (mph) 5 9
Aver age ch ange in Acc uracy 15% 25%
Experiment 3 For two weeks, a third group of 100 right-handed tennis players worked on their basis serves. 50 players received no v erbal instruction; t hey watched the same video of the right handed tennis coach, who also physically guided them through the motions of the serve. T he other 50 players did not observe the video but received verbal instructi on from the coach, who then physically guided them thought the motions of the serve. The results are shown in Table 3.
Tabl e 3 Guided Plus Video Verbal Coaching
Av erag e Chan ge i n speed (mph) 7 10
10. Which of the following results would the expected if Experiment 3 were repeated using left-handed tennis players and a left -handed coach– (A) The average servic e accuracy of all the play ers would increases by at least 30%. (B) The average servic e speed of all the players would decrease slightly. (C) Verbal coaching would improve average service speed less than would watching the v ideo. (D) The average servi ce speed of the players who watched the video would increase by at l east 8 mph. 11. Whi ch of the foll owing conclusions could NOT be supported by the results of Ex periment– (A) Imitating someone whose handedness is the opposite of one’s own will cause one’s skill s to deteriorate. (B) Left-handed people are better than right-handed people at imitating the movement of someone with similar handedness. (C) People learn more easily by observing someone with simi lar handedness than by observi ng someone with handedness opposite their own. (D) Right-handed people are better than lef t-handed people at im itating the m ovement of someone whose handedness is opposite their own. 12. Which of the following hypotheses is best supported by the results of Ex periment 2– (A) Instructional v ideos are more helpful for right-handed tennis players than is verbal instructi on. (B) Instructional videos are m ore helpful f or left-handed tennis players than for right-handed tennis players. (C) Physical guidance by a coach improves both speed and accuracy of service for right-handed tennis players. (D) Physical guidance by a coach improv es service accuracy f or right-handed tennis players more than for left-handed players. 13. Suppose 50 left-handed tennis players watch a video of a left- handed coach and are also physically gui ded by that coach. The results of t he experiments suggest that the player’s average change in service speed will most closely approximate– (A) –1 mph (B) +5 mph (C) +8 mph (D) +12 mph 14. Which of the following hypotheses is best supported by the results of Experiment 1 alone – (A) Tennis players improv e less by observi ng coaches whose handedness is the opposite of their own than by observing those with simi lar handedness. (B) Right-handed tennis players are coached by left-handed coaches more frequently than left-handed players are coached by right-handed coaches. (C) Right-handed coaches are better models for all tennis players than are lef t-handed coaches. (D) People learn much better from physical contact plus a vi sual stimulus than from the vi sual stimulus alone.
THE NEXT QUESTIONS REFER TO THE FOLL OWING PASSAGE It is known that if a steel ball and a feather are dropped from the same height, the steel ball f alls faster than the feather. The following scientists have two different vi ews on fallin g bodies: Scientist 1 : The force of gravity makes things fall. The greater the gravitational force on an object, the faster it falls. The steel ball falls f aster because it is more massive than the feather, and thus is attracted more strongly by the earth’s gravitatio nal field. Scientist 2 : The mass of an object does not determine how fast an object f alls, but shape does. Two identical pieces of paper will fall at dif ferent speeds, if one of them is crumpled into a small ball. That is because gravity is not the only force acting on a f alling object. Air buoys up objects falling through it. Since the shape of the steel ball giv es it less air resistance than a feather, it f alls faster. 15. Accordi ng to Scientist 1, the propert y that det erm ines how fast an o bject wil l fall is it s– (A) chemical composition (B) mass (C) shape (D) gravity 16. Accordi ng to Scientist 1, the speed of the fa lling ball could be in creased by– (A) dropping the ball from a greater height (B) making the ball out of alumi num (C) reshaping the ball (D) using a larger, steel ball 17. Accordi ng to Scientist 2, a crumpled piece of paper fall s at a dif ferent speed from a fl at pie ce of paper because of its– (A) mass (B) air resistance (C) gravitational attraction (D) texture 18. Both scientists agree that the rate at which an object falls is affect ed by the– (A) force of gravity (B) mass of the object (C) object’s resistance of air (D) shape of object 19. Scientist 2 would predict that in a v acuum, two objects would fall at the same speed if they had the same– (A) shape and different masses (B) air resistance and shape (C) composition and air r esistance (D) mass and diff erent shapes 20. Suppose a small ball and a f eather having the same weight are dropped from the same height. Which of t he following would Scientist 1 predict? (A) the feather would fall f aster than the ball (B) the ball and feather would fall at the same speed. (C) the ball would fall faster than the feather (D) none of the above. 21. Two balls of identical shape, one made of lead and one m ade of aluminum, are both dropped from t he same height. W hich scientist’s prediction and reasoning is incorrect based on the arguments presented? (A) Scientist 1 : the lead ball f alls faster because the balls have different m asses. (B) Scientist 1 : the l ead ball falls f aster because the balls have dif ferent gravit ational attraction. (C) Scientist 2: The lead ball fal ls faster because the balls have the same air resistance. (D) Scientist 2: The ball s fall at t he same rate because they have the same air resistance. THE NEXT QUESTIONS REFER TO THE FOLL OWING PASSAGE A student perfo rms an experim ent to determi ne how the range of a ball depends on the velo city with which it is released. The “range” i s the distance between where the ball lands and where it was released, assuming it lands at the same height f rom which it was released. In each trial , the student uses the same baseball, and launches it at th e same angle. Table 1 shows the experimental results.
Table 1 Trial 1 2 3 4
Launch speed (m/s) 10 20 30 40
Range (m) 8 31.8 70.7 122.5
Based on this data, the student then hypothesizes that the range, R, depends on the initial speed, 0 , according to the following to the following equation: R
n
Cv 0 , where C is a constant, and n is another
constant. 22. Based on this data, the best guess for the value of n is– 1
(A)
(B) 1
2
(C) 2
(D) 3
23. The student speculates that he constant C depends on: I. The angle at which the ball was launch. II . The ball’s mass III. The ball’s diameter If we neglect air resistance, then C actually depends on– (A) I only (B) I and II (C) I and III
(D) I, II and III.
24. The student performs another trial in which the ball is launched at speed 5.0 m/s. Its range is approximately – (A) 1.0 meters (B) 2.0 meters (C) 3.0 meters (D) 4.0 meters
25. Let denote the angle of the ball’s initial v elocity, as measured from the horizontal. Neglect air resistance. At the pea k (highest po int) of its t rajectory, th e bal l’s spe ed is– (B) 0 sin
(A) 0
(C) 0 cos
(D) 0
26. For trial 2, which of the fol lowing graphs best represents the vertical component of the ball’s ve locity as a function of time, assuming upward is positive? + vy
(A)
0
+ vy time
(B)
0
+ vy time
(C)
0
+ vy time
(D)
0
time
THE NEXT QUESTIONS REFER TO THE FOLL OWING PASSAGE Aristotle dev eloped a syst em of phy sics based on what h e tho ught occurred in nature. For e xampl e, he thought that if a stone is released from rest, i t instantaneously reaches a speed that remains constant as the stone falls. He also believed that the speed attained by a stone falling in air varies directly with the weight of the stone. A 5–pound stone, for example, f alls with a constant speed 5 times as great as that o f a 1-pound stone. Aristotle also noted that stones dropped into water continue t o fall, but at a slower rate than stones falling through air. To account for t his, he explained that the resistance of the medium t hrough which an object falls also aff ects the speed. Therefore, he said, the speed of a fall ing object also v aries inversely with the resistance of the medium, and this resistance is the same for all objects. Galielo disagreed with Aristotle’s explanation. He generated the following arguments to refute. Aristotle. Consider a stake partially driv en into the ground and a heavy stone falling f rom vari ous heights onto the stake. If the stone falls f rom a height of 4 cubits, the stake will be driven into the ground, say, 4 finger breadths. But if the stone falls from a height of 1 cubit, the stake will be driven in a m uch smaller amount. Certainly, Galileo argued, if the stone is raised above the stake by only the thickness of a leaf, then the eff ect of the stone’s falling on the stake will be altogether unnoticeable. On the basis of a careful set of exp eriments, Galie lo argued that the speed of an object released from r est varies directly with the tim e of fall. Also, the distance the object fall s varies directly with the square of the time of f all if the effect of air resistance on the object is negligible. Thus, according to Galileo, objects actually fall with constant acceleration, and if air resistance is negligible, all object exactly the same acceleration. 27. Which graph accurately represents Galil eo’s theory of the relationship between speed and time for an object falling from rest under conditions of negligible air resistance –
(A)
d e e p S
(B)
d e e p S
Time
(C)
d e e p S
Time
(D)
d e e p S
Time
Time
28. A book dro pped from a height of 1 meter fal ls to the floo r in t seconds. To be consistent with Aristotl e’s views, from what height, in meters, should a book 3 times as heavy be dropped so that it will fall to the f loor in the same amount of time– (A) 1 9
(B) 1 3
(C) 1
(D) 3
29. Suppose a heavy object fal ls to the ground in t seconds when dropped from shoulder height. According to Galileo, if air resistance were negligible, h ow many seconds would it take an object half as heavy to fall t o the ground from the same height– (A) 0.5 t (B) 1.0 t (C) 1.5 t (D) 2.0 t 30. A piece of putty weighing 2 pound s is dropped down a shaft f rom the top of a tall building. 1 second later, a 3 pound piece of putty si dropped down the shaft. According to Aristotle, what happens to the 2 pieces of putty if they fall for a relativel y long time– (A) The separation between the 2 piece constantly i ncreases until they strike the ground. (B) The separation between the 2 pieces is constant until the strike the ground. (C) The heavier piece catches up to the smaller piece. and the 2 pieces travel together with the speed of the heavier piece. (D) The heavier piece catches up to the smaller piece, and the 2 piece trav el together with a speed faster than the speed of either. THE NEXT QUESTIONS REFER TO THE FOLL OWING PASSAGE Abhi shek a nd Sweta Bachhan are two friends Abhishek is Joint to IIT-P awai and Sweta com e to see off him at Indore railway station. Non Abhishek drops ball 1 from 1 m t height, just when the train starts to move. Abhishek observ es that ball doesn’t hit the floor point exac tly below the dropp ed point, i t gets dev iated by 20 cm. For 50 seconds train accelerates and attains the maximum speed and starts moving with thi s velocity.
50 m
50 m
Initial
Final
Now Abhishek drops ball 2 at th e same tim e (t = 50 sec.) Abhishek throw ball 3 such t hat it just rea ch Sweta. 31. Path of ball 1 seen by Sweta will be (A) Parabolic (B) Straight line
(C) Circle
(D) Can’t predict
32. Path of bal l 1 seen by Abhishek will be – (A) Parabolic (B) Straight line
(C) Circle
(D) Can’t predict
33. The acceleration of train is (A) g/2 (B) g/4
(C) g/5
34. The path of ball 2 as seen by Sweta will be (A) Parabolic (B) Straight line
(C) Circle
(D) Can’t predict
35. The path of bal l 2 as seen by Abhishek will be (A) Parabolic (B) Straight line
(C) Circle
(D) Can’t predict
36. The speed with which Abhishek throw ball 3 will be (A) 30 (B) 20
(D) Can’t predict
(C) 15
(D) None of these
THE NEXT QUESTIONS REFER TO THE FOLL OWING PASSAGE A cir cus wishes to develop a new clown act. Figur e 1 shows a diagram of the pro posed setup. A clown will be shot out of a cannon with velocity v0 at a tr ajectory that makes an angle 45 with the ground. At this angle, the clown will trav el a maximum hori zontal distance. The cannon will accelerate the clown by applying a constant force of 10,000 N ov er a v ery short time of 0.24 sec. The height above he ground at which the clown begins his trajectory is 10 m. A large hoop is to be suspended f rom the ceiling by a massless cabl e at just the right place so tha t the clown will be able to dive thro ugh it when he reaches a maximum height abov e the ground. After passing through the hoop he will then continue on his trajectory until arriving at the safety net. 42
vy (m/s)
28 21
4 0 k g
6 0 k 8 0 g k g
45°
t(s) 2.1 2.8
Figure 1
4.2
Figure 2 Figure 2 shows a graph of the ver tical component of t he clown’s velocity as a function of ti me between the cannon and the hoop. Since the v elocity depends on the mass of the particular cl own performing the act, the graph shows data for sev eral diff erent masses. 37. If the angle the cannon makes with the horizontal is increased from 45°, the hoop will have to be: (A) moved f arther away from the cannon and lowered. (B) moved f arther away from the cannon and raised. (C) moved closer to the cannon and lowered. (D) moved closer to the cannon and raised. 38. If the clown’s mass is 80 kg, what initial velocit y v0 will he have as he leav es the cannon? (A) 3 m s
(B) 15 m s
(C) 30 m s
(D) 300 m s
39. The slope of the line segments plotted in figure 2 is a constant. Which one of t he following physical quantities does this slope represent? (A)
g
(B) v0
(C) y y0
(D) sin
40. From Figure 2, approxim ately how much time will it ta ke for clown with a mass of 60 kg to reach the safety net located 10 m below the height of the cannon? (A) 4.3 s (B) 6.4 s (C) 5.9 s (D) 7.2 s 2
41. If the m ass of a clown doubles, his initial kineti c energy, mv0 2 , will: (A) remain the same (C) double
(B) be reduced to half (D) four times
42. If a clown holds on to the hoop instead of passing through it, what is the expression for minimum length of the cable so that he doesn’t hit is head on the ceili ng as he swings upward? 2
(A)
2v0 g
2
(B)
v0 g
2
(C)
v0
2g
2
(D)
v0
4g
Level # 2 MOTION IN ONE DIMENSION 1.
Two intersecting straight lines moves out parallel t o themselves with the speed v1 and v 2. Calculate the speed of the point of intersection of the lines if the angle between them is .
2.
A base bal l player hits a pitched ball so that i ts velocity reverses di recti on and i ts speed chan ges fr om 30 m/s to 40 m/s. The bat moves at an average velocity of 30 m/s. and it is in contact with the ball for a distance of 0.05 m. (A) For how long are the bat and ball i n contact ? (B) What is the av erage acceleration of the ball while it i s in contact with the bat ?
3.
A car sta rts f rom rest with an acceleration of 6m/s2 which decreases to zero linearly with time, i n 10 sec., after which car continues at a constant speed. Find the time required f or the car to travel 400 m from the start.
4.
The speed of a motor launch with respect to the water is v = 7 m/s, the speed of the stream u = 3 m/s. When the launch began traveling upstream, a float was dropped from it. The launch travelled 4.2 km upstream, turned about and caught up with the float. How long is it before the launch reaches the float again ? Assume that f loat is moving with the speed of stream.
5.
The speed of a train increases at a constant rate from zero to v and then remains constant for an interval and finally d ecreases to zero at a constant rate . If
is
the total distance described, prove that the total
v 1 1 . At what value of v is the time of trav el the shortest ? What is the value of the v 2
time taken is t = shortest time ? 6.
A passenger is running at his max imum velocity of 8 m/s to cat ch a train. When he is a distanc e d from the nearest entry to the train, the train starts from rest with constant acceleration a = 1 m/s2 away from the passengers. (A) If d = 30 m and he keeps running, will he be able to jump onto t he train. (B) Sketch the position function x (t) f or the train, choosing x = 0 at t = 0. On the same graph sketch x(t) for the passenger for various value of initial separation distance d, including d = 30 m and the critical separation distance dc, such that he just catches t he train, (C) For the critical separation d istance d0, what is the speed of the train when the passenger catches it ? What is its av erage speed for the tim e interval f rom t = 0 until he catches it ? What is the val ue of dc.
7.
The acceleration of a particl e is giv en by a = 4t – 30, where a is in meters per second squared and t is in seconds. Determine the velocity and displacement as functions of time. The initi al displacement at t = 0 is 2 s0 = –5m, and the initial velocit y is v0 = 3 m/s. a ,m/s x
8.
The acceleration of a particle that moves in the positive x-direction v aries with its position as shown. If the v elocity of the particle is 0.8 m/s when x = 0, determine the velocity v of the particl e when x = 1.4 m.
0.4
0.2
0 0
9.
A particle st arts from rest at x = –2 m and m oves along the x-axis with the veloci ty history shown. Plot the corresponding acceleration and di splacement histories for the two seconds. Find the tim e t when the particle crosses the origin.
0.4
0.8
1.2
v, m/s 3
0 0
2.0 0.5
1.0
1.5
-1
MOTION UNDER GRAVITY 10.
One student throws a ball vertically upward with an initi al speed of 9.8 m/s. Another student standing. 5m away starts running towards the ball on release and catches it at t he same height. What was the students acceleration ? (Assume a uniform acceleration).
11.
An alum inium bal l with a mass of 4 kg and an ir on ball o f the same size with a mass of 11.6 kg are dr opped simultaneously from a height of 49 m. (A) Neglecting air resistance, how long does it take the aluminium ball to gall to the ground. (B) How much later does the heavier iron ball strike t he ground ?
12.
Ball A is dropped from t he top of a building at the same instant that ball B i s thrown vertically upward from the ground. When ball s collide, they are mov ing in opposite directions and the speed of A is twice the speed of B. At what fracti on of the height of the building did the collision occur ?
13.
A ball is thrown ve rtically upward fr om the 12 m level in an elev ator shaft with an initi al velocity of 18 m/s. At the same instant an open platform elev ator passes the 5m l evel, mov ing upward with a constant velocity of 2 m/s. Determine (A) W hen and where the ball will hit the elevator,(B) the relative v elocity of the ball with respect to the elevator when the ball hits the elevator.
14.
An obje ct is thrown upwar d with an initial velocity v0. The drag on the object is assumed to be proportional to the velocity. What time will it take the object to move upward and what maximal alti tude will it reach ?
15.
A nut come s loose fr om a bolt on the bottom of an el evator as the elev ator is mov ing up th e shaft at 3 m/ s. The nut strikes the bottom of t he shaft in 2 sec. (A) How far from the bottom of the shaft was the elevator when the nut fell off. (B) How far above the bottom was the nut 0.25 s after it fell of f ?
16.
A heli copter is descending vertically downward with a unif orm v eloci ty. At a certain instant, a fo od packet is dropped from it which takes 5 seconds to reach the ground. As this packet strikes the ground, another f ood packet is dropped from it, which takes 4 seconds to reach the ground. Find the velocity with which the helicop ter is descending and its height, when second packet is dropped. Also find the distance travell ed by the helicopter during the interval of dropping the packets.
RELATIVE MOTION 17.
Two bodies moves in a straight line towards each other at initial velocities v1 and v 2 and with constant accelerations a1 and a2 directed against the corresponding veloc ities at the initial instant. What must be the maximum initial separation max. between the bodies for which they meet during t he motion ?
18.
The velocity of a ship in still water is 20 km/h. What is the velocit y of a motor boat approaching the ship at right angle to its course if it appears to people on board the ship, that the motor boat heads towards the ship at 60° ?
19.
A man runnin g on a horizontal roa d at 8 km/h, find the rain fall ing vertically. He increase the speed to 13 km/ h and find the drops make angle 30° with the vertical. Find the speed and the direction of the rain with respect to the read.
20.
A person riding in the back of a pickup truck tr avel ing at 60 km/h. On a strai ght, lev el road throws a ball with a speed of 20 km/h relativ e to the truck in the directi on opposite to its motion. (A) What i s the velocity of the ball relativ e to a stationary observer by the side of the read ? (B) What is the v elocity of the ball relativ e to the driver of a car mov ing in the same direction as the truck at a speed of 90 km/h. ?
21.
The speed of a boat in still water is v. The boat is to make a round trip in a riv er whose current tr avels at speed u. Derive a formul a for a time needed to make a round trip of total distance D if the boat makes the round trip by moving (A) Upstream and back down stream. (B) Directly across the river and back. W e must assume u < v ; W hy ? O
22.
In the adjacent figure, a v ertical cemented circular plane of radius R having frict ionless slots along the chords OA, OB, OC, OD and DE is shown. OGC is the diameter of t he circle. Five Boys starts sliding down from rest along the slots OA, OB, OC, OD and O E simultaneously. Find the ti me taken by each boy to reach at the point A, B, C, D and E.
PROJECTILE
G A
E D
C
B
MOTION
23.
A body is dropped from a station ary balloon at a height h above the ground. At the same time a bullet is fir ed from a gun on the ground with a velocit y u. If the angle of elevation of the balloon from th e position of the gun is , in which direction should the bullet be fired so that it strikes the body before reaching the ground. Also find the minimum v alue of u required for this.
24.
Particl es P and Q of mass 20 gms and 40 gms respectiv ely are simultaneously proje cted from po ints A and B on the ground. The initial velo cities of P and Q makes 45° and 135° angles respectively with the horizontal as shown in the figure. Each particle has in initial speed of 49 m/s. The separation AB is 245 m. Both particles travel in the same v ertical plane and under go a collision. After the col lision P retraces its 135° pat h. Det erm ine the posi ti on of Q whe n i t hi ts the 45° ground. How much time after the collision does the A B 245 m particle Q take to r each the ground. Take g = 9.8 m /s2.
25.
A boy throws a bal l horizontally with a speed of v 0 = 12 m/s from the gandhi setu bridge c of patna in an effort to hit the top surface AB of a truck travelling directly underneath the boy on the bridge. If the truck maintains a constant speed u = 15 m/s, and the ball is projected at the instant B on the top of the truck appears at point C, determi ne the position s where the ball strikes the top of the truck.
v0=12 m/s
8m
s u
B
A C
10m u
26.
Two inclined planes intersect in a horizontal plane. Their inclinations to the horizontal being and . If a particle is projected at right angle to the former from a point in it so as to strike the other at right angles, then f ind the velocity of projection.
P a
O
27.
A cannon fires from under a shelte r incl ined at an angle to the horizontal. The canon is located at a point at a distance from the base of the shelter. The initial velocity of the shell is v0 and its trajectory lies in the plane of the fi gure. Determine the maximum range of the shell.
28.
A juggler manages to keep five balls in motion, throwing each sequentially up a distance of 3m. (a) Determine the time interv al between successive throws. (B) Give t he positions of the other balls at the in stant when one reaches her hand. (Neglect the time taken to t ransfer balls from one hand t o the other.)
29.
A ball is shot fr om the gro und into the air. At a height of 9.1 m., its velocity is observ ed to be v
7.6 ˆi 6.1 jˆ
in water per second ( ˆi horizontal, jˆ upward). (A) To what maxi mum height does the ball rise ? (B) What total horizontal distance does the ball travel ? W hat are (C) The magnitude and (D) The direction of the ball’s velocity j ust before it hits the ground ? 30.
A bomb bursts on conta ct wit h the ground and pieces fly of f in al l directions with speed up to 30 m/ s. A girl is standing 40 m away. W hat is the tim e duration over which she can be hit by a piece ?
31.
A boy sitti ng at t he rear end of a railway compartmen t of a train running at a constant acceleration a on horizontal rai ls fires a shot towards the fore end of the compartm ent with a muzzle velocit y u = 20 m/s at an angle = 37° above the horizontal when the train’s velocity v = 10 m /s. If the boy catches the shot without moving f rom his seat at the same height as that of projection f ind (a) Speed of the train at the t ime when he catches the shot and (B) the acceleration of the train (g = 10 m/s2). y
32.
33.
34.
35.
A proj ectil e of m ass m is fired into a li quid at an angle 0 with an initial velocity v 0 as shown. If the liquid develop s a frictional or drag resistance on the projectile which is proporti onal to its velocit y. i.e., F = – kv, where k is a positive constant, determine the x and y components of its velocit y at any instant. Also, what is the maximum distance x max. that it travel s ?
v0 0
x
An elevator is going up with an upward acceleration of 1 m/s2. At the instant when its velocity is 2 m/s, a stone is projected upward from its floor with a speed of 2 m/s relativ e to the elevator, at an elevation of 30°. (A) Calculate the tim e taken by the stone to return the fl oor. (B) Sketch the path of t he projectile as observed by an observer outside the elevator. (C) If the elevator was movi ng with a downward acceleration equal to g, how would the motion be al tered ? A particle is moving in the plane wit h vel ocit y giv en by, u u0 ˆi a cos t jˆ , where ˆi and ˆ are unit vectors along x and y axes respectiv ely. If particle is at the origin at t = 0 ; 3 (A) Calculate the trajectory of the particle. (B) Find its distance from the origin at time . 200
A man is t rav ell ing on a flat car which is moving up a plane incli ned at cos –1 (4/5) to the horizontal with a speed 5 m/s. He throws a ball towards a stationary hoop located perpendic ular to the i nclined in such a way that the ball mov es parallel to the slope of the incline while going through the hoop. If the distance d of the hoop from the level of the man’s hand is such that its component perpendicular to the incline is 4 m, calculate the tim e taken by the ball to reach the hoop.
36.
A rifleman on a train movi ng with a speed of 60 km/ hr. fires at an object running away fr om the train at righ t angles with a speed of 45 km/hr. If the line connecting the man and t he object makes an angle of 30° to the train at the instant of shooting to what angle to the train should be aim in order to hit the object if the muzzle velocity is 850 km/hr. ?
37.
Water is ejected from the water nozzle with a speed v0 = 14 m/s. For what value of the angle will the water land closest to the wall after clearing the top ? Neglect the effects of wall thickness and air resistance. Where does the water land ?
38.
A bullet is proj ected so as to graze the tops of two walls each of hei ght 20 m located at distances of 30 and 170 m in the same line from the point of projection as shown in figure. Find the angle and the speed of projection of the bullet. P1
V0
P2 O
20m
30m
20m 170m
B
30°
39.
Two guns are projected at each o ther, one upward at an angle of 30 ° and the other at the same angle of depression, the muzzles being 30 m apart as shown in figure problem. If t he guns are shot with velocities of 350 m/s upward and 300 m/s downward respectiv ely, find when and where the bullets may meet. A
30m P y 30° x Bomber
40.
A bomb er is flyi ng horizontally at a speed of 500 km/h at an altitude of 3 km such that a ship lies in a verti cal plane through the line of sight as shown in figure. Determine the angle of the line of sight of the bomber with the ship at the instant a bomb is relea sed so as to hit the ship. W here would the bomber be at the instant the ship is wrecked ?
Line of Flight
Trajectory
3km Line of Sight
ship
41.
Two guns are pointed at each ot her one upwards at an angle of elev ation 30° and the other downwards at the same angle of depression, the muzzles being 42 m apart. If the charges leave the guns with vel ocities 400 m/s and 300 m/s r espectively, find when and where they meet.
42.
A parti cle is projected up an i nclined plane of inc lination with an initial velocity u at an angle to the horizontal. Find the maximum distance of the projectile f rom the inclined plane.
43.
On a fricti ons horizontal surface, assumed to be the x-y plane, a small trolley A is moving along a straight line parallel to the y-axis (see figure) with a constant veloci ty of
y
3 1 m s . At a particular instant, when the
A
line OA makes an angle of 45° with the x-axis, a ball is thrown along the surface from the origin O. Its veloci ty makes an angle with the x-axis and it hits t he trolley. (a) The motion of the ball is observed from the frame of the t rolley. Calculate the angle made by the velocity vector of the ball with the x-axis in this frame. 4 (B) Find the speed of the ball with respect to the surface, if . 3
45° O
x
[IIT 2002]
44.
Two guns, situated on the top of a hill of height 10 m, f ire one shot each with the same speed 5 3 ms –1 at some interval of time. One gun fires horizontall y and other fires upwards at an angle of 60° with the horizontal. The shots collide in air at a point P. Find (a) the time-interval between the firings, and(b) the coordinates of the point P. [IIT 1996] Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in x-y plane.
45.
Two towers AB and CD are situated a distance d apart as shown in figure AB is 20 m high and CD is 30 m high from the ground. An object of m ass m is thrown from t he top of AB horizontally with a
2m C 60°
velocity of 10 m s towards CD. Simultaneously another object of mass 2 m is thrown from the top of CD at an angle of 60° to the horizontal towards AB with the same magnitude of initial v elocity as that of the first object. The two objects move in the same v ertical plane, collide in mid-air and stick to eac h other. (i) Calculate the di stance ‘d’ between the towers and, (ii) Find the position where the object s hit the ground.
A
d
B
D
Answer Key Assertion & Reason
Q. Ans. Q. Ans. Q. Ans. Q. Ans.
Que.
1
2
3
4
5
6
7
8
9
10
11
Ans .
C
D
A
A
A
A
C
B
D
E
E
1
2
3
Lev el – 1 4 5
A
A
A
A
C
D
B
AB
C
10
11
12
13
14
15
16
17
18
C
B
C
B
C
B
C
B
A
19
20
21
22
23
24
25
26
27
C
B
A
B
B
A
B
A
B
28
29
30
31
AD
ABC
ABC
A BD
6
7
8
9
Fill in the Blanks / True–False / Match Table 1. 2 r , r
2. d v
3. 0.6
5. T
6. T
7. F
4. a 2 4b , 8. F
tan
9. A
1
a
KINEMATICS
Passage Type Que. Ans. Que. Ans. Que. Ans. Que. Ans.
1 B 12 C 23 A 34 A
2 C 13 D 24 B 35 B
3 D 14 A 25 C 36 D
4 A 15 B 26 A 37 D
5 A 16 D 27 A 38 C
6 A 17 B 28 D 39 A
7 B 18 A 29 B 40 C
8 C 19 B 30 D 41 B
9 A 20 B 31 B 42 D
10 D 21 C 32 B
11 A 22 C 33 C
Level # 2 MOTION IN ONE DIMENSION 1. V
4.
v 12 v 22 2 v 1v 0 cos . cos ec 2. (a) 0.001667 s (b) 42000 m/s 2 2
( )
5. tmin. =
8. V = 1.166 m/s
6. (a) yes (b) 8 m/s, 4 m/s 32 m
2
(v u) 15 t 2 + 2/3 t 3
3. 16.678
7. v = 3 – 30t + 2t 2, S = –5 + 3t –
9. t = 0.917 s
MOTION UNDER GRAVITY 10. 2.5 m/s2
11. 3.2 s, O
r v 0 v 20 , h = 14. tm = n 1 r g 2g 1
12. 2/3
13. (a) 3.65 s, 12.30 m (b) – 19.81 m/s
15. (a) 13.6 m (b) 14 m
16. 11.04 m/s downward at 177.8 k height , 55.2 m downward.
RELATIVE MOTION 17.
max
( v 1 v 2 )2 2 (a1 a 2 )
20. (a) 40 km/h (b) – 50 km/h
22. t = 2
18. 34.6 km/h 21. (a)
DV v2
u
19. 4 7 km/h
D
2 (b)
v2
u2
R , every boy will take same time g
PROJECTILE 23. , v min. =
gh .cosec 2
2 ag 26. u sin sin sin cos( )
MOTION
24. 122.5 m, 3.53 sec. 2
v0
27. g
sin 2 sin
25. 3.84 m
1
g sin 2 v0
