MOTION IN A PLANE AND RELATIVE MOTION 1

EXERCISE – 3 : ADVANCED OBJECTIVE QUESTIONS

Single choice correct. g u2 1

1. There are two values of time for which a projectile is  b  
2u 2 cos 2  g 2b cos 2 
at the same height. The sum of these two times is
equal to u 2 sin 2 
H
(a) 3T/2 (b) 4T/3 2g
(c) 3T/4 (d) T sin 2  tan 2 
 
(T = time of flight of the projectile) 2(2b cos 2 ) 4b
Ans. (d)
 a2 
Sol. H  
 4b 

3. A particle moves in the x-y plane according to the

law x = kt and y =kt (1 – at), where k and a are
positive constants and t is time. What is the equation
of trajectory of the particle?
x 2
(a) y = kx (b) y  x 
k
1
y  sin θ(t)  gt 2 x 2
2 (c) y  (d) y  x
k
v1 sin α  u sin θ  gt1 Ans. (b)
 v1 sin α  u sin θ  gt 2 Sol. x  kt, t  x / k
0  2u sin θ  g  t1  t 2 
x x
y  kt(1  t)  k  1    
g  t1 + t 2  = 2usin θ k k

 2usinθ   x 2 
 t1 + t 2 = = T y  x  
 g   k 
2. The trajectory of a projectile in a vertical plane is 4. The equation of motion of a projectile is
y = ax – bx2, where a and b are constants and x and y 3
y  12x  x 2 . Given that g =10 ms–2, what is the
are respectively horizontal and vertical distance of 4
the projectile from the point of projection. The range of the projectile
maximum height attained by the particle and the (a) 12.4 m (b) 16 m
angle of projection from the horizontal are (c) 30.6 m (d) 36.0 m
b2 a2 Ans. (b)
(a) , tan 1  b  (b) , tan 1  2a 
2a b  x
2 2
Sol. y  x tan   1  
a 2a  R
(c) , tan 1  a  (d) , tan 1  a 
4b b 3  3 x
 12x  x 2  12x 1  
Ans. (c) 4  4 12 
g
Sol. y  x tan   x 2  xa  bx 2  x 
2u cos 
2 2  12x 1  
 16 
 tan   a   tan 1

(a)  (R  16m)
MOTION IN A PLANE AND RELATIVE MOTION 2

5. A ball is dropped from the top of a tower in a high- 

 
1/ 2
AB u sin  3cos   1
2 2
speed wind. The wind exerts a steady force on the  g
Vagg   
ball. The path followed by the ball will be  t AB 2g usin

(a) Parabola (b) Circular arc
u 
(c) Elliptical arc (d) Straight line  1  3cos 2  
2 
Ans. (d)
Sol. since the force by wind is steady it means the path of
the ball will be a straight line making some angle 7. A particle A is projected from the ground with an
with the vertical. initial velocity of 10 m/s at an angle of 60o with
horizontal. From what height h should another
particle B be projected horizontally with velocity 5
6. A particle is projected from the ground with an initial
m/s so that both the particles collide in ground at
speed of u at an angle  with horizontal. The
point C if both are projected simultaneously
average velocity of the particle between its point of
(g = 10 m/s2)
projection and highest point of trajectory is
5 m/s
u u B
(a) 1  2cos 2  (b) 1  cos2 
2 2
u
(c) 1  3cos 2  (d) u cos  h
2 10 m/s
Ans. (c)
60°
Sol. A C
(a) 10 m (b) 30 m
(c) 15 m (d) 25 m
Ans. (c)
Sol. For A –
u 2 sin(2  60 ) 10 10  3 / 2
R   5 3m
g 10

 R  2 
1/ 2 For B –
AB     H 2  R  ux t
 2  
R 5 3
2
 2u 2 sin  cos    u 2 sin 2  
2
t   3sec
AB  2
   ux 5
 2g   2g 
1 1
h  u x t  gt 2  0  gt 2
u 4 sin 2  cos 2  u 4 sin 4  2 2
AB2  
g2 4g 2 1
h  10  ( 3)2  5(3)  (15 m)
u sin   2
4 2
sin   2 2
  cos   
g2 4  8. A particle is projected at an angle of 60o above the

horizontal with a speed of 10 m/s. After some time
u 4 sin 2  the direction of its velocity makes an angle of 30o
  4 cos 2   sin 2  
4g 2  above the horizontal. The speed of the particle at this
u 4 sin 2  instant is
AB2 
4g 2
 
3cos 2   1  cos 2   sin 2   1
5
(a) m/s (b) 5 3 m / s
3
u sin 
2
u sin 
 
1/ 2
AB  3cos 2   1 & t AB  10
2g g (c) 5 m/s (d) m/s
3
Ans. (d)
MOTION IN A PLANE AND RELATIVE MOTION 3

Sol. For A its time of flight must be greater than 5 sec

2u A
 5  u A  25
g

11. A projectile is fired at an angle of 30o to the

horizontal such that the vertical component of its
initial velocity is 80 m/s. Its time of flight is T. Its
v cos30  u cos 60 velocity at t = T/4 has a magnitude of nearly
(a) 200 m/s (b) 300 m/s
3 1
v  10  (c) 140 m/s (d) 100 m/s
2 2
Ans. (c)
10
v m/s ux
3 Sol.  cot 30  3
uy

9. In projectile motion, the modulus of rate of change of u x  80 3 m / s

speed 2u y
(a) is constant T
g
(b) first increases then decreases
T  16 sec.
(c) first decreases then increases
(d) none of these T
At t   4 sec, vx  80 3 m / s
Ans. (a) 4
Sol. We know that speed at any instant is the magnitude v y  80  10  4  40 m / s
of instantaneous velocity.
80 3 
2
  40 
2
v 140 m / s
And, the modulus of rate of change of speed is equal
to modulus of rate of change of velocity.
In projectile motion, rate of change of velocity that is 12. A particle A is projected vertically upwards. Another
acceleration is a  g ˆj  10jˆ / s2 that is downward particle B of same mass is projected at an angle of
Hence the modulus of rate of change of speed is 45o. Both reach the same height. The ratio of the
10 m/s2 = constant. initial kinetic energy of A to that of B is [given
1
KE  mv 2 ]
10. Two particles A and B are projected simultaneously 2
from a point situated on a horizontal plane. The (a) 1:2 (b) 2:1
particle A is projected vertically up with a velocity (c) 1 : 2 (d) 2 :1
uA while the particle B is projected up at an angle of Ans. (a)
30o with horizontal with a velocity uB. After 5 sec
u 2A u 2 sin 2 45
the particles were observed moving mutually Sol. H1  & H2  B
2g 2g
perpendicular to each other. The velocity of
projection of the particle uA and uB respectively are according to question, H1  H2
(a) 50 ms–1, 100 m/s u 2A 1 u 2B u 2A 1
  
(b) 100 ms–1, 50 ms–1 2g 2 2g u 2B 2
(c) uA > 25 m/s and uB <50 m/s  KE (1/ 2)mu A2 u A2 1 
(d) none of these So,  A
   
 KE B (1/ 2)mu 2B u 2 2 
Ans. (a)  B 
u B sin 30 u  1/ 2 u B
Sol. For B, t  5  5 B  13. A body of mass m is thrown upwards at an angle 
g 10 20
with the horizontal with velocity v. While rising up
 u B  100m / s  the velocity of the mass after t seconds will be
MOTION IN A PLANE AND RELATIVE MOTION 4

(a) (v cos )2  (vsin )2 1

and H  gt 2
2
(b) (v cos   vsin )2  gt 9.8 19.6 19.6
19.6   
2 u u
(c) v2  g 2 t 2  (2 vsin ) gt
u 2  4.9 19.6  (u  9.8 m / s)
(d) v  g t  (2 v cos ) gt
2 2 2

Ans. (c) 15. A particle is projected with a speed V from a point O

Sol. let W be the velocity after t seconds making an angle of 30o with the vertical. At the same
instant, a second particle is thrown vertically
upwards with a velocity v from a point A. The two
particles reach H, the highest point on the parabolic
V
path of particle simultaneously. Then ratio is
v

Wx  v cos 
Wy  v sin   gt

So, W  Wx2  Wy2

W2  v2 cos2   (vsin   gt)2

(a) 3 2 (b) 2 3
W2  v2 cos2   v2 sin 2   2(vsin )gt  g 2 t 2 2 3
(c) (d)
W  v2  g 2 t 2  (2vsin)gt  3 2
Ans. (c)
Sol. H1  H2
14. From the top of a tower 19.6 m high, a ball is thrown
horizontally. If the line joining the point of V 2 sin 2 60 v 2

projection to the point where it hits the ground makes 2g 2g
an angle of 45o with the horizontal, then the initial
V2 1 V 1 2 
velocity of the ball is     
v 2 sin 2 60  v sin 60 
3
(a) 9.8 ms–1 (b) 4.9 ms–1
(c) 14.7 ms –1
(d) 2.8 ms–1
16. A projectile is thrown in the upward direction
Ans. (a)
making an angle of 60 with the horizontal direction
Sol.
with a velocity of 147 ms–1. Then the time after
which its inclination with the horizontal is 45o is
(a) 15 s (b) 10.98 s
(c) 5.49 s (d) 2.745 s
Ans. (c)
Sol. v cos 45  u cos 60

H  19.6 m v 147  147  u

 v   or
2 2  2  2
x  H  19.6 m
So, x  ut  19.6 So, v y  u y  gt
(t  19.6 / u) vsin 45  u sin 60  gt
MOTION IN A PLANE AND RELATIVE MOTION 5

3 147 1 147 3  147 u 2 sin150 u 2  1  u2

gt  147    R3      (0.5)
2 2 2 2 2g 2g  2  2g

gt 
147
( 3  1)
 R1 : R 2 : R3  1: 2 :1
2
147  0.732 19. A projectile is thrown at an angle of 40o with the
t  (5.49sec)
2  9.8 horizontal and its range is R1. Another projectile is
thrown at an angle 40o with the vertical and its range
17. From the top of a tower of height 40 m a ball is is R2. What is the relation between R1 and R2?
projected upwards with a speed of 20 m/s at an angle (a) R1 = R2 (b) R1 = 2 R2
of elevation of 30o. Then the ratio of the total time
(c) R2 = 2 R1 (d) R1 = 4 R2/5
taken by the ball to hit the ground to its time of flight
Ans. (a)
(time taken to come back to the same elevation) is
(take g = 10 ms2) Sol.
(a) 2:1 (b) 3:1 u 2 sin 2θ1 u 2 sin 80
R1  
(c) 3:2 (d) 4:1 2g 2g
Ans. (a) u 2 sin 2θ 2 u 2 sin100
R2   [ θ 2  50 ]
2usin30 2g 2g
Sol. t AB 
g sin 80  sin100   R1  R 2 
 2  20 
 t AB   25sec 
 10  2  20. A cricketer hits a ball with a velocity 25 m/s at 60 o
1 2 above the horizontal. How far (approximately) above
h  u y t AC  gt AC the ground it passes over a fielder 50 m from the bat
2
(assume the ball is struck very close to the ground)
1
40  20  sin 30  t AC  10  t 2 (a) 8.2 m (b) 9.0 m
2 AC
(c) 11.6 m (d) 12.7 m
40  10t AC  5t 2
AC Ans. (a)
5t 2
 10t AC  40  0 Sol.
AC

t 2AC  2t AC  8  0
 t AC  4sec
So,  t AC / t AB  4 / 2  2 /1
18. Three identical balls are thrown with same speed at
angles of 15o, 45o and 75o with the horizontal gx 2
y  x tan  
respectively. The ratio of their distances from the 2u cos 2 
2

point of projection to the point where they hit the x  50

ground will be
  60 , u  25 m / s,g  9.8 m / s2
(a) 1: 2 :1 (b) 1 : 2 : 1
9.8  50 2
(c) 2 : 4 : 3 (d) 1: 2 : 3 y  50 tan 60 
2  252  cos 2 60
Ans. (b)
9.8  2500
u 2 sin 30 u 2  1  u2  50 3 
Sol. R1      (0.5)
1
2g 2g  2  2g 2  625 
4
u 2 sin 90 u 2 u2 2  9.8  2500
R2   (1)  (1)  50 1.73   86.6  78.4
2g 2g 2g 625
(y  8.2 m)
MOTION IN A PLANE AND RELATIVE MOTION 6

21. From a point on the ground at a distance 2 metres Ans. (c)

from the foot of a vertical wall, a ball is thrown at an Sol. for same range angles should be complimentary if it
angle of 45o which just clears the top of the wall and projected with same initial speed.
afterward strikes the ground at a distance 4m on the
  15 other angle  90    75
other side. The height of the wall is
24. A fielder in a cricket match throws a ball from the
2 3
(a) m (b) m boundary line to the wicket keeper. The ball
3 4
describes a parabolic path. Which of the following
1 4 quantities remains constant during the ball’s motion
(c) m (d) m
3 3 in air? (neglect air resistance)
Ans. (d) (a) its kinetic energy
Sol. (b) its speed
(c) the horizontal component of its velocity
(d) the vertical component of its velocity
Ans. (c)
Sol. as there is no horizontal force therefore, its
horizontal component of velocity will remain
R  2  4  6m constant.

 x
y  x tan   1   25. The height y and the distance x along the horizontal
 R
plane of a projectile on a certain planet (with no
 2
y  2 tan 45 1   surrounding atmosphere) are given by y = (8t – 5)
 6 metre and x = 6t metre where t is in seconds. The
 2 4  velocity of projection is
 y  2   m (a) 8 m/sec
 3 3 
(b) 6 m/sec
22. Two projectiles A and B are projected with same (c) 10 m/sec
speed and angle of projection 30° for the projectile A (d) not obtained from the data
and 45° for the projectile B. If RA and RB are the Ans. (c)
horizontal ranges for the two projectiles, then Sol. y  8t  5, x  6t
(a) RA = RB
u sin   u y  8, u cos   u x  6
(b) RA > RB
(c) RA < RB  u  u 2x  u 2y  82  62  10m / s
(d) the information is insufficient to decide the
relation of RA and RB 26. A body is projected horizontally with speed 20 m/s
Ans. (c) from top of a tower. What will be its speed nearly
u 2 sin 2 after 5 sec? Take g = 10 m/s2
Sol. R (u is same)
g (a) 54 m/s (b) 20 m/s
(c) 50 m/s (d) 70 m/s
 R B >R A [ sin 90  sin 60 ]
Ans. (a)
Sol. v x  u x  20m / s
23. A projectile is projected at an angle of 15° to the
horizontal with some speed v. If another projectile is v y  u y  gt  0  10(5)  50m / s
projected with the same speed, then it must be
 
1/ 2
projected at what angle (other than 15°) with the v  v2x  v2y  202  502
horizontal so as to have the same range.
(a) It is never possible (b) 12.5°
v  400  2500  (2900)1/ 2  54m / s 
(c) 75° (d) 65°
MOTION IN A PLANE AND RELATIVE MOTION 7

27. A body is projected horizontally with speed 20 m/s

from top of a tower, what will be the displacement of
the body if it hits the ground after 5 sec and doesn’t
bounce (quote nearest integer)
(a) 100 m (b) 125 m
(c) 160 m (d) 225 m
Ans. (c)
(a) 2 (b) 0.5
Sol. x = u x t  20  5  100m
(c) 3/ 2 (d) 1
1 1
y = u y t  gt 2  0  10  25  125m Ans. (b)
2 2
Sol.
S = x 2 + y2  1002  1252  160m

28. A body is projected at an angle of 30° with the

horizontal with speed 30 m/s. What is the angle with
the horizontal after 1.5 seconds? Take g = 10 m/s2.
(a) 0° (b) 30°
(c) 60° (d) 90° t AC  t BC
Ans. (a) v1 sin 30 v 2
 
Sol. vx  u x  30cos30  15 3 g g
v y  u y  gt  30sin 30  10(3 / 2) v1 1 2
 
30 30 v2 sin 30 1
    Vy  0 
2 2 v 
Or  2  0.5 
Vy  v1 
 tan   0
Vx 31. An aeroplane is flying at a constant horizontal
  0  velocity of 600 km/h at an elevation of 6 km towards
a point directly above the target on the earth’s
surface. At an appropriate time, the pilot released a
29. From certain height, two bodies are projected ball so that it strikes the target on the earth. The ball
horizontally with velocities 10 m/s and 20 m/s. They will appear to be falling
hit the ground in t1 and t2 seconds. Then (a) on a parabolic path as seen by pilot in the plane
(a) t1 = t2 (b) t1 = 2 t2 (b) vertically along a straight path as seen by an
(c) t2 = 2 t1 (d) t1  2 t 2 observer on the ground near the target
(c) on a parabolic path as seen by an observer on the
Ans. (a) ground near the target
2h (d) on a zig-zag path as seen by pilot in the plane
Sol. t (same h for both)
g Ans. (c)
 t1  t 2 Sol. Ball also acquire the velocity of aeroplane, so ball
also have a horizontal velocity same as aeroplane and
pilot.
30. A body is projected with velocity v1 from the point A
The pilot will see the ball falling freely under gravity
as shown in figure. At the same time, another body is in a straight line and as the ball is moving with the
projected vertically upwards from B with velocity v2. same horizontal velocity so, the observer at rest will
The point B lies vertically below the highest point. see the ball falling in parabolic path.
v
For both the bodies to collide, 2 should be
v1
MOTION IN A PLANE AND RELATIVE MOTION 8

32. Three particles A, B and C are thrown from the top (a) 20.2 m (b) 12 m
of a tower 100 m in height with the same speed 10 (c) 31.2 m (d) 62.4 m
m/s. A is thrown straight up, B is thrown straight Ans. (d)
down, and C is thrown horizontally. They hit the
3
ground with the speeds vA, vB and vC respectively. Sol. u y  20sin 37  20   12 m / s
5
Then
(a) vA > vB = vC (b) vB > vC > vA 4
u x  20 cos 37  10  20   10  26 m / s
5
(c) vA = vB = vC (d) vA = vB > vC
2u 2 sin 37 cos 37 2 12  26
Ans. (c) R   62.4 m
g 10
Sol. A  v2A  102  2 10 100
35. Two men A and B, A standing on the extended floor
 100  2000  2100  vA  45.8 nearby a building and B is standing on the roof of the
B  v2B  102  2 10 100 building. Both throw a stone towards each other.
Then which of the following will be correct.
 2100  vB  45.8
(a) stone will hit A, but not B
C  vx  10m / s
(b) stone will hit B, but not A
v  0  2 10 100  2000
2
y (c) stone will not hit either of them, but will collide
with each other
vC  v 2x  v 2y  100  2000  2100  45.8
(d) none of these
  vA  vB  vc  Ans. (d)
33. A body is thrown horizontally with a velocity 2 gh Sol. no information is given about speed of projection and
height of the roof so, we cannot say anything.
from the top of a tower of height h. It strikes the
level ground through the foot of the tower at a
distance x from the tower. The value of x is 36. A particle is projected from a point (0, 1) on Y–axis
(a) h (b) h/2 (assume + Y direction vertically upwards) aiming
towards a point (4, 9). It fell on ground along x axis
(c) 2h (d) 2h/3
in 1 sec.
Ans. (c)
Taking g = 10 m/s2 and all coordinate in metres. Find
Sol. x  ut
the x–coordinate of the point where it fell.
2h (a) 3 (b) 4
t
g
(c) 2 (d) 2 5
2h Ans. (c)
x  2gh   2h
g Sol.

8 2
tan   
4 1
34. Consider a boy on a trolley who throws a ball with 1 2
cos   ,sin  
speed 20 m/s with respect to ground at an angle 37° 5 5
with vertical and trolley is moving with a speed 10
t Ar  1sec
m/s in horizontal direction then what will be
maximum distance travelled by ball parallel to road :
MOTION IN A PLANE AND RELATIVE MOTION 9

1 x y
h  u y t  gt 2
2 u x  u cos30 u y  u sin 30
1
1  u sin (1)  10  (1)2 a x  gsin 30 a y  g cos 30
2
u sin   5  1  4 2usin30 2  20 1/ 2 4
T   sec
2 g cos 30 10  3 / 2 3
u  4 (u  2 5)
5 v y  u y  a y t  u sin 30  g cos 30  T
So, x  u x t  u cos (t) 1 3 4
 20   10  
1 2 2 3
x  2 5 (1)  2
5 v y  10  20  10 m / s
(x  2)(y  0)

39. A particle P is projected from a point on the surface

37. The position vector of a particle is given as of smooth inclined plane (see figure).
   
r  t 2  4t  6 ˆi  t 2 ˆj. The time after which the Simultaneously another particle Q is released on the
velocity vector and acceleration vector becomes smooth inclined plane from the same position. P and
perpendicular to each other is equal to Q collide on the inclined plane after t = 4 second.
The speed of projection of P is nearly:
(a) 1 sec (b) 2 sec
(c) 1.5 sec (d) not possible
Ans. (a)
Sol. v  (2t  4)i  2t j

a  2i  2 j
If v and a are to be , then
av  0
(a) 5 m/s (b) 10 m/s
 2(2t  4)  2(2t)  0
(c) 15 m/s (d) 20 m/s
4t  8  4t  0
Ans. (b)
8t  8 (t  1sec)
Sol. For Q –

38. A particle is projected up an inclined plane with

initial speed v = 20 m/s at an angle   30o with
plane. The component of its velocity perpendicular to
plane when it strikes the plane is
(a) 10 3 m / s (b) 10 m/s
(c) 5 3 m / s (d) data is insufficient
Ans. (b)
Sol. 1
l  0(4)  g sin 60 (4)
2
1 3
l  10  4
2 2
(l  10 3 m)
For P
MOTION IN A PLANE AND RELATIVE MOTION 10

v
2
2u y 2  2v
T 
ay g g
2
1 v 2v 1 g 4v 2
R  u x T  a x T2      2
2 2 g 2 2 g

2v 2 2v 2  2v 2 
R   2 
g g  g 

41. Position vector of a particle moving in x-y plane at

X Y time t is: r  a 1  cos t  ˆi  a sin t ˆj. The path of
u x  u cos  u y  u sin 60 the particle is
(a) a circle of radius a and centre at (a, 0)
a x  gsin 60 a y  g cos 60
(b) a circle of radius a and centre at (0, 0)
2u sin 60 (c) an ellipse
T  4sec 
g cos 60 (d) neither a circle nor an ellipse
Ans. (a)
2 u  3 / 2
4
10 / 2 Sol. r  a(1  cos t)iˆ  a sin tjˆ
 20  v  a(sin t)iˆ  a cos tjˆ (velocity)
u   11.5  10 m / s 
 3  a  a2 cos ti  a2 sin t j (acceleration)

40. A ball is projected horizontally with a speed v from

v  a  a 2 3 cos t sin t  a 2 3 sin t cos t  0
the top of a plane inclined at an angle 45° with the  circular motion
horizontal. How far from the point of projection will 42. A particle moves in x-y plane. The position vector of
the ball strike the plane?   
particle at any time t is r   2t  ˆi  2t 2 ˆj m. The
v2 v2 rate of change of  at time t = 2 second. (where  is
(a) (b) 2
g g the angle which its velocity vector makes with
2 v2  2 v2  positive x-axis) is
(c) (d) 2 
g  g  2 1
(a) rad / s (b) rad / s
17 14
Ans. (d)
Sol. 4 6
(c) rad / s (d) rad / s
7 5
Ans. (a)
ˆ
Sol. r  2ti  2t 2 j
dr
v  2i  4t j
dt
X Y
v v
2 2
g -g
2 2

 is angle made by v with x-axis

MOTION IN A PLANE AND RELATIVE MOTION 11

4t 45. Let v and a denote the velocity and acceleration

 tan    2t
2 respectively of a particle moving in a circular path
Differentiate w.r.t. time then,
d (a) v . a  0 all the time
sec2  2
dt (b) v . a  0 all the time
d
 2 cos 2  (c) v . a  0 all the time
dt
(d) (a),(b) & (c) all are possible depending upon the
2
where cos   direction of net acceleration.
22  (4t) 2 Ans. (d)
2 Sol. If the particle is moving with constant speed then,
at t  2, cos  
68 v and a will be perpendicular to each other but if
d  4  2 speed is not constant, then option-(a) and option-(b)
  2    rad / sec both are possible depending upon whether speed is
dt  68  17
increasing or decreasing

43. A particle has an initial velocity of 3iˆ  4 ˆj and an

46. A person walks up a stationary escalator in time t1. If
acceleration of 0.4iˆ  0.3 ˆj . Its speed after 10 s is: he remains stationary on the moving escalator, then it
(a) 10 unit (b) 7 unit can take him up in time t2. How much time would it
(c) 7 2 unit (d) 8.5 unit take him to walk up the moving escalator.
Ans. (c) t1  t 2
(a) (b) t1  t 2
Sol. vx  3  0.4(10)  7 2
v y  4  0.3(10)  7 t1 t 2
(c) (d) t1  t 2
t1  t 2
v  v  v  7 2 unit
2
x
2
y
Ans. (c)
Sol. Say, height covered is h
44. Velocity and acceleration of a particle initially are h
So, t1  ( vm  man’s speed)
 
v  3iˆ  4 ˆj m/s and 
a   6 ˆi  8 ˆj m/s2 vm
respectively. Initially particle is at origin. maximum h
and t 2 = ( v e = escalator’s speed)
x–coordinate of particle will be: ve
(a) 1.5 m (b) 0.75 m
On moving escalator, if he moves-
(c) 2.25 m (d) 4.0 m
v g  v me +v m ( his capability remains same)
Ans. (b)
Sol. u x  3,a x  6 So, v g  man’s new speed wrt ground

1 h h h
x  ux t  ax t2 So, t   
2 vg v me  ve h  h
For x  maximum t1 t 2
dx  t1 t 2 
 vx  u x  a x t  0 t  
dt  t1  t 2 
ux 3 1
t  
ax 6 2 47. A horizontal wind is blowing with a velocity v
1 1 1
2
towards north-east. A man starts running towards
 x max  3    (6)   north with acceleration a. The time after which man
2 2 2
will feel the wind blowing towards east is :
 3 3 3 
 x max     0.75 m 
 2 4 4 
MOTION IN A PLANE AND RELATIVE MOTION 12

v 2v u2 1
(a) (b) (c) 1  (d)
a a v2 u2
1 2
v 2v v
(c) (d) Ans. (b)
2a a
2
Ans. (c) Sol. T0 
v
Sol.
vbW  vb  v w
vb  vbw  vw  v  u (Forward journey)
vb  v  u (Backward journey)
1 1 1 1
T    
vb vb v  u v  u
vw v
v m  v w cos 45  
2 2 vuu 2 vl
 l 2 2 
 2
 v u  v u
2
vm  0  at
 

t 
vm

v 
  

 T 2l v / v 2  u 2
 2
1

1 

 a 2a 
 T0 2l / v v u 2
u2 
 1 2 
 v 2
v 
48. Two trains are each 50 m long starts moving parallel
towards each other at speeds 10 m/s and 15 m/s
50. A river is flowing from West to East at a speed of 5
respectively, after how much time will they pass
metres per minute. A man on the south bank of the
each other?
river, capable of swimming at 10 metres per minute
(a) 8s (b) 4s in still water, wants to swim across the river in
(c) 2s (d) 6s shortest time. He should swim in a direction
Ans. (b) (a) due North (b) 30° East of North
Sol. (c) 30° West of North (d) 60° East of North
Ans. (a)
Sol. Due north, to utilize his full capacity to the direction
of crossing the river

51. A river is flowing from west to east at a speed of 20

Crossing is over when A & D are at same point m/min. A man on the south bank of the river, capable
v12  v1   v2   10  15  25 m / s of swimming at 10 m/min in still water, wants to
swim across the river without any drift. He should
 2l 2  50  swim in a direction:
t    4sec 
 v12 25  (a) due north
(b) 30° east of north
49. On a calm day a boat can go across a lake and return (c) 30° west of north
in time T0 at a speed v. On a rough day there is (d) zero drift is not possible
uniform current at speed u to help the onward Ans. (d)
journey and impede the return journey. If the time Sol. for no drift displacement along the direction of river
taken to go across and return on the rough day be T, flow should be zero, i.e, net velocity along the river
then T/T0 is: flow direction is zero.
let the man swims at an angle  from south towards
u2 1
(a) 1  (b) west.
v2 u2
1 2 then, vman sin   vriver  0
v
MOTION IN A PLANE AND RELATIVE MOTION 13

vriver 20 Multiple choice correct.

 sin     2(not possible)
vman 10 54. An observer moves with a constant speed along the
line joining two stationary objects. He will observe
that the two objects.
52. The rowing speed of a man relative to water is 5
(a) have the same speed
km/h and the speed of water flow is 3 km/h. At what
angle to the river flow should he head if he wants to (b) have the same velocity
reach a point on the other bank, directly opposite to (c) move in the same direction
starting point: (d) move in opposite direction
(a) 127° (b) 143° Ans. (a,b,c)
(c) 120° (d) 150° Sol. with respect to observer the stationary objects will
Ans. (a) move with the same velocity directed opposite to the
Sol. velocity of the observer

55. A particle is projected at an angle  from ground

with speed u (g = 10 m/s2)
(a) if u = 10 m/s and  = 30°, then time of flight will
be 1 sec.
(b) if u = 10 3 m/s and  = 60°, then time of flight
VR 3 will be 3 sec.
sin θ= 
VmR 5 (c) if u = 10 3 m/s and  = 60°, then after 2 sec
θ =37 0 velocity becomes perpendicular to initial velocity.
(d) if u = 10 m/s and  = 30°, then velocity never
 (angle is 90  37  127  becomes perpendicular to initial velocity during its
flight.
53. Two cars are moving in the same direction with the Ans. (a,b,c,d)
same speed of 30 km/h. They are separated by 5 km. 2usinθ
What is the speed of the car moving in the opposite Sol. For a projectile T 
g
direction if it meets the two cars at an interval of 4
minutes? 2  10  sin 30
(a) T   1sec
(a) 15 km/h (b) 30 km/h 10
(c) 45 km/h (d) 60 km/h 2×10 3×sin60
(b) T   3sec
Ans. (c) 10
Sol.
(c) if u  10 3m / s and θ  60
 u  10 3 cos60 ˆi  10 3 sin 60 ˆj
u  5 3iˆ  15jˆ
After 2sec, v  5 3i  (15  gt) j

 5 3iˆ  5jˆ
vca  vcs  vc   vA 
u  v  5 3(5 3)  15  5
vCA  u  v  u  30
 75  75  0
5 km
vCA   u  30 i.e. u  v
4 / 60hr
(d) If u  10 m / s,   30
u  30  75
(u  45 km / h) u  5 3i  5j
At any time v  5 3iˆ  (5  gt)jˆ
MOTION IN A PLANE AND RELATIVE MOTION 14

For v to be perpendicular y
vu  0 A
75  5(5  gt)  0
H
100  5gt  0 B
x
O
100
t  2sec R1 R2
5  10
(a) t1 will decrease while t2 will increase
v will not be perpendicular to u during its flight
Hence, (a),(b),(c)and (d) are correct option. (b) H will increase
(c) R1 will decrease while R2 will increase
56. A particle leaves the origin with an initial velocity (d) T may increase or decrease
 
u  3iˆ m/s and a constant acceleration Ans. (a,d)
Sol. T  may increase or decrease, depending upon
a   1.0 ˆi  0.5 ˆj m/s . its velocity
2
v and position strength of air drag
vector r when it reaches its maximum x-co-ordinate t1  will decrease as gravity & air drag both will
are: retard the body
(a) v  2jˆ  
(b) v  1.5jˆ m / s t 2  opposite to t1 situation

 
(c) r  4.5iˆ  2.25jˆ m (d) r   3iˆ  2ˆj m
58. From an inclined plane two particles are projected
Ans. (b,c) with same speed at same angle  , one up and other
Sol. u  3im/s down the plane as shown in figure. Which of the
following statement(s) is/are correct?
a  1i  0.5j
x-co-ordinate will be maximum at a time when
vx  0  u x  a x t  0
3 1 t  0
t  3sec
(a) the particles will collide the plane with same
 vy  u y  a y t
speed
 0  0.5  3jˆ (b) the times of flight of each particle are same
v y  1.5jˆ m / s (c) both particles strike the plane perpendicularly
(d) the particles will collide in mid-air if projected
1
Also x  u x t  a x t 2 simultaneously and time of flight of each particle is
2
less than the time of collision.
1 Ans. (b)
 3  3  (1)32
2 Sol. For body A
 9  4.5  4.5 m
X Y
1
y  uyt  ayt2 u x =u cosθ u y =u sin θ
2
1 1 2 a x  gsin θ a y  g cos θ
 0     3  2.25 m
2 2  2usinθ 
 T1 = 
 r  4.5iˆ  2.25jm  gcosθ 

57. In a projectile motion let tOA = t1 and tAB = t2. the 1 1

x1  u cosθt  g sin θt 2 , y1  u sin θt  g cosθt 2
horizontal displacement from O to A is R1 and from 2 2
A to B is R2. Maximum height is H and time of flight
is T. If air drag is to be considered, then choose the
correct alternative (s)
MOTION IN A PLANE AND RELATIVE MOTION 15

 u2 
 R max   2 h   (a)
 g 
R  nH
2u 2 sinθcosθ u 2 sin 2 θ
 n
g 2g
4   4 
tan θ   θ  tan 1     (b)
n   n 
2u 2 sinθcosθ 2usinθ
for body B R and T 
g g
X Y
R 2u 2 sinθcosθ g2 gcosθ g
u x  u cosθ u y  u sin θ  2
  2 2  
T g 4u sin θ 2sinθ 2tanθ
a x  g sin θ a y  gcosθ
gT2  2R tan θ  (c)
 2usinθ 
 T2 =    T1 =T2  2u 2u sinθ
 gcosθ  t1  1 & t 2  2
g g
1 1
x 2  x 0  u cos θt  g sin θt 2 , y2  u sin θt  g cosθt 2 &t1  t 2  u1  u 2 sin θ 
2 2
where x 0 is the initial distance between A and B u12 u 2 sin 2 θ
So, h1  and h 2  2
2g 2g
After time ‘t’ when they collide
 y1 = y2 , and x1 = x 2 h 
  1  1  (d)
 u cos θt  x 0  u cos θt  h2 
x0
t=
2u cos θ 60. Two particles A and B are located in x-y plane at
Time of collision depends upon initial distance points (0, 0) and (0, 4 m). They simultaneously start
between then and if x 0 is large enough they may not moving with velocities.

collide in the air. v  2jˆ m/s and v  2iˆ m/s. Select the correct
A B

So, only option-b is correct. alternative(s)

(a) the distance between them is constant
59. Choose the correct alternative(s) (b) the distance between them first decreases and
(a) If the greatest height to which a man can throw a then increases
stone is h, then the greatest horizontal distance up to (c) the shortest distance between them is 2 2 m
which he can throw the stone is 2h (d) time after which they are at minimum distance is
(b) The angle of projection for a projectile motion 1s
whose range R is n times the maximum height H is Ans. (b,c,d)
tan–1 (4/n) Sol.
(c) The time-of-flight T and the horizontal range R of
a projectile are connected by the equation
gT2 = 2R tan θ where θ is the angle of projection
(d) A ball is thrown vertically up. Another ball is
thrown at an angle θ with the vertical. Both of them
remain in air for the same period of time. Then the
ratio of heights attained by the two balls is 1 : 1.
Ans. (a,b,c,d)
u2
Sol. h
2g
MOTION IN A PLANE AND RELATIVE MOTION 16

relative to river. The width of the river is 100 m

along y-direction. Choose the correct alternative(s)
(a) the boatman will cross the river in 25 s
(b) absolute velocity of boatman is 2 5 m/s
(c) drift of the boatman along the river current is 50
m
(d) the boatman can never cross the river
y'2 x 2  l2 Ans. (a,b,c)
Sol.
(4  2t)2 (2t)2  l2
l2  16  16t  4t 2  4t 2
l2  8t 2  16t  16

l  8t 2  16t  16
dl 1
 
1/ 2
 (16t  16)  8t 2  16t  16 0
dt 2  vm  velocity of boat  vR  velocity of river 
16(t  1)  0
vm  vmR  vR
 (t  1sec)
v m  2i  4 j  4i
l2min  8(1)2 16(1)  16  8
v  2i  4j 
l 
m

min  2 2m
 vm  y  4 ,  vm x  2
61. The co-ordinate of the particle in x-y plane are given  d 100 
 t    25sec 
  m  y
as x = 2 + 2t + 4t2 and y = 4t + 8t2 the motion of the v 4

particle is
 v  22  42  20  2 5 
(a) along a straight line  m 
(b) uniformly accelerated
 x   vm x  t  2  25  50 m
(c) along a parabolic path
(d) nonuniformly accelerated
Ans. (a,b) Numeric questions
63. A particle of mass m = 2 kg is projected along X–
Sol. x  4t 2  2t  2, y  8t 2  4t  (i) –1
axis with velocity V0 = 5 ms . It is acted on by a
 
2x  8t  4t  4  (ii)
2
variable force acting along Y–axis as shown in
2 x = y+4 figure. What is the magnitude of its velocity at 2
–1
(y  2x  4)  Linear (straight line) seconds? (in ms )

v x  8t  2 , v y  16t  4
ax  8 , a y  16

 a  a 2x  a 2y  constant 
 

62. River is flowing with a velocity vR  4iˆ m/s. A boat Ans. (7.5)
is moving with a velocity v BR  
 2iˆ  4ˆj of m/s Sol. y  mx
F  5t {t  [0,1]}
MOTION IN A PLANE AND RELATIVE MOTION 17

5 5  4 4 
a t t  v RM  v m   12  16 km/h 
m 2  3 3 
dv
dt
 2.5t 65.   
VA  x ˆi  2 ˆj m/s and VB  3iˆ  2 ˆj  m/s find x

v 1
such that, the relative speed of A with respect to B
v0
dv  2.5 tdt
0
becomes 5 m/s.
Ans. (8)
2.5 2 1 2.5
v  v0  t   m
2  0 2 Sol. V AB  5
s
 5 5 25 m 
 v  v0   5    V AB  V A  V B  (x  3)i  0j
 4 4 4 s 
F  5t  5 {t  [1, 2]} V AB  (x  3)  5
5 5 (x  8)
a t
2 2
66. A particle is projected up an inclined plane of
dv 5 5 inclination  at an elevation  to the horizontal.
 t
dt 2 2
Find the ratio between tan  and tan  , if the particle
v1 5 2 5 2
v
dv   tdt   dt
2 1 2 1
strikes the plane horizontally.
Ans. (2)
5 2 2 5 2 H
v1  v   t   [t]1
4  1 2 2H
Sol. tan β  R 
5 5 15 5 2 R
v1  v  (4  1)  (2  1)  v  
4 2 4 2
15  10 5
v1  v   v
4 4
 1 25 5 30 m
v     7.5 
 4 4 4 s 

64. A man standing on a road has to hold his umbrella at

37° with the vertical to keep the rain away. He u 2 sin 2 α
throws the umbrella and starts running at 12 km/h. 2g
He finds that raindrops are hitting his head vertically. tan β  2  2
2u sinαcosα
Find the speed (in km/hr) of raindrops with respect to g
the moving man.
Ans. (16) sinα 1
tan β   tanα
2 2
Sol.
cosα
 tanα 
 =2 
 tanβ 
67. A train takes 2 minutes to acquire its full speed 60
kmph from rest and 1 minute to come to rest from
the full speed. If somewhere in between two stations
1 km of the track be under repair and the limited
speed on this part be fixed to 20 kmph, find the late
running of the train ( in sec) on account of this repair
From triangle
work, assuming otherwise normal at running of the
vm 3 train between the stations.
tan 37  
v RM 4 Ans. (160)
MOTION IN A PLANE AND RELATIVE MOTION 18

60  0 km
Sol. α  60  1800 2
2 h
0  60 km
β  60  3600 2
1 h
Total time taken when there is no repair work, train g
from 0 to 60m and from 60 to 0 kmph.
t = (2 + 1) = 3 min.
Now, if the repair worker is on progress that the
journey will be like this pattern.
(i) Train will accelerate from 0 to 60 kmph in time t1
= 2 min.
(ii)Train will deaccelerate from 60 to 20 kmph in Assertion Reason
 60  20   60 2
time t 2   min
3600 3 (A) If both Assertion and reason are true and reason is
(iii)Train will be on uniform speed, 20 kmph up to 1 the correct explanation of the assertion.
1 60 (B) If both assertion and reason are true but reason is not
km in time, t 3   3min the correct explanation of the assertion.
20
(C) If assertion is true but reason is false.
(iv)Train will accelerate from 20 to 60 kmph in time
(D) If assertion is false but reason is true.
 60  20   60 4
t4   min (E) If both assertion and reason are false.
1800 3
(v)Train will decelerate from 60 to 20 kmph in time
68. Assertion: For a particle moving along a straight line
t 5 = 1 min. or in a plane, the average velocity vector over a time
So, total time, t’ = t1 + t2 + t3 + t4 + t5 = 8 min. interval can be equal to instantaneous velocity at the
for normal run, end of the interval, even if velocity of particle is not
distance travelled during duration of t 2 constant.
r2  r1 d r
602 -202 3200 4 Reason: 
= s2 = = = km , t 2  t1 d t
2×3600 7200 9
distance travelled during duration of t 3 (a) A (b) B
s3 =1km (c) C (d) D
(e) E
distance travelled during duration of t 4
Ans. (c)
602 -202 3200 8
s4 = = = km r2  r1
2×1800 3600 9 Sol.  avg. vel.
t 2  t1
489 7
total distance  s 2  s3  s 4   km
9 3 dr
 Instantaneous velocity
total time taken for these duration dt
7 60 7 69. Assertion: Two stones are simultaneously projected
t   min
3 60 3 from level ground from same point with same speeds
So, delay time = t2 + t3 + t4 – t but different angles with horizontal. Both stones
move in same vertical plane. Then the two stones
2 4 7 8
=  3    min  160s = 5 min. may collide in mid-air.
3 3 3 3
Reason: For two stones projected simultaneously
from same point with same speed at different angles
with horizontal, their trajectories may intersect at
some point.
MOTION IN A PLANE AND RELATIVE MOTION 19

(a) A (b) B 72. Assertion: When a body is dropped or thrown

(c) C (d) D horizontally from the same height, it would reach the
(e) E ground at the same time.
Ans. (c) Reason: Horizontal velocity has no effect on the
Sol. For simultaneous projection with same speeds vertical direction.
x1  u cos α.t, x 2  u cosβ.t (a) A (b) B
(c) C (d) D
1 1
y1  u sin .t  Rt 2 , y2  u sin .t  gt 2 (e) E
2 2
Ans. (a)
for collision (mid-air) if they are projected
Sol. in either case there is no initial velocity in the
simultaneously
vertical direction.
x1  x 2 and y1 = y2
2h 1
u cos αt  u cosβt  cos α  cosβ so, t1  h  0  gt 2
g 2
Or α  β
73. Assertion: In order to hit a target, a man should point
hence, reason is false, his rifle in the same direction as target.
although if the projectiles are projected at different Reason: The horizontal range of the bullet is
instances, they may collide mid-air independent of the angle of projection with
horizontal.
(a) A (b) B
70. Assertion: In a plane to plane projectile motion, the (c) C (d) D
angle between instantaneous velocity vector and (e) E
acceleration vector can be anything between 0 to  Ans. (e)
(excluding the limiting case). Sol. in order to hit a target, the man should point his rifle
Reason: In plane to plane projectile motion, slightly above the direction of the target to
acceleration vector is always pointing vertical compensate the distance travelled downwards by
downwards. (neglect air friction). gravity and horizontal range is indeed dependent on
(a) A (b) B the angle of projection.
(c) C (d) D
(e) E Match the column
Ans. (b)
Sol. acceleration is always in downward direction, and 74. A ball is projected from the ground with velocity v
direction of velocity keeps changing in a projectile such that its range is maximum.
so, angle between them can be anything between 0 to Column–I Column–II

(A) Velocity at half of the (P) 3 v/2

71. Assertion: Two particles of different mass, projected maximum height

with same velocity and angle of projection, the v
(B) Velocity at the maximum (Q)
maximum height attained by both the particle will be 2
same. height
Reason: The maximum height of projectile is (C) Change in its velocity when (R) v 2
independent of particle mass.
it returns to the ground
(a) A (b) B
(c) C (d) D v 5
(D) Average velocity when it (S)
(e) E 2 2
Ans. (a) reaches the maximum height

u 2 sin 2 θ Ans.  A  P;B  Q;C  R;D  S

Sol. H
2g
 H  m0 
MOTION IN A PLANE AND RELATIVE MOTION 20

Vcos Vcos
Sol. θ  450
H v 2 sin 2 45° v 2
y  
2 2×2g 8g v Vsin Vsin

Now, w  v  2 gy  v  2gy
2 2 2
Velocity just Components of velocity Components of velocity
2 2
before collision just before collision just after collision
v 3v
 v2  2 g   The other parallel component of velocity will remain
8g 4
constant if wall is given smooth.
 3v  Now let us take a problem. Three balls ‘A’ and ‘B’ &
 w=  [(A)  (P)]
 2  ‘C’ are projected from ground with same speed at
same angle with the horizontal. The balls A, B and C
v
v cos 45  [( B)  (Q)] collide with the wall during their flight in air and all
2 three collide perpendicularly with the wall as shown
in figure.
vf  vcos 450 ˆi  vsin 450 ˆj
A
1 C
| v | -2vsin45° ˆj  2  v  B
2
 2v[(c)  (R)] 75. Which of the following relation about the maximum
2
v sin 45 2 0
v R v sin 90 2
v 2 2  2 height H of the three balls from the ground during
H  ,   their motion in air is correct:
2g 84g 2 2g 2g
2
(a) HA = HC > HB (b) HA > HB = HC
R v4 v4 5 v4
S2  H2   ,  (c) HA > HC > HB (d) HA = HB = HC
4 16 g 2 4 g 2 16 g 2
Ans. (a)
 5 v2   v sin 45 v 
 S    t    u 2 sin 2 θ
 4 g   g 2 g  Sol. HA  HC 
2g
5 v2 2g 5 2 5 u 2 sin 2 θ
Vavg    v  v HB  [due to collision before max. height]
4g v 16 8 2R
 v 5  H A  H C  H B 
 Vavg   [(D)  (s)]
 2 2 
76. If the time taken by the ball A to fall back on ground
Paragraph is 4 seconds and that by ball B is 2 seconds. Then the
Using the following comprehension, solve Q. 75 to Q. 78 time taken by the ball C to reach the inclined plane
after projection will be:
PASSAGE - 1
(a) 6 sec (b) 4 sec
We know how by neglecting the air resistance, the
problems of projectile motion can be easily solved (c) 3 sec (d) 5 sec
and analysed. Now we consider the case of the Ans. (c)
collision of a ball with a wall. In this case the Sol.
problem of collision can be simplified by considering
the case of elastic collision only. When a ball
collides with a wall, we can divide its velocity into
two components, one perpendicular to the wall and
other parallel to the wall. If the collision is elastic,
then the perpendicular component of velocity of the
ball gets reversed with the same magnitude.
MOTION IN A PLANE AND RELATIVE MOTION 21

77. The maximum height attained by ball ‘A’ from the

ground is
(a) 10 m
(a) 30o (b) 45o
(b) 15 m
(c) 60o (d) None of these
(c) 20 m
Ans. (c)
(d) insufficient information
Sol.
Ans. (c)
2usinθ
Sol. TA  4 
g
(usinθ = 2 g)
 (usinθ)2 2 g  2 g 
 HA    2 g  20 m 
 2 g 2 g 
78. The maximum height attained by ball B from ground
is: y = 6000 m
(a) 20 m (b) 5 m
vg cosθ = vP
(c) 15 m (d) none of these
Ans. (c) v P 250 1
cos θ=  
Sol. vg 500 2
For ball B
θ = 60  °

t = 1 sec to reach maximum height.

1
 h max  u sin (1)  g(1)2 80. The time after which the aircraft is hit is:
2
(a) 20 3 s (b) 15 3 s
1
 2g  g  1.5g  15
2 (c) 20 s (d) 10 3 s
Ans. (d)
Using the following comprehension, solve Q. 79 & 80
Sol. θ  60
PASSAGE - 2
1
An aircraft moving with a speed of 250 m/s is at a  
y  6000  vg sin60° t   gt 2
2
height of 6000 m, just overhead of an anti-aircraft
gun.  3
6000   500   t  5t
2

79. If the muzzle velocity of the shell is 500 m/s, the  2 

firing angle  should be
On solving (t  10 3sec)