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KINEMATICS PROBLEM AND THEIR SOLUTION
Q.1.
A projectile is fired from the top of a tower 40 meter high with an initial speed of 50 m/s at an
unknown angle. Find its speed when it hits the ground.
Q.2.
An aeroplane is flying in the horizontal direction with a velocity 540 km/hr at a height of
2000 m. When it is vertically above the point 'A' on the ground, a body is dropped from it. The
body strikes the ground at point B. Calculate the distance AB.
Q.3.
The co-ordinates of a moving particle at any time t are given by x = ct2 and y = bt2. Find initial
speed of the particle.
Q.4.
A ball is thrown at a speed of 50 m/s at an angle of 600 with the horizontal. Find
(a) the maximum height reached.
(b) the range of ball. (Take g = 10 m/s 2)
Q.5.
Two cars are moving in the same direction with the same speed 30 km/hr. They are separated by a
distance of 5 km. What is the speed of a car moving in the opposite direction if it meets these two
cars at an interval of 4 minutes?
Q.6.
A stone is projected with a speed of 40 m/s at an
angle of 300 with the horizontal from a tower of
height 100 m above ground. Find
(a) the maximum height attained by the stone.
(b) the horizontal distance from the tower where it
hits the ground.
40 m/s
300
100 m
Q.7.
A projectile is fired from the top of a tower 40 meter high with an initial speed of 50 m/s at an
unknown angle. Find its speed when it hits the ground.
Q.8.
An aeroplane is flying in a horizontal direction with a velocity 600 km/hr at a height of 1960 m.
When it is vertically above the point 'A' on the ground, a body is dropped from it. The body strikes
the ground at point B. Calculate the distance AB.
Q.9.
Two cars are moving in the same direction with the same speed 30 km/hr. They are separated by a
distance of 5 km. What is the speed of a car moving in the opposite direction if it meets these two
cars at an internal of 4 minutes?
Q.10. A man standing on a road has to hold his umbrella at 300 with the vertical to keep the rain away. He
throws the umbrella and starts running at 10 km/hr. He finds raindrops are hitting his head vertically.
Find the speed of raindrops with respect to
(a) the road (b) the moving man.
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Q.11. Find the relation between the acceleration of rod A and
wedge B in the arrangement shown in the figure. Assume all
the surfaces to be smooth.
A
B

Q.12. The position of a particle at time t = 0 is P = (-1, 2, -1). It starts moving with an initial velocity

u  3iˆ  4ˆj and with uniform acceleration 4iˆ  4ˆj . Find the final position and the magnitude of
displacement after 4 sec.
Q.13. A particle is projected with velocity u and angle  with the horizontal. Find the time after which the
velocity will be perpendicular to the initial velocity.
Q.14. A particle moves in x-y plane with constant acceleration ‘a’ directed along the negative y-axis. The
equation of motion of the particle has the form y = x - x2, where  and  are positive constants.
Find the velocity of the particle at the origin.
Q.15. From the velocity time graph shown in figure.
Find the distance travelled by the particle during
the first 40 sec. Also find the average velocity
during this period.
5 m/s
20
40
sec.
-5 m/s
Q.16. A train travels from one station to another at a speed of 40 km/hr and returns to the first station at
the speed of 60 km/hr. Calculate the average speed and average velocity of the train.
Q.17. A body travels 200 cm in the first two seconds and 220 cm in the next four seconds. What will be
the velocity at the end of seventh second from start ?
Q.18. A man standing on a hill top projects a stone
horizontally with speed v0 as shown in figure. Taking
the co-ordinate system as given in the figure find the coordinates of the point where the stone will hit the hill
surface.
y
v0
x
(0, 0)

Q.19. A particle of mass 3 kg moves under a force of 4 î + 8 ĵ + 10 k̂ . Newton. Calculate the
acceleration (as vector) to which the particle is subjected to. If the particle starts from rest and was
at
origin
initially,
what
are
its
new
coordinates
after
3
seconds?.
Q.20. In a car race, car A takes a time t sec less than car B at the finish and passes the finishing point
with speed v m/s more than the car B. Assuming that both the cars starts from rest and travel with
constant acceleration a1 and a2 respectively. Show that v=( a1a2 ) t

Q.21. On a cricket field the batsman is at the origin of co-ordinates and a fielder stands in position (46 i +

28 j ) m. The batsman hits the ball so that it rolls along the ground with constant velocity
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

(7.5 i  10 j ) m/s. The fielder can run with a speed of 5 m/s. If he starts to run immediately the ball
is hit, what is the shortest time in which he could intercept the ball ?
Q.22. A particle moves in the x - y plane with velocity v x = 8t - 2 and vy = 2. If it passes through the point
x = 14 and y = 4 at t = 2s, Find the equation of the path.
Q.23. (a) 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.
Q.24. A farmer has to go 500 m due north, 400 m due east and 200 m due south to reach his field. If he
takes 20 minutes to reach the field,
(a) what distance he has to walk to reach the field ?
(b) what is his displacement from his house to the field ?
(c) what is the average speed of farmer during the walk ?
(d) what is the average velocity of farmer during the walk ?
Q.25. Two cars are moving in the same direction with the same speed 30 km/hr. They are separated by a
distance of 5 km. What is the speed of a car moving in the opposite direction if it meets these two
cars at an internal of 4 minutes?
Q.26. The equation of motion of a particle moving along a straight line is given as x = ½ vt where x, v, t
have usual meaning. Draw its approximate acceleration time graph.
Q.27. A river 400 m wide is flowing at a rate of 4 m/s. A boat is sailing at a velocity of 20 m/s with respect
to the still water in a direction making an angle 370 with the direction of river flow.
(a) Find time taken by the boat to reach the opposite bank.
(b) How far from the starting point does the boat reach on the opposite bank.
Q.28. An object projected with the same speed at two different angles covers the same horizontal range
1
R. If the two times of flight be t1 and t2, prove that R = gt1t2.
2
v (m/s)
Q.29. Velocity-time graph of a particle moving in a
straight line is shown in figure. Plot the
corresponding displacement time graph of a
particle if at t = 0 displacement s = 0.
C
20
A
10
B
D
0
2
4
6
8
t (sec.)
Q.30. A particle is projected with velocity v and at angle  from the horizontal. Find the instantaneous
power delivered by the gravity at the highest point.
Q.31. Two particles move in a straight line towards each other with initial velocities v 1 and v2 and with
constant accelerations a1 and a2 directed against the corresponding velocities at the initial instant.
What must be the initial maximum separation Smax between the two particles for which they meet
during the motion?
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Q.32. If an object travels one half of its total path in the last second of its fall from the rest then find
(a) the time and
(b) the height of its fall.
Q.33. A truck moving with constant acceleration covers the distance between two points 180 m apart in 6
seconds. Its speed as it passes the second point is 45 m/s. Find
(a) its acceleration
(b) its speed when it was at the first point.
Q.34. A body undergoing uniformly accelerated motion starts moving along +x-axis with a velocity of 5
m/s and after 5 seconds its velocity becomes 20 m/s in the same direction. What is the velocity of
the body 10 seconds after the start of the motion ?
Q.35. What is the speed with which a stone is projected vertically upwards from the ground if it attains a
maximum height of 20 m?
Q.36. A stone is projected from the ground with a velocity of 20 2 m/s at an angle of 450 with the
horizontal? What is the maximum height from the ground attained by the stone?
Q.37. The velocity of a car changes from 72 km/hr due east to 72 km/hr due north in 10 seconds. What is
the average acceleration of the car over this duration of time?
Q.38. A train moving along a straight road with a speed of 108 km/hr is brought to stop to next station
within 120 sec. after applying the brakes for the next station. What is the magnitude of the
retardation of the train.
Q.39. A car moving along a long straight road with a speed of 10 m/s is brought to rest within
10 seconds after applying the brakes. What is the magnitude of the retardation of the car ?
Q.40. A box is sliding on a smooth frictionless surface as shown in
the figure. A particle is projected at any unknown angle w.r.t.
box. At the same time another particle of mass m is released
from the ceiling of the block. Find out the relative acceleration
of the two particles.
u
m
2m
Q.41. A stone is projected from a balloon which is ascending with a velocity 2 m/s. The velocity of the
stone w.r.t. balloon is 2 m/s at an angle of 450. Find out the velocity of stone with respect to
ground.
Q.42. A body is projected with velocity 5 3 m/s at an angle of 600 with the horizontal. Find the angle
between the initial velocity vector and the velocity vector at a height of 2.5 m.
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Q.43. The velocity – time graph of a particle moving
along X – axis is shown. Find the displacement
and the distance travelled in 10 s.
v 2
(m/s)

0
10
5
t(s)
-1
Q.44. A particle of mass m is projected horizontally from certain height with a velocity v0. Find kinetic
energy of the particle after t seconds, assuming it is in the air.
Q.45. A particle moves in x-y plane such that x = kt and y = mt2. Where k and m are constants and t is
the time. Find the velocity and the equation of trajectory of the particle.
Q.46. A car starting from rest moving on a straight
line has acceleration – time graph as shown
in the figure. Draw the velocity – time
graph.
2
1
a
m/s
2
1
2
3
4
5
t (sec.)
-1
-2
Q.47. The acceleration of a particle varies with time
as shown. If velocity at t = 0 is v 0 then
calculate velocity as a function of time.
a
450
t
Q.48. A stone is dropped from a tower of height H. If distance covered during last second is half of
total height H. Find the total time taken by stone to reach the ground. Also find the value of H.
Q.49. A particle moves in a circle of radius 20 cm at a speed given by v = 1 + t + t2 m/s where t is time in
s. Find (a) the initial tangential and normal acceleration. (b) the angle covered by the radius in first 2
s.
Q.50. A particle of mass m is projected at angle  with the horizontal. The speed of a particle, when it is
at the greatest height is (2/5)1/2 times its speed when it is at half of its greatest height. Determine its
angle of projection.
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Q.51. A particle starts from origin at t = 0 along +ve x axis.
It’s velocity –time graph is shown in the figure. Draw
(i) a, t graph
(ii) x, t graph
v
4
2
4
O
t
-4
Q.52. The motion of a particle along a straight line is described by the function, x = 6 + 4t2 - t4 where x is
in meters and t is in seconds. Find the position, velocity and acceleration at t = 2 sec.
Q.53. A boat travels downstream from point A to point B in two hours and upstream in four hours. Find the
time taken by a log of wood to cover the distance from point A to point B.
Q.54. A particle slides down a smooth inclined plane of
elevation  fixed in the elevator going up with an
acceleration a0 as shown in figure. The base of the
incline has a length L. Find the time taken by the particle
to reach the bottom.
a0
m

L


Q.55. Two particles of masses m1 and m2 in projectile motion have velocities v1 and v 2 respectively at


time t = 0. They collide at time t0. Their velocities become v1 and v2 at time 2t0 while moving in air.




Find the value of | (m1v 1  m 2 v 2 )  (m1v 1  m2 v 2 ) | .
Q.56. A particle moving with uniform acceleration describes distances S1 and S2 meters in successive
intervals of time t1 and t2 seconds. Prove that the acceleration is
2 (S2t1 - S1t2) / t1t2 (t1 + t2)
Q.57. Two cars having masses m1 and m2 move in circles of radii r1 and r2 respectively. If they complete

the circles in equal time, find the ratio of the their angular speeds 1 .
2
Q.58. A body of mass m is projected vertically upwards with a velocity v 0. It goes up and comes back to
the same point. For this motion draw displacement-time, velocity-time, acceleration-time and
velocity-displacement graphs.
0
Q.59. A man standing on a road has to hold his umbrella at 30 with the vertical to keep the rain away.
He throws the umbrella and running at 10 kmph. He finds that rain drops are hitting his head
vertically. find the speed of raindrops with respect to
(a) the road
(b) the moving man.
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Q.60. A projectile travelling in a direction at 300 to the horizontal after 2 seconds of its start. It is travelling
horizontally after one more second. Calculate the speed and angle of projection of the projectile.
Q.61. Find the speed of two objects if, when they move towards each other, they get x meter closer every
second and when they move uniformly in the same direction with their original speeds, they get y
meter closer every two seconds.
Q.62. From the velocity time graph shown in figure. Find the
distance travelled by the particle during the first 40 sec.
Also find the average velocity during this period.
5 m/s
20
40
sec.
-5 m/s
Q.63. A particle A is moving along a straight line with velocity 3 m/s and another particle B has a velocity
5 m/s at an angle 300 to the path of A. Find the velocity of B relative to A.
Q.64. One end of a massless spring of spring constant 100 N / m and natural length 0.5 m is fixed and the
other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The
spring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find the
elongation of the spring.
Q.65. A ball takes t second to fall from a height h1 and 2t second to fall from a height h 2 then what is the
ratio of h1/h2.
Q.66. A projectile is projected with a unknown velocity at an unknown angle .
4 sec. What will be the maximum height reach by the projectile.
If time of flight is
Q.67. A stone is projected from the ground with a velocity of 10 m/s in the vertically upward direction.
How long does it remain in the air ?
Q.68. Two particles A and B are moving in a horizontal plane
anticlockwise on two different concentric circles with
different constant angular velocities 2 and  respectively.
(a) Find the relative velocity of B w.r.t. A after time t = /.
(Initial position of particles A and B are shown in figure.)
(b) Also find the relative position vector of B w.r.t. A.
Y
r
2r
Q.69. The velocity-time graph of a particle is given as shown in
the figure. Find the distance travelled by the object in 7 th
second.
A
B
X
7 m/s
4 m/s
v
3m/s
O
1 sec 4 sec
7 sec t
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Q.70. A plate is moving with a constant velocity v 0 in a
horizontal plane. A small particle is moving in a
circular (horizontal) path of radius  with constant
angular velocity. Find velocity of particle with
respect to ground when line OP makes angle 
with x-axis.
y

O
P
v0 

x
Q.71. A particle is projected with a velocity u at an angle  with an
inclined plane which makes an angle   45 with the
horizontal. Calculate the radius of curvature of the path of
projectile when velocity of projectile becomes parallel to the
plane.
v


Q.72. A man standing on a road has to hold his umbrella at 300 with the vertical to keep the rain away. He
throws the umbrella and starts running at 10 km/hr. He finds raindrops are hitting his head vertically.
Find the speed of raindrops with respect to
(a) the road (b) the moving man.
Q.73. Two bodies are projected from the same point with equal speeds and different angle of projection.
If they both strikes at the same point on an inclined plane whose inclination is . If  be the angle
of projection of the first body with the horizontal show that the ratio of their times of flight is
sin(  
cos 
Q.74. The velocities of particles P and Q are in direction
inclined at an angle  and  with the line segment
PQ and if the distance between P and Q remains
constants and given velocity of P is u. Find angular
speed of Q with respect to P.
u
v

P


Q
Q.75. (a) From an elevated point A, a stone is projected vertically upwards. When the stone reaches a
distance h below A, its velocity is double of what it was at a height h above A. Show that the
5
greatest height attained by the stone is h .
3
(b) The dependence of the x coordinate of two
b
x
bodies moving in a straight line (x - axis) is
given by curves a and b, respectively. Which
curve corresponds to the accelerated motion
a
and which curve to decelerated motion ?
Explain .
t
Q.76. An open elevator is ascending with zero acceleration . The speed v = 10m/sec. A ball is thrown
vertically up by a boy when he is at a height h = 10m from the ground. The velocity of projection is v
= 30m/sec with respect to elevator . Find
(a) the maximum height attained by the ball
(b) the time taken by the ball to meet the elevator again.
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(c) time taken by the ball to reach the ground after crossing the elevator
Q.77. A projectile is launched from on inclined plane with an
initial velocity v 0 as shown in the figure. Find the time
after which the projectile hits the plane for the first
time.
900

Q.78. A particle is projected up with a speed of 25 m/s from the ground. What is the maximum height
attained by the stone ? What is the distance travelled by the stone during 3rd second?
Q.79. A car moving with constant acceleration covers a distance of 24 m in first 2 seconds and
51 m in the next 3 seconds. What is the velocity of the car after next 5 seconds.
Q.80. A person walks with constant speed of 5 km/hr. He walks for 1hr. due east, then for 2 hrs. due
north, then for 1 hr. again due east and finally for 2 hr. due south west.
(a) What is the displacement of the person ?
(b) What is the total distance travelled by him?
Q.81. A car starts moving from rest with an acceleration whose value linearly increases with time from
zero to 6 m/s 2 in 6 sec after which it moves with constant velocity. Find the time taken by the car to
travel first 72 m from starting point.
Q.82. In the pulley-block system shown, find the
accelerations of A, B, C and the tension in the
string. Assume the friction to be negligible and
the string to be light and inextensible. The
masses of the blocks are m, 2m and 3m
respectively.
C
A
0
30
B
Q.83. A farmer has to go 500 m due north, 400 m due east and 200 m due south to reach his field. If he
takes 20 minutes to reach the field,
(a) what distance he has to walk to reach the field ?
(b) what is his displacement from his house to the field ?
(c) what is the average speed of farmer during the walk ?
(d) what is the average velocity of farmer during the walk ?
Q.84. A man wants to reach point B on the opposite bank of a
river flowing at a speed u as shown in the figure. What
minimum speed relative to water should the man have so
that he can reach point B ? In which direction should he
swim ?
B
45
A
0
u
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Q.85. Two Blocks A and B of masses 10 kg and 6 kg
respectively are connected through pulley as
shown in figure. Find
(a) acceleration of A and B.
(b) friction force acting on block A and B.
(c) tension between A and B. You may assume
the string was initially in just taut position.
2kg
Q.86. Two particles projected vertically upward from
point (0, 0) and (1, 0) with uniform velocity 10 m/s
and v m/s respectively, as shown in the figure. It is
found that they collide after time t in space. Find v
and t.
A
B
10kg
6kg
=0.2
=0.5
4kg
[1+2+1=4]
Y
v m/s
10 m/s
30
(0, 0)
45
(1, 0)
x
Q.87. Two smooth wedges of equal mass m are placed as
shown in figure. All surfaces are smooth. Find the
velocities of A & B when A hits the ground.
A
B

h
Q.88. A particle moves in the x - y plane with velocity v  p î  qx ĵ where î and ĵ are unit vectors in the
direction of x and y-axis, p and q are constants. At the initial moment of time, the particle was
located at the point x = y = 0. Find the equation of the trajectory of the particle.
Q.89. The velocity of a boat in still water is n times less than the velocity of flow of a river. At what angle
to the stream direction must the boat move so that drift is minimised ? If n = 2, show that the angle
 = 1200.
Q.90. The velocity of a particle, when it is at the greatest height is (2/5)1/2 times its velocity when it is at
half of its greatest height. Determine its angle of projection.
Q.91. On frictionless 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 velocity of (31) m/s. At a particular
instant, when the line OA makes an angle of 45 with the x-axis, a
ball is thrown along the surface from the origin O. Its velocity
makes an angle  with the x-axis and it hits the trolley.
y
A
45
(a) The motion of the ball is observed from the frame of the trolley.
Calculate the angle  made by the velocity vector of the ball
with the x-axis in this frame.
(b) Find the speed of the ball with respect to the surface, if  = 4  3
O
x
Q.92. A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10 m/sec w.r.t the
water in a direction perpendicular to river, find
(a) the time taken by the boat to reach the opposite bank.
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(b) how far from the point directly opposite to the starling point does the boat reach on opposite
bank.
Q.93. Two guns, situated at the top of a hill of height 10 m, fire one shot each with the same speed 53
m/s at some interval of time. One gun fires horizontally and the other fires upwards at an angle of
600 with the horizontal. The shots collide in air at a point P. find (I) the time interval between the
fringes and (ii) the co-ordinates of the point P, take origin of the co-ordinate system at the foot of the
will right below the muzzle and trajectory in x-y plane.
Q.94. A man can row a boat in still water at 3 km/h He can walk at a speed of 5 km/h on the shore. The
water in the river flows at 2 km/h. If the man rows across the river and walks along the shore to
reach the opposite point on the river bank find the direction in which he should row the boat so that
he could reach the opposite shore in the least possible time. The width of the river is 500 m.
Q.95. There are two parallel planes each inclined to the horizontal at an angle . A particle is projected
from a point mid-way between the two planes so that it grazes one of the planes and strikes the
other at right angle. Find the angle of projection.
Q.96. A body falling freely from a given height H hits an inclined plane in its path at a height h. As a
result of this impact the direction of the velocity of the body becomes horizontal. For what value of
h/H the body will take maximum time to reach the ground ?
Q.97. The velocity time graph of moving object is given in the figure. Draw the acceleration versus time
and displacement versus time graph. Find the distance travelled during the time interval when the
acceleration is maximum. Assume that the particle starts from origin.
80
60
V
(m/s
40
20
O
10
20
30
40
50
60
70
80
Time (s)
Q.98. A ball is thrown from the origin in the
x - y plane with velocity 28.28 m/s at an
angle 45 to the x - axis. At the same
instant a trolley also starts moving with
uniform velocity of 10m/s along the
positive x - axis. Initially, the trolley is
located at 38m from the origin. Determine
the time and position at which the ball hits
the trolley.
y
v0
10m/s
B
45
O
A
38m
3m
C
D
2m
x
Q.99. A particle is suspended from a fixed point by a inextensible string of length 5m. It is projected from
its lowest position in the horizontal direction in a vertical plane with such velocity that the string
slackens after the particle has reached a height 8m above the lowest position. Find the velocity of
the particle, just before the string slackens. Find also to what height the particle can rise further.
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Q.100. A bus is running along a highway at a speed of v1 = 16 m/s. A man is at a distance a = 60 m from
the highway and at a distance b = 400 m from the bus. In what direction should the man run to
reach any point of the highway, before or at the same time as the bus. The man can run at a speed
of v2 = 4 m/s.
Q.101. An aircraft flies at 400 km/hr in still air. A wind of
N
B
200 2 km /hr is blowing from the south. The pilot
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. (Given cos 150 = 0.9659)
0
va
45
C
0
45
A
vw

vaw
E
Q.102. A insect moves with constant speed of 10 m/s. At t = 0. It moves for 3 second due to east, next 3
second due to North and finally for 3 2 second due south west.
(a) What is the displacement of the insect ?
(b) What is the total distance travelled by insect ?
Q.103. A stone ‘A’ is dropped from the top of a tower 20 m high simultaneously another stone ‘B’ is thrown
up from the bottom of the tower so that it can reach just on the top of the tower. What is the
distance of the stones from the ground while they pass one another.
Q.104. A car moving with constant acceleration covers a distance of 100 m in 3 sec and then next 100 m in
2 sec. Find the acceleration of the car.
Q.105. A cyclist moves with constant speed 5 m/s along eastward for 2 seconds, then along southward for
2 seconds, then he moves along west for one second and finally along North- west for 2 seconds.
Find
(a) Distance and displacement of cyclist for whole journey.
(b) Average speed and Average velocity for whole journey
(c) Average acceleration of cyclist for whole journey.
Q.106. A car starts from rest and moves with a constant acceleration of 2.0 m/s2 for 30 seconds. The
brakes are then applied and the car comes to rest in another 60 seconds. Find
(a) total distance covered by the car.
(b) Maximum speed attained by the car
(c) Find shortest distance from initial point to the point when its speed is half of maximum speed.
Q.107. A particle is projected with speed 60 m/sec at an angle 300 from the horizontal. Find
(a) Minimum time taken to reach a height of 25 m.
(b) Vertical and horizontal component of velocity at the time found in part (a)
(c) The horizontal displacement covered by particle in the time calculated in part (a).
Q.108. How much high above the ground a monkey can throw a mango if he is able to throw the same
mango upto a maximum horizontal distance of 100 m ?
http://www.rpmauryascienceblog.com/
m
Q.109. A body of mass ‘m’ is dropped freely from a
height ‘h’. Then draw the following graphs
for the given body ?
(a) Displacement - Time graph
(b) Kinetic energy - Time graph
(c) Total energy - Time graph
(Assume displacement to be zero at ground
level.)
h
Q.110. A batsman hits a ball at a height of 1.22 m above the ground so that ball leaves the bat an angle
45o with the horizontal. A 7.31 m high wall is situated at a distance of 97.53 m from the position of
the batsman. Will the ball clear the wall if its range is 106.68m.
Take g = 10 m/s2.
Q.111. A particle is projected from point 0 on the
ground with velocity u=55 m/s at angle
  tan1 (0.5). It strikes at a point C on a
fixed smooth plane AB having inclination
of 370 with horizontal as shown in figure.
Calculate
(a) Coordinates of point C in reference to
coordinate system as shown in figure
(b) Velocity of particle with which it strikes
inclined plane AB.
B
C
y
55
37

O
x
0
A
10/3 m
Q.112. A motor car is travelling at 30 m/s on a circular road of radius 500 m. It is increasing in speed at the
rate of 2m/s 2. What is its acceleration?
Q.113. An object A is kept fixed at the point x =
3m and y = 1.25 m on a plank P raised
above the ground.
At time t = 0 the
plank starts moving along the +x direction
with an acceleration 1.5 m/s2. At the
same instant a stone is 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 450 to the
horizontal. All the motions are in the x-y
plane. Find u and the time after which the
stone hits the object.
y
A
P
P
1.25m
u
Q.114. Two masses ‘m’ and ‘2m’ are connected by a massless
string which passes over a light frictionless pulley as
shown in fig.1. The masses are initially held with equal
lengths of the strings on either side of the pulley. Find the
velocity of the masses at the instant the lighter mass
moves up a distance of 6.54 m. The string is suddenly cut
at that instant. Calculate the time taken by each to reach
the ground. (g = 9.81 m/s2)
x
3.0 m
O
2m
m
13.08 m
ground
Fig. 1
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Q.115. A particle moves in a circle of radius 20 cm at a speed given by v = 1 + t + t2 m/s where t is time in
s. Find
(a) the initial tangential and normal acceleration.
(b) the angle covered by the radius in first 2 s.
Q.116. A car starts moving from rest with an acceleration whose value linearly increases from zero to 6
m/s2 in 6 sec after which it moves with constant velocity. Find the time taken by the car to travel first
72 m from starting point.
Q.117. A particle is projected vertically with velocity v0. Wind is
blowing and is providing a constant horizontal
acceleration a0. There is a vertical wall at some
distance from point of projection. If particle strikes the
vertical wall perpendicularly then calculate
(i) Time of flight.
(ii) Velocity with which particle strikes the vertical wall.
(iii) Distance x and y.
(iv) If collision at vertical wall is perfectly elastic will
particle retrace its path ?
(v) Is path of particle parabolic?
Q.118. A particle is projected up a large inclined plane
from a point O on it as shown in the figure. The
projection velocity has a magnitude of 5.5m/s and
its direction makes an angle 37° with the inclined
plane. The inclination of the plane is also 37°. The
inclined plane starts moving towards left with an
acceleration a0 = 5 m/s2 at the moment the
particle is projected. The particle strikes the
inclined plane at a point P. Find the time taken by
the particle to move from O to P. Also find the
magnitude of displacement along the inclined
plane as it moves from O to P. (Take sin 370 =
3/5)
`
y
v0
x
P
a0 = 5 m/s2
u = 5.5 m/s
370
O
370
Q.119. Two inclined planes of inclinations 300 and 600
respectively meet at 900 as shown in figure. A particle is
projected from point P on the first inclined plane with a
velocity u = 10 3 m/s in a direction perpendicular to the
inclined plane and it is observed to hit the other inclined
plane at 900.
Find (a) the height of point P from ground
(b) the length of PQ .
B
u
A
Q
P
h
60
30
O
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Q.120. A sleeve ‘A’ can slide freely along the smooth rod bent in the
shape of a half circle of radius R as shown in figure. The system
is set in rotation with a constant angular velocity  about the
vertical axis OO. Find the angle  corresponding to the steady
position of the sleeve.
O
R

A
O’
Q.121. A large heavy box is sliding without friction down a
smooth inclined plane of inclination . From a point
P on the bottom of the box a particle is projected
inside the box, with speed u (relative to box) at
angle  with the bottom of the box.
u


Q
P
(a) Find the distance along the bottom of the box between the point of projection P and the point Q
where the particle lands. The particle does not hit any other surface of the box.
(b) If horizontal displacement of the particle with respect to ground is zero. Find the speed of the
box at the moment when particle was projected.
Q.122. A stone is dropped from the top of a tower 20 m high. Simultaneously another stone is thrown up
from the bottom of the tower so that it can reach the top of the tower. What are the speeds of the
stones while they pass one another ?
Q.123. Two rockets are fired vertically from launching pads which are side by side. The first rocket moves
vertically upwards with an acceleration of 6g and second with an acceleration of 8g. If the second
rocket is fired 1 sec. after the first, find how long after its launching the second rocket overtakes the
first.
Q.124. A particle is moving along a vertical circle of
radius R = 20 m with a constant speed v = 31.4
m/sec. as shown. Line ABC is horizontal and
passes through the centre of the circle. A shell
A
is fired from point A at the instant when particle
is at C. If distance AB is 20 3 m and shell
collides with the particle at B, calculate
(a) Smallest possible value of the angle  of projection.
(b) Corresponding velocity u of projection
u
20 m

C
B
Q.125. A projectile is fired with velocity v0 at an angle  with the horizontal on a horizontal plane, Find
(a) The average velocity of projectile in half of time of flight.
(b) The time in which the velocity of projectile becomes perpendicular to its initial velocity.
(c) The radius of curvature of projectile at the instant when it is at its maximum height.
Q.126. A stone is projected from a point of ground in such a direction so as to hit a bird on the top of a
telegraph post of height h and then attain the maximum height 2h above the ground. If at the
instant of projection the bird were to fly away horizontally with a uniform speed, find the ratio
between the horizontal velocities of the bird and the stone, if the stone still hits the bird while
descending.
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Kinematics Solution
Q.1.
1
1
m.u2  m.502
2
2
1
Final K.E. = mv 2
2
Work done by gravity = +mgh = mg. 40
Initial K.E. =
From W-E principle
mg . 40 = kf - kI =

Q.2.
1
m(v2 - 502)
2
v = 57.4 m/s.
For plane,
5
m/s; vertical velocity = 0
18
2h
2  2000
= 20 sec.

g
10
Horizontal velocity = 540 
Time of flight t =
 Horizontal displacement = 20  540 
Q.3.
zero
Q.4.
(i) Maximum height H =
5
= 3000 m
18
u2 sin 2
2g
2
(50)2  3 
H=

 m
2  10  2 
H = 93.75 m
u2 sin 2 (50)2  sin120

g
10
R = 216.5 m
(ii) Range R =
Q.5.
4
5

60 v c  30
 vc = 45 km/hr.
Q.6. (a) Maximum height above the tower, using v2 = u2 + 2as in vertical direction.
(u sin 300)2 = 2gh As u = 40 m/s,  = 300
40  40  1
1600
= 2  10  h  h =
 20m
4
80
 height above ground = 100 + 20 = 120m.
1
(b) Range, time of flight = t, H = u sin t - gt 2 , H = - 100 m,
2
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1
1
)t -  10  t 2
2
2
- 100 = 20 t - 5t2
t2 – 4t – 20 = 0, t = 6.9 sec.
R = ucos  t,
R  distance from tower
3
R = 40 
 6.9 = 238.9 m.
2
- 100 = (40 
1
1
m.u2  m.502
2
2
1
Final K.E. = mv 2
2
Work done by gravity = +mgh = mg. 40
From w~E principle
1
mg . 40 = kf - kI = m(v2 - 502)
2

v = 57.4 m/s.
Q.7.
Initial K.E. =
Q.8.
For plane,
Horizontal velocity = 600 
Time of flight t =
2h

g
5
m/s; vertical velocity = 0
18
2 1960
= 142sec
10
 Horizontal displacement = 142  600 
5
18
= 3299.83 m  3300 m.
Q.9.
Q.10.
4
5

60 v c  30
 vc = 45 km/hr.
(a) vmg = vrg sin 300
vrg = 20 km/hr.
vm, g
0
30
(b) v rm = v rg cos 30
vrm = 10 3 km/hr
Q.11.
y
 tan 
x

vr, m
aA = aB tan .
Q.12. Initial position vector of particle

 ˆi  2ˆj  kˆ

Final position of the particle after 4 seconds
 1
S f  Si  ut  at 2
2
vrg
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1
  ˆi  2ˆj  kˆ  3iˆ  4ˆj  4   4iˆ  3ˆj  16
2
final position = 21iˆ  42ˆj  kˆ , Displacement = - 20 î + 40 ĵ .

 


Magnitude of displacement =

2
 20    40 
2
 20 5 m
Q.13. v  u cos  î  (u sin   gt ) ĵ
u  u cos  î  u sin  ĵ
u.v  0 = u2 cos2  + u2 sin2  - (u sin)gt
u
 t=
g sin 
Q.14. Comparing this equation y = x - x2 with equation of projectile.
ax 2
y = x tan  ,
2u 2 cos 2 
a
we get tan  =  &  =
sec 2 
2u 2
a
or,
u2 =
(1  tan 2 )
2

u=
a
(1   2 )
2
Q.15. Distance = 100 m
Velocity = zero
2s
2v 1v 2
2  40  60


s
s
v1  v 2
40  60

v1 v 2
= 48 km/hr
Average velocity = 0
Q.16. Average speed =
1
a(2)2

2 = 2u + 2a
2

1=u+a
. . . . (i)
1
2.2 + 2 = u(6) + a(6)2
2
4.2 = 6u + 18 a

2.1 = 3u + 9a

0.7 = u + 3a
. . . . (ii)
Solving (i) and (ii) we get
a = -0.15 m/s 2. , u = 1.15 m/s.
v = 1.15 - 0.15  7 = 0.01 m/s.
Q.17. 2 = u(2) +
Q.18. x = v0 cos  t
1
y = - gt 2
2
|y| = |x| tan 
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y=-
1
x2
g 2
2
v 0 cos2 
1
x2
g 2
= x tan 
2 v 0 cos2 
 x = 0 or x =
x=

y= 
=-
2v 02 cos 2  tan 
g
v 20 sin 2
g
1 v 04 sin2 2
g
2 g2 v 02 cos2 
v 20 sin2 2
2g cos2 

Q.19. Taking F as the net force
 4 8 10
a  î  ĵ  k̂ m/s 2
3 3
3
  1 2
s  ut  a t
2
1 4
8
10 
= 0   î  ĵ 
k̂  9
2 3
3
3 
= 6 î  12 ĵ  15k̂
sx = 6m;
sy = 12m
Q.20.
sz = 15m
Let time taken by the car and their final velocities are t1, t2 and v 1, v2 respectively.
Given t1 = t2 - t and v 1 = v2 + v
1
1
s1 = a1t 12  s 2  a 2 t 22 = s (say)
2
2
2
2
 a1t 1  a 2 t 2 = 2s
also v 1 = a1t1, v2 = a2t2
 v1t1 = a1t12 = 2s and v2t2 = a2t22 = 2s
2s
2s
 t1 =
and t2 =
v1
v2
 1
1
so t2 - t1 = 2s
 t
 v 2 v1 
 v  v2 
 2s 1
 t
 v 1v 2 
 2s
v
t
v 1v 2
v v 
v =  1 2 t 
 2s 
v 12 v 22
 2s 2
t
http://www.rpmauryascienceblog.com/
=
v 1v 2
t  a1a 2 t
t 1t 2
Q.21. The ball's position is at time t,
(7.5)t. î + (10)t ĵ
Suppose the fielder runs from his position with constant velocity 5 [( î cos   ĵ sin  ] m/s. relative to
the wicket. At interception of the ball by the fielder the position must coincide so equating the
components we get
7.5 t = 46 + 5t cos 
. . . (I)
and 10t = 28 + 5t sin 
. . . (ii)
2
2
 7.5t  46 
 10t  28 
These give 
 
 1
5t


 5t

116
which simplifies to t = 4 sec, or
sec.
21
Hence for earliest interception t = 4s.
Q.22. vx = 8 t - 2
x = 8  t2/2 - 2  t + c 1
at t = 2
x = 14 so c1 = 2
so x = 4t2 - 2t + 2
vy = 2
so y = 2  t + c 2
at t = 2, y = 4 so c2 = 0
y = 2t
eliminatating t form (I) and (ii)
x = y2 - y+ 2
Q.23. (a) s = ut +
1 2
at
2
. . . (i)
. . . (ii)
table
for vertical motion
v
4m
1
4 = 0 + (10) t2
2
5m

4
5
4
t=
5
t2 =
v = 5/t =
5
4/5
5=vt

=
5 5
4

5 5
m/sec.
2
Q.24. (a) Distance
= 500 + 400 + 200 = 1100m


(b) Displacement = 500 j  400 î  200  ĵ = 300 j  400 î

Magnitude of displacement =

400
 
2
 300 = 500m
N
y
x
2
E
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1100
11
Total distance
=
=
m/s
20  60
12
Total time
D is placement
500
5
(d) Average velocity =
=
=
m/s
Time
20  60
12
(c) Average speed =
Q.25.
4
5

60 v c  30
 vc = 45 km/hr.
dx t
.
dt 2
 x = kt2
k some constant
d2 x
 2k .
dt 2
Q.26. x =
Q.27. (a) Resultant velocity of the boat is
v = (vR + vB cos 370 ) i + vB sin 370 j
4
3
4i + 20  i + 20 
j
5
5
v =20 i + 12 j m/s
time taken by boat to cross the river
dis tan ce travelledin y direction
=
velocity in y direction
400
100
=
sec.
12
3
(b) Displacement along x = v t
100
2000
= 20 
=
m
3
3
distance from starting point
 t=
2
400 109
 2000 
m.
( 400)  
 
3
 3 
2
Q.28. For same range, angles are  and 900 - 
u2 sin 2

R=
g
2u sin 
2u sin( 90   2u cos 
t1 =
, t2 =

g
g
g

t1t2 =

R=
4u2 sin  cos  2R

g
g2
1
gt1t 2 .
2
a
t
vB
400 m
Y
0
37
vR
X
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s(m)
Q.29.
D
80
60
30
10
0
2
4
6
8
t (sec.)
Q.30. Pins = F.v
F  mg ĵ, v  u cos  î
Pins = 0
Q.31. | Relative velocity | of approach = v 1 + v2
| Relative acceleration | = a1 + a2
Acceleration is directed against the initial velocity. For meeting of the two bodies the maximum
separation Smax be such that relative velocity is just reduced to zero.

v r2  ur2  2ar Smax

Smax =
ur2
( v  v 2 )2
 1
2ar 2(a1  a 2 )
Q.32. Let the time of fall be t and height of fall be h, then
1 2
gt  h
… (i)
2
1
h
g(t – 1)2 =
… (ii)
2
2
2
1
t 1
1
 t  1
 


t=
 
2
t
2
 t 
1
and h =
 10  ( 2 + 2)2 = 58.2 m
2
1
 a  36 = 180
2

6u + 18 a = 180
u + 6a = 45
from (i) and (ii)
a = 5 m/s2 and u = 15 m/s.
Q.33. 6u +
… (i)
… (ii)
Q.34. u = 5 m/s, v = 20 m/s, t = 5 sec.
v  u 20  5
a=

 3 m/s 2
t
5
v = u + at = 5 + 3  10 = 35 m/s
Q.35. h = 20m, a = - g, v = 0
thus 0 = u2 - 2gh
2 + 2 seconds
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u=
2gh =
2  10  20 = 20 m/s
Q.36. uy = u sin  = 20 2 sin 45-0 = 20 m/s
u2 sin2  20  20
maximum height =

= 20 m.
2g
2  10



v 2  v1 
Q.37. aav 
, v 2  20 m / s ĵ
t

v1  20 m / s î

20 ĵ  20 î
Thus aav 
 2 î  2 ĵ
10
hence a = 2 2 m/s due north west.
Q.38. vi = 108 km/hr =
108  1000
= 30 m/sec.
60  60
t = 120 sec.
vf = 0
vf = vi + at
or,
0 = 30 m/s + 100 sec,
30 m / sec
1
or,
a=  m / s2
120 sec .
4
or
a = -(0.25) m/sec 2
magnitude of retardation = 0.25 m/s 2
Q.39. u = 10 m/s, v = 0, t = 10 s
vu
10
a=

 1m / s2
t
10
|a| = 1m/s 2
Q.40. Acceleration of particle of mass m w.r.to ground = g
Acceleration of particle mass (2m) w.r.to ground = g
Thus ar = g – g = 0

Q.41. v SB  v cos 450 ˆi  v sin 450 ˆj

2
1 ˆ
1 ˆ
i 2
j
2
2
 (iˆ  ˆj)m / s

vBG  2ˆj m / s



Thus v S,G  v SB  v BG
= 2 ˆj  (iˆ  ˆj) = ( (iˆ  3ˆj)
v=
12  32  10 m / s and tan =
 = tan-1(3).
3
3
1
ĵ

î
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Q.42. For vertical component of motion
v1
vy2 = uy2 + 2gh
2
2
2
2
 v sin 1 = u sin 1  2gh
2
 v 2sin22 = (5 3 ) 2 sin260  2 10 (2.5)
 v sin 2 =  5/2
1
5 3
15 
Initially velocity vector u = 
î 
ĵ
O
 2
2 

5 3
5 3
5
5 
Velocities vector at height 2.5 m v1 = 
î  ĵ  and v 2 = 
î 
 2

 2
2
2



Now the angle between initial velocity and velocities at height 2.5 m is
75 75
75

2
u.v 1
3
4
4
4
cos1 =
=
=

| u || v 1 |
75 225 75 25
300 10
2 3



4
4
4
4
2
2
3
 cos 30
2
75 75

4
4
cos2 =
0
300 100
4
4
cos1 =

 1  30

 2  90
1  10 
1
10
15
 2   5   1  5  1     m
2  3 
2
3 
 6 
1  10 
1
10 
 55 
Distance travelled =    2    5 
1  5  1    m
2 3 
2
3 
 6 
Q.43. Displacement =
For plane,
5
m/s; vertical velocity = 0
18
2h
2  2000
= 20 sec.

g
10
Horizontal velocity = 600 
Time of flight t =
 Horizontal displacement = 20  600 
5
10000
=
m
3
18
Q.44. vx (horizontal) = v 0
vy (vertically downward) = gt
v2 = v 2x  v 2y = v 02  (gt )2
K.E. of particle after time ‘t’ =
Q.45. vx =
t=
x
k
dx
k;
dt
vy =
1
1
mv2 = m( v 02  (gt )2 )
2
2
dy
 mt
dt
2
mx 2
x
& y = mt2 = m    2
k
k

 v  k ˆi  2mtjˆ
2
v2

ĵ 


2.5m
http://www.rpmauryascienceblog.com/
 Equation of trajectory, y =
mx 2
k2
Q.46.
5
4
v (m/s)
3
2
1
1
v
Q.47.

2
3
4
5
t (sec.)
t

dv  tdt
v0
0
 v = v 0 + t2/2
1 2
gt
2
H 1
2
 g  t  1
2 2
Q.48. H 
t  2t  2
t
H=
2



t
 2
t 1


H

2 1 t  2
sec =3.4 sec
2 1
1 2
gt  57.8m
2
dv
= 2t +1
dt
v2
normal acceleration an =
 (at)t=0 = 1 m/s2
R
2
v
1
(an)t=0 = 0 
= 5 m/s 2
R
0.2
d
(b) v = R
dt
Q.49. Tangent acceleration at =
'
2
R d = (1+t+t )dt
 R d 

0
2
 1  t  t dt
2
0
 = 33.3 rad
Q.50. Let u and  are projection speed and angle of projection respectively.
vx = u cos  and vy = u sin 
t
http://www.rpmauryascienceblog.com/
u2 sin2 
2g
vertical velocity at half the greatest height
v2 = v 2y  2gh / 2
At greatest height h =
v2 = u2 sin2 v=
gu 2 sin 2  u 2 sin 2 

2g
2
u sin 
2
Resultant velocity at half the greatest height
vR =
v 2  (u cos )2 
 u cos  =
tan  =
3,
2
5
u2 sin 2 
 u 2 cos 2 
2
u2 sin2 
 u2 cos2 
2
 = 600
Q.51. (i) Velocity is decreasing
so, a = -4/2 = -2
a

t
O
-2
(ii)
4
x
t
Q.52. At, t = 2 sec, x = 6 + 16 - 16 = 6 m
dx
velocity, v =
= 8t - 4t3
dt
at t = 2 sec, v = 16 - 32 = -16 m/sec.
dv
Acceleration,
= 8 - 12 t2
dt
at t = 2 sec. Acceleration = 8 - 12 (2)2 = -40 m/s2
Q.53. 8 hrs.
4
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1/ 2


2L
Q.54. 

 (g  a0 ) sin  cos  
Q.55. 2 (m1 + m2)gt0
1 2
f t1 where f is the acceleration . In the next t2
2
seconds the point moves to S2. It moves through distances S1 + S2 in (t1 + t2) seconds
1
S1+S2 = u(t1 + t2 )+ f (t2 + t1)2
2
1 2
1
=(ut1 + f t1 ) + ut2 + ft2 (t2 + 2t1)
2
2
1
S2 = ut2 + ft2 (t2 + 2t1)
2
1
Hence S2t1 - S1t2 = ft1t2 (t1 + t2 )
2
2(S2 t1  S1t 2 )
or f =
t1t 2 ( t1  t 2 )
Q.56. Let u be the initial velocity. Then S1 = ut1 +
Q.57. 1
Q.58. displacement time
s
velocity displacement
v
t
d
velocity time
v
acceleration time
a
t
t

Q.59. Velocity of rain w.r.t. road is v r and

velocity of rain w.r.t. moving man is vrm
but

 
v rm = v r  v m
vr
vrm
0
30
=-vr sin 30 î - vr cos 30 ĵ - 10 î
= (- v r sin 30 – 10 ) î - v r cos 30 ĵ
But - vr sin 30 – 10 = 0

vr sin 30 = -10
 10
vr =
= - 20 m/s.
sin 30
vm
But vr is not negative 

v m  10 î
http://www.rpmauryascienceblog.com/

and vrm  [ 20 cos 30]
= 20 cos 30 ĵ
= 10  3 ĵ .
Q.60. After 2 seconds, vertical component of velocity, v y = v0 sin  - g(2) = (v 0 sin  - 2g)
Horizontal component of velocity = v0 cos 
v sin   2g
1
Hence, 0
= tan 300 =
. . . (i)
v 0 cos 
3
Again, velocity becomes horizontal when projectile reaches maximum height. Hence, time to reach
maximum height = 3 seconds.
v sin 
Therefore, 0
=3

v0 sin  = 3g
. . . (ii)
g
From (i) and (ii), v0 cos  = 3 g, v0 sin  = 3g
Hence, squaring, and adding, v0 = 2g3 = 20 3 m/sec.
Dividing, tan  = 3 = tan 600

 = 600
Q.61. Let vA and vB are the speeds of the objects
While moving towards each other
vA + vB = x (m/s)
. . . (i)
While moving in same direction (Assuming vA > vB )
vA - vB = y/2 (m/s)
. . . (ii)
Solving (i) and (ii),
1
vA = (2x  y )
4
1
vB = (2x  y ) .
4
Q.62. Distance = 100 m
Velocity = zero
-vA
Q.63. |vB - vA| = v B2  v 2A - 2vAvB cos 300
= 52 + 32 - 2 5  3  (3/2)
= 8.02
3
2.832
using sine rule

  = 320
sin  sin 300

1 2
1
h2 = g( t 22 )  2gt 2
gt ,
2
2
h1 (1/ 2)gt 2 1


h2
4
2gt 2
Q.65. h1 =
2v 0 sin 
g
0
30
3m/s
Q.64. kx = m2 (0 + x)
 x  1 cm.
Q.66. Time of flight T =
5 m/s
vB - vA
… (i)
vB
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Maximum height H =
v 20 sin2 
2g
… (ii)
eliminating v0 sin 
2
 Tg 
 
T 2 g2
gT 2
2
H=    f
=
2g
8g
8
At T = 4 sec.,
gT 2 10  4  4
H=
= 20 m

8
8
Q.67. u = 10 m/s, t = ? y = 0 thus
1
1
0 = 0 + 10t - gt 2  10t =
 10  t2  t = 2 seconds.
2
2

= 2, vA = 2r ĵ


B =  = , vB = 2r(- ĵ )




(a) v BA  v B  v A  2r(  ˆj)  2r( ˆj) = -4r ĵ

 
(b) rBA  rB  rA = 2r (  î ) - (r) î = - 3r î .
Y
Q.68. A = 2
Q.69.
A
r
2r
X
B
dv 7  4

 1m / s2
dt
4 1

 a
st 7  u  (2n  1)
2
1
 3  (2  7  1)  9.5 m .
2
Q.70. Velocity of particle w.r.t. ground
= v0 î   sin (  î ) +  cos  ĵ
O

= (v0 -  sin ) î +  cos  ĵ
Q.71. Along y,
u sin   gcos t  0
u
 t  ta n 
g
Along x,
u
v x  u cos   g sin .t  u cos   gsin . tan 
g
=
u cos   u sin  ucos 2

cos 
cos 
P


 v0
X
v
Y

g sin

g cos
http://www.rpmauryascienceblog.com/
 v 
r x
an
2

u2 cos2 2
gcos3 
vm, g
Q.72. (a) v rm = v rg cos 30
vrm = 10 3 km/hr
(b) vmg = vrg sin 300
vmg = 20 km/hr.
0
30
vr, m
Q.73. Let  be the angle of projection of the second body
u2
R=
[sin( 2  )  sin 
g cos 2 
since range of both the bodies is same. Therefore
sin (2 - ) = sin (2 - )
or 2 -  =  - (2 - )

 = - ( - )
2
2u sin(  
Now T =
g cos 
2u sin(   
T =
g cos 
T sin(  

T
cos 
vrg
u



Q.74. Let v be the velocity of Q
if  is constant
u cos  = v cos 
 v=
u cos 
cos 
v sin   u sin 

u(tan  cos   sin  )
=

=
Q.75. (a) If B & D be the points h above and h below A, then in the stone’s downward motion
vD = 2vB
C
2
2
v D  v B  2 g2 h


4 v 2B  v 2B  4gh
4
v 2B  gh
3
B
h
A
h
D
If C be the highest point attained by the stone then,
[when h = BC]
v 2C  v 2B  2gh'
4
2

v 2B  2gh'

h'  gh  gh
6
3
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5
h
3
(b) The acceleration of a body moving along straight line is the second derivative of its position
coordinate. For a concave surface , the second derivative is +ive while it is negative for the convex
surface. Hence ‘a’ corresponds to accelerated while ‘b’ corresponds to decelerated motion.
Greatest height = h + h =
Q.76. (a)
Velocity of ball relative to elevator = 30 m/s
Velcoity of ball relative to ground = 10 + 30 = 40 m/s
Maxmum height attained by ball = 10 
(b)
(c)
40 2
= 90m
2  10
When they meet again their displacement is same
1
10t = 40t -  10  t2
2
t = 6 sec
1
70 = 40t +  10  t2
2
5t2 + 40t - 70 = 0
Q.77. Let the projectile hit the plane after time t.
The horizontal displacement x = (v0 sin) t
The vertical displacement y = (v 0 cos) t -
Y
1 2
gt
2
y = -(tan) x for the plane
2v 0
t=
g cos 
u2
625

= 31.25 m
2g 2  10
25
time after which v = 0, t =
= 2.5 second.
10
Distance travelled in 3rd second = |y1| + |y2|
where,
y1 = y(2.5) - y(2)
y2 = y(3) - y(2.5)
1
y(2) = 25  2 -  10  22  50  20  30m
2
1
y(3) = 25  3 -  10  32  75  45  30m
2
y(2.5) = 31.25 m
thus distance during 3rds = 1.25 + 1.25 = 2.5 m
Q.78. Maximum height h =
Q.79. Let initial speed of the car is u and acceleration be a. then
1
2u +
 a  22 = 24
2
2u + 2a = 24
4 + a = 12
…(i)
V0
O

X
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1
 a  52 = 75
2
 10 u + 25 a = 150
…(ii)
2
from (i) and (ii) a = 2 m/s s and u = 10 m/s
after 10 s = 10 + 2  10 = 30 m/s
and 5u +
Q.80. Unit vector due east be î and due west be ĵ . then
net displacement = 5 î  10 ĵ  5 î  [5 2 cos 45(  î )  5 2 sin 45( ĵ)]
= (10 – 5) î  (10  5) ĵ
= 5 î  5 ĵ
(a) thus net displacement = 5 2 due north east
(b) distance = 5 + 10 + 5 2 + 5 = 20 + 5 2 m
Q.81. Since acceleration varies linearly so
a  t m/s2 by given condition
6
 a = kt

6
 da  k  dt
0
k=1
0
dv
t2
t  v=
m/s
dt
2
ds t 2
t3
then,

or, s =
dt
2
6
at the end of t = 6 sec. Acceleration becomes zero.
Distance moved by car at t = 6 sec is
6 66
S1 =
= 36 m
6
66
Speed of the car =
= 18 m/s
2
Remaining distance = 72 – 36 = 36 m.
36
so time taken to cover this distance = t2 =
sec .  2 sec .
18
Total time = 6+2 = 8 sec.
then,
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Q.82. The FBD of A,B,C are shown
T = ma1
. . . (1)
2mg – 2T = 2ma2
. . . (2)
T + 3 mg sin 300 = (3m)a3
. . . (3)
constraint relation :
a1+ a3 = 2a2
. . . (4)
solving the equations
9
11
13
9
a1 =
g , a2 =
g , a3 =
g, T =
mg
20
20
20
20
a1
mg

T
N1
2T
N3

T
3mg
Magnitude of displacement =
400
 
2

2mg
Q.83. (a) Distance = 500 + 400 + 200 = 1100m


(b) Displacement = 500 j  400 î  200  ĵ = 300 j  400 î

a3
a2

N
x
 300 = 500m
2
E
1100
11
Total distance
=
=
m/s
20  60
12
Total time
D is placement
500
5
(d) Average velocity =
=
=
m/s
Time
20  60
12
 300 
 = tan-1 
= 370 due North of East.

 400 
(c) Average speed =
Q.84. vx = u – v sin 
vy = v cos 
vy
tan 450 =
vx
u
 45
0
v
=1
 v y = vx
u – v sin  = v cos 
u
v=
sin   cos 
u
=
2 sin(  450 )
clearly minimum value of v =
u
2
0
for  = 45 .
Q.85. fmax = 1000.5+600.2 = 62 N
Fx = 40 – 20 = 20 N
fmax > Fx
So acceleration of Block = zero.
fB = 12 N
40 – T – 12 = 0
T = 28 N
fA + 20 = T
fA = 8 N
y
20
10kg
6kg
fA
fB
40
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Q.86. x1  10 cos 30t
x 2  v cos 45t
1 2
gt
2
1
y 2  v sin 45t  gt 2
2
For collision
y1  y 2
1
v
10  
2
2
y1  10 sin30t 
 v  5 2 m/s
 x1  x 2  1
10 cos 30t  5 2 cos 45t  1


t 5 3 5 1
and
1
t=
5


sec
3 1
Q.87. Writing constraint relation
yA = yB tan 
differentiate w.r.t. t we get
vA = vB tan 
using COE
yA
… (i)

yB
1
1
m( v B )2  mv 2A
… (ii)
2
2
putting value vA in equation (ii) and solving we get
mgh =
2gh cos , vA =
vB =
2gh sin 
Q.88. v  p î  qx ĵ
It is clear that p is the x-component of velocity and qx is the y-component

dx
 p or dx = p.dt
dt
or

x
0
t
dx   p  dt
0
or x = pt
and
… (1)
dy
= qx or dy = qx.dt
dt
or dy = q.p.t dt  x = p.t

y
0
t
dy   p.q.t dt
y=
0
p.qt 2
2
… (2)
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Combining (1) and (2)
 x2  x2 q
2y = pq  2  
.
p
p 
v
Q.89. Given vb = R
n

=
(-v
sin
 ) î  ( v b cos ) ĵ
vb
b
Resultant velocity of boat


= v b  vR
B
vb
= (vR - v b sin ) î + (vb cos ) ĵ
If w = width of the river, time for crossing is
W
T=
v b cos 
Drift during time T is (v R - vb sin ) T
w

Drift x = v b(n - sin )
= w(n sec  - tan  )
v b cos 
dx
For x to be minimum,
= 0 lead to  = sin-1 (1/ n)
d
Direction of boat w.r.t. stream is
900 +  = 900 + sin-1 (1/n)
For n = 1/2, the required angle = 900 + 300 = 1200
Q.90. Let u and  are projection velocity and angle of projection respectively.
vx = u cos  and vy = u sin 
u 2 sin 2 
At greatest height h =
2g
vertical velocity at half the greatest height
v2 = v 2y  2gh / 2
v2 = u2 sin2 v=
gu2 sin2  u2 sin2 

2g
2
u sin 
2
Resultant velocity at half the greatest height
vR =
v 2  (u cos )2 
given, u cos  =
tan  =
3,
2
5
 = 600
u2 sin 2 
 u 2 cos 2 
2
u2 sin 2 
 u2 cos 2 
2
vR

A
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
Q.91. (a) From the diagram v BT makes an angle of 45 with
the x-axis.
(b) Using sine rule
vB
vT

sin135 sin15
B
vT

vB
v B  2m / s
A
vBT
0
450 60
O
Q.92. Let VB,R = velocity of boat w.r.t river = 10 m/sec
VR,G = velocity of river w.r.t ground = 2 m/sec
VB,G = velocity of boat w.r.t ground.
From the figure,
width of river
Time t =
VB,R
(a) t =
`
d
VR
400 m
VB,R
400
 40 sec
10
(b) Let the drift be d, then from geometry
VR
d
 tan =
width of river
VB,R
VR
2
 400 = 80 m
 width of river =
10
VB,R
Alternatively d = (VR,G) (t) = 80 m.

Q.93. Let the 1st and 2nd shorts take time t and t
0
60
nd
respectively to collide at point P(x, y).
2 shot
st
I shot
st
x = 53 t
. . . (I)
I shot
1
10 m
y1 = 10 - gt2
. . . (ii)
P(x, y)
2
For 2nd shot
x2 = (53 cos 600)  t
. . . (iii)
1
y2 = 10 + (52 sin 600)t - gt2
. . . (iv)
2
For collision, we have x1 = x2 and y1 = y2.
From (I) and (iii), t = 2t and from (ii) & (iv), t = 1 sec, t = 2 sec.
Now (I) Time interval = 2 - 1 = 1 sec.
(ii) x1 = 53  t = 53
1
y1 = 10 - (10) (1)2 = 10 - 5 = 5
2
 Required co-ordinates = (5 3 m, 5m).

VB,G
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Q.94. Let the points towards B and reches at C
B
t1 : the time taken by the boat to reach C
AD
t1 =
CD = (v - u sin )t1
u cos 
1
500  103
t1 =
hr =
3 cos 
6 cos 
1
CD = (-3 sin  + 2)
6 cos 
1
= - 0.5 tan  +
3 cos 
t2 : time taken by the man from C to D
CD
0.5 tan 
1
1
1
t2 =
=
=
tan  +
vs
5
3 cos   5
10
15 cos 
sin 
1
=
10 cos  15 cos 
( 3 sin   2)
=
30 cos 
1
 3 sin   
7  3 sin 
total time t = t1 + t2 =

=
6 cos 
30 cos 
30 cos 
7
1
=
sec  tan
30
10
for minimum t
dt
7
1
0

sec tan  sec 2 = 0
d
30
10
1
7


sec   tan   sec   = 0
10
3

7

tan  - sec  = 0
3
7 sin   
=0 
 = sin-1 (3/7)
3 cos 
C
D
v
u

A
(i)
Q.95. Let the angle of projection be  and velocity
be u.
The velocity parallel and
perpendicular to the planes are u cos( ) and u sin ( - ). The component u sin
 -
( - ) becomes zero at the first plane




where as the component ucos ( - ) at the
second plane.
If the time required in the first case is t1 and the second is t2.
u sin(   )
Then 0 = u sin ( - ) – g cos t1  t1 =
. . . (i)
g cos 
u cos(   
and 0 = u cos ( - ) - g sin t2  t2 =
. . . . (ii)
g sin 
As the particle is mid way between the planes
1
1
u sin ( - )t1 - g cos  t12 = - {(u sin ( - ) t2 - g cos  t 22 }
2
2
(ii)
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1
g cos  ( t12  t 22 )
2
putting (i) and (ii) in (iii) we obtain
 u sin(    u cos(    

u sin ( - ) 

g sin  
 g cos 
 u sin ( - ) (t1 + t2) =
=
. . . (iii)
 u2 sin2 (      u2 cos2 (    
1


g cos  
2
2
2
2
2
 g cos    g sin  
canceling u2/g from both the sides and rearranging we get
sin2 (    sin(    cos(    sin2 (    cos2 (    cos 
=0



cos 
sin 
2 cos 
2 sin2 
 sin2 ( - ) sin2 + 2 sin ( - ) cos ( - ) sin  cos  - cos2 ( - ) cos2  = 0
dividing each side by cos2 ( - ) sin2
tan2 ( - ) + 2 tan ( - ) cot  - cot2  = 0
 2 cot   4 cot 2   4 cot 2 
=  cot   2 cot 
2
since ( - ) is an acute angle tan ( - ) is + ve
  -  = tan1(cot  (2 – 1))
 =  + tan1 {cot  (2 – 1)}.
tan ( - ) =
Q.96. Let t1 be the time taken to fall through a distance (H - h)
1
2(H  h)
 (H - h) = g t12

t1 =
2
g
Further, the time taken by the body to fall through a distance h is given by t2 =
Total time taken by the body = t = t1 + t2
2
=
Hh  h
g


dt
=0
dh
1
1

(H  h)1 / 2  h1 / 2  = 0
2
2

For the time to be maximum,
2

g 
H = 2h
1
or h/H = .
2
2h
g
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Q.97.
80
60
40
20
O
10
20
30
40
50
60
70
80
Time (s)
6
a
m/s
2
1
O
10
20
30
40
50
60
70
80
10
20
30
40
50
60
70
80
Time (s)
2
4100
900
x
(m)
400
200
O
Time (s)
y
v0
10m/s
B
45
O
Q.98.
A
38m
Let t be the instant at which the ball hits rear face AB of the trolley.
Then (v0cos45 - u0)t = 38
3m
C
D
2m
x
http://www.rpmauryascienceblog.com/
or
t=
38
Q.99. (a)
(b)
Q.101.
vQ cos
Equation of motion along radius at point Q.
mv 2
T + mg sin =

When the string slackens i.e. T = 0
mv 2
0 + mg sin =

3
vQ = g sin   10  5  = 30 = 5.48 m/s
5
Vertical velocity at point Q = vQ cos
v2 = u2 + 2gh
0 = v Q2 cos2 + 2(g)h
h=
Q.100.

38
 38
.s
v 0 cos 45  u 0 28.28 cos 45  10
At
t = 3.8 s, the y - coordinate of the ball is
1
y = (v0sin45)t - gt2 = 20t - 5t2
2
or
y = 20(3.8) - 5(3.8)2 = 3.8 m
Since 3.8m > 2m, therefore, the ball cannot hit the rear face of the trolley.
Now, we assume that the ball hits the top face BC of the trolley , and let t be that instant. Then,
y = 2 = 20t - 5 t  2
or
t 2 - 4 t + 0.4 = 0
t = 3.9 s
Let d be the distance from the point B at which the ball hits the trolley. Then,
d = (v0cos45 - u0) (t - t) = (20 - 10) (3.9 - 3.8) = 1m

90
T
mg sin
mg

5m
P
v 2Q cos 2 
30  16 25
24
=
=
= 0.96 m
2g
2  10
25
AD
BD

sin  sin 
vt
t
v b
 sin  = 1 1 sin   1  2 sin 
t 2 v1 a
v 2t 2
av1
since t1  t2 . Then sin  
bv 2
0
 sin   0.6  37    1430
AC
BC

  = 300
0
sin 
sin 45
v0
400
 v0 = 546.47 km/hr

sin105 sin 450
AB
1000
t=

= 1.83 hr.
va
546.47
A
Q

(Bus)
O
 v1

v2
B (Man)
D
3m
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N
Q.102. (a) displacement = zero
(b) distance = OA + AB + BO
3+3+3 2
B
= 6 + 3 2 cm
W
E
O
A
S
Q.103. Let t be the time when they pass one another
1
for stone B, y= v Bt + ( g)t 2
… (i)
2
1 2
for stone A, H – y =
… (ii)
gt
2
from (i) and (ii),
H = v Bt
… (iii)
Stone B can reach just one the top of tower. We can calculate
the velocity of stone B,
v 2f  v i2  2a y y
A
y
H- y
H
C
y
B
g
O
uf = 0, for ymax = H = 20 m
vi = vB
ay = -g
vB = 20 m/s
20m
from (iii) t =
 1sec .
20m / s
from equation (i), the required distance (BC) from ground = 20  1 Q.104. If u is initial velocity, a is acceleration
1
9a
 100 = u  3 + a(3)2 = 3u +
2
2
1
25
a
200 = 5u + a(5)2 = 5u +
2
2
25a
18a

 5u +
 64 
2
2
7a

u
2
Now, putting value of  v in equation (i)
7a 9a 30a
100 = 3 


2
2
2
100
2
 a=
ms
15
= 6.67 m/s2
…(i)
…(ii)
1
 10  12 = 15 m
2
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y
Q.105. (a) In figure shown final displacement

OD  5 ĵ m
Distance = OA + AB + BC + CD
= (25 + 5 2 ) m
North
A
10 m
x
O
D
10 m
52
B
C 5 m
(b) Average speed =
Average velocity =
total dis tan ce
total time
 5 ĵ
( 5  2)

25  5 2
5 
= 5 m/s.
2
m/s
(c) For average acceleration =
v f  vi
t
v f  5 î  5 ĵ, v i = 5 î
average =
 5 î  5 ĵ  5 ĵ
(5  2 )
=
 10 î  5 ĵ
5 2
m/s 2
Q.106. Final velocity at A vA = 2  t1 = 2  30 = 60 m/sec.
For AB, Let the retardation be ‘b’
v
60
 0 = vA + bt2  b = - A = = -1 m/s2
t
60
(a) total distance = OA + AB
1
1
OB = at 12  ( v A t 2  bt 22 )
2
2
1
1
= (  2  30  30  60  60  60   1 60  60)
2
2
= 900 + 3600 – 1800 = 2700 m.
(b) Maximum speed vA = 60 m/s.
(c) v 2 = 2  a  s
( v / 2)2 30  30
s= A
= 225 m.

2a
22
A
Q.107. (a) Analyzing the vertical motion
1
1
y = uyt + ( g)t 2 ,
Uy = u sin 30 = 60 
= 30 m/s
2
2
1
25 = 30 t -  10t2  5t2 - 30 t + 25 = 0
2
solving for t  t = 1 sec. & 5 sec
so minimum time = 1 sec.
(b) vertical velocity vy = uy – gt = 30 – 10  1 = 20 m/s
horizontal velocity vx = u cos 30 = 60 
3
= 30
2
B
O
3 m/s
East
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3
 1 = 30 3 m
2
u2 sin 2 60  60  sin 60
3
Range =

= 60  6 
= 180 3 m
g
10
2
(c) Horizontal distance covered = ux t = 60
Q.108. Max. range, R =

u2
g
[ R =
u2 sin 2
g
u2
 100
g
Now, max. height for a projectile is H =
H will be max. for  = 900, Hmax =
 Hmax =
u2 sin2 
2g
u2
2g
1  u2 
1
  = (100) = 50 m.
2  g 
2
Q.109. (a)
S = 0 + ½ gt2
 St
h
S
O
(b)
2
KE = ½ mv  KE  v
v= 0 + gt
 KE  t2
T
2
mgh
KE
O
(c) TE = constant always
T
TE
mgh
O
v 20 sin 2
g
Rg
= 45o.
v 02 
 Rg as
sin 2
v o = Rg
. . . (1)
T


Equation of trajectory
y
vo
Q.110.  R(range) =
45o
1.22 m
x
A
B
106.68 m
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gx 2
gx 2
=
x
using (1)
1
2v 20 cos2 45o
2Rg
2
Putting x = 97.53, we get
Y = xtan45o -
2
10  97.53 
 8.35
106.68  10
Hence height of the ball from the ground level is h= 8.35 + 1.22 = 9.577m.
as height of the wall is 7.31m so the ball will clear the wall.
y = 97.53 -
Q.111. Let (x,y) be the coordinates of point C.
C
Now x component of point C=OA+AD
y
55
10
10  4y
x 
 y cot 370 
(1)
3
3
370

As point C lies on the trajectory of a parabola we have O
D
A
gx 2
y  x tan   2
(2)
24 cos2 
1
2
Given that tan   ,cos  
2
5
From (1) & (2) we get y=1.25 m
Hence x=5 m
Hence the coordinates of point C are (5m,1.25m)
(b) Let vy be the vertical component of velocity of the particle just before collision at C. from
V=u+at
 x 
v y  u sin   g 

 u cos  
1
5
vy  5 5 
 10 
0
5
5 5 cos 
2
Thus at C the particle has only horizontal component of velocity, v x  u cos   5 5 
 10m / s
5
Q.112. Instantaneous speed of the car = 30 m/s = V
Radius = 500 m
dV
= 2 m/s2 = at (Tangential acceleration)
dt
Normal or radial acceleration or centripetal acceleration
v 2 (30)2 9

 m/s 2 = an
r
500 5

1/ 2
 Total acceleration a an2  at 2
=

 2  9 2 
= (2)    
 5  


81 
25 
1/ 2

1/ 2
181
1/ 2
= 
= 2.7 m/s 2


 25 
Let direction of this acceleration make an angle  with the velocity then
= 4 
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tan =
an 9 / 5 9


at
2
10
 9 

 10 
  = tan–1 

Q.113. Let u  ux î  u y ĵ
At the instant of impact (x)stone = (x0) plank + (x) plank
uxt = x0 + (1/2) at2

uxt = 3 + 0.75 t2
. . . (i)
In y-direction
uyt – (1/2) gt2 = y0
 uyt – 5t2 = 1.25
. . . (ii)
also (vy/ux) = tan (-450)
{(uy – gt)/ux} = -1
. . .. (iii)
solving (I), (ii) and (iii), we get
t = 1 s, ux = 3.75 m/s, uy = 6.25 m/s
 u = 7.28 m/s and 0 = tan1 (5/3).
T
Q.114.
FBD of m;
and T – mg = ma
…(1)
and 2mg – T = 2ma
…(2)
mg
T
FBD of 2m;
2mg
Adding (1) and (2) mg = 3ma
 a = g/3
Hence velocity of ‘m’ after moving up 6.54 m is;
 9.81 
 (6.54) = 2(3.27) (6.54)
 3 
V2 = 2 (g/3) (6.54) = 2 
 V = 6.54 m/s upwards
Velocity of ‘2m’ at that instant = 6.54 m/s downwards.
When string is cut ‘m’ falls to the ground from a height of 13.08 + 6.54 = 19.62 m
 19.62 = – 6.54t + 1/2 (9.81) t2 where ‘t’ is the time taken to reach the ground
t = 2.7 seconds
‘2m’ falls a distance of 13.08 – 6.54 = 6.54 m
t = 0.6 seconds.
dv
Q.115. Tangent acceleration at =
= 2t +1
dt
v2
normal acceleration an =
R
 (at)t=0 = 1 m/s2
2
v
1
(an)t=0 = 0 
= 5 m/s 2
R
0.2
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(b) v = R
d
dt
'
R d = (1+t+t2 )dt
 R d 

0
2
 1  t  t dt
2
0
 = 33.3 rad
Q.116. Since acceleration varies linearly so
a = t m/s2 by given condition
dv
or,
t
dt
t2
v=
m/s
2
ds t 2
t3
or
or, s =

dt
2
6
at the end of t = 6 sec. Acceleration becomes zero.
Distance moved by car at t = 6 sec is
666
S1 =
= 36 m
6
66
Speed of the car =
= 18 m/s
2
Remaining distance = 72 – 36 = 36 m.
so time taken to cover this distance
36
= t2 =
sec .  2 sec .
18
Total time = 6+2 = 8 sec.
Q.117. v0 – gt = 0
 t = v 0/g
v = 0 + a0t = a0v0/g
1
a v2
1 v2
v2
x = a0t2 = 0 0 ; y = v 0t - g 02 = 0
2
2g
2 g
2g
yes it will retrace its path, path is not parabolic.
Q.118.  Let us take x – y axes as shown in the figure
The magnitude of pseudo force acting on the
particle has a magnitude of ma0 and its direction
will be towards right as shown in the free body
diagram
The components of the acceleration of
the particle are
ma o cos 37   mg sin 37 
ax =
m
4
3
= 5   10   2m / s 2

5
X
Y
P
u = 5.5 m/s
0
37
O
0
37
Y
X
37
37
0
ma0
0
mg
F.B.D. of the particle
2
a0 = 5 m/s
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 (mg cos 37 o  ma o sin 37 o )
m
4
3

=  10   5    -11m/s2
5
5

4
 ux = ucos37° = 5.5 
= 4.4 m/s.
5
3
uy = usin37° = 5.5 
= 3.3 m/s
5
 Displacement of the particle along y – axis
1
y = uy t + ayt2
2
1
 y = 3.3t  11t2
2
When the particle strikes the plane y = 0
2u y 2  3.3
t=
=
= 0.6 sec.
11
 ay
ay =
 OP = x = uxt +
= 4.4  0.6 -
1 2
axt
2
1
 2  (0.6)2 = 2.28 m.
2
Q.119. (a)
We observe the motion of projectile fixing Y-axis with OP and x-axis with OQ. Hence;
velocity at any instant t along x-axis:
vx =10 3 - (g sin 600)t
vy = 0 – (g cos 600)t
As v x = 0 at the time of hitting;
Time of flight = T = 2 sec.
Displacement OP during this time
1
1
1
= (g cos 600)t2 =
 10 
 4 = 10 m
2
2
2
1
Hence; h = OP sin 300 = 10 
= 5m
2
1
3
(b) Similarly displacement OQ = (10 3 )(2)  10 
 4= 10 3 m
2
2
Hence PQ =
OP 2  OQ 2 = 20 m.
Q.120. N sin  = m2 R sin  … (i)
and N cos  = mg
….(ii)
from (i) and (ii)
 g 
 = cos-1  2 
  R
g can not be greater than 2R.
If 2 R > g then there are two equilibrium position.
O

O’
N
A
mg
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 g 
1 = 0 and 2 = cos-1  2 
  R
Q.121. For the part (a) choose the co-ordinate axes parallel to the plane and perpendicular to the inclined
plane and for the part (b) choose the horizontal axis and vertical axis as co-ordinate axes.

X
(a)
ab   g sin  î
Y

v p / b( t  0 )  u cos  î  u sin  ĵ


a P  g sin  î  g cos  ĵ  a P / b  g cos  ĵ

1
RP / b ( t )  u cos t î  (u sin t  g cos t 2 ) ĵ
2
1
At Q, u sin t  g cos  t2 = 0
2
u2 sin 2
2u sin 
t=
;  PQ =
.
g cos 
g cos 

(b) Let v b / G   v cos  î  v sin ĵ

v p / b  u cos(  )î  u sin(  ) ĵ
Y

 v p / G  [u cos(     v cos ] î  [u sin(  )  v sin ] ĵ
For x to be 0
u cos ( + ) = v cos 
u cos(  )
v=
.
cos 
Q.122. Let they meet after time t, and let the speed with the stone is thrown up be u. then
u2
h=
 u = 2gh = 2  10  20 = 20 m/s
2g
1
1
Now 20 t - gt 2  gt 2  20
2
2

t = 1s.
Hence they pass other after 1s
speed of the stone projected up = 20 – 10  1 = 10 m/s
speed of the stone dropped = 0 + 10  1 = 10 m/s.
Q.123. Let the second rocket overtakes the first t second. after it is projected
 When they meet their displacement are equal
1
1

(8g)t 2  (6g)(t  1)2
2
2
2
 t 1 
 8 4
 
 t 
6 3


t 1
2


t
3

t
3

t 1
2
X
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 t=
 3 /2
3
2
1
 t=
3

 3
2 3
sec .,
 3
2 3
2 3
1.732
=
 6.46 sec.
0.268
sec ., (not possible)
 2n  1
Q.124. Before collision number of revolution completed by the particle is odd multiple of half or 

 2 
where n is an integer.
2R
Period of revolution of the particle =
= 4 seconds.
v
2n  1
Time of flight of the shell =
( 4) = (4n - 2) second
2
u 2 sin 2
Range is
= 20 3
g
Hence tan  =
( 2n  1)2
3
For  to be smallest; n = 1 ; therefore  = 300 hence u = 20 m/sec.

displaceme nt
x î  yĵ
Q.125. (a) v av 

v 0 sin 
time
2g
x = half of range =
y = Max. height =
v av
y
v 20 sin 2
2g
v 20
2
sin 
2g
y ĵ
x
î
v 20 sin 2
v 2 sin2 
î  0
ĵ
2g
2g
v sin 2
v sin2 

= 0
î  0
ĵ
v 0 sin 
sin 
sin 
2g
= 2v0 cos  î + v0 sin  ĵ

(b) Let the perpendicular velocity be v
v  v = 0  v  v = (v0 cos  î + v0 sin  ĵ ). (v 0 cos  î + (v0 sin  - gt) ĵ
 v 20 cos2   v 0 sin ( v 0 sin   gt )  0
v 20  v 0 sin gt  0
v0
 t=
g sin 
(c) radius of curvature =
v 2 v 20 cos 2 

g
g
x
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Q.126. Velocity of projection of stone = v0
Angle of projection = 
Velocity of bird = u
u
Required ratio =
v 0 cos 
Bird is hit at Q.
1
For the stone; h = v0 sin . t - gt2
2
t=
Y
Q
P
v0
 h
O
X
2v 0 sin   4v 20 sin 2   8gh
2g
Maximum height of stone =
v 20 sin 2 
= 2h
2g
Therefoere ; v0 sin  = 2 gh
Hence time of flight for both stone and bird= t = (2 +
Horizontal displacement of bird = ut = u(2 +
2)
2)
h
g
h
g
But, for the stone ; displacement PQ needs a time = t2 – t1 = 2 2
Therefore, for the projectile; PQ = (v0 cos ) (2 2 )

u
2 2


v 0 cos  2  2
2
2 1
h
g
h
h
= u (2+ 2 )
for bird
g
g