# FRANK Solutions for Class 9 Physics Chapter 3 - Laws of Motion

Practise TopperLearning’s Frank Solutions for ICSE Class 9 Physics Chapter 3 Laws of Motion for revision. Understand Newton’s Laws of Motion with the detailed answers for important textbook questions. These expert answers will also take you through concepts like inertia, acceleration, momentum, velocity, and more.

In addition, study the ICSE Class 9 Physics Frank textbook solutions to revise MCQs on gravitational force. At our study portal, explore additional online resources for Physics learning such as video lessons, revision notes, sample papers, etc. for exam preparation.

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## Chapter 3 - Laws of Motion Exercise 113

Solution 1
The property by which a body neither changes its present state of rest or of uniform motion in a straight line nor tends to change the present state,is known as inertia.
Solution 2
A book lying on a table will remain placed at table unless it is displaced by some external force. This is an example of inertia of rest.
A ball rolling on the ground  will continue to roll unless the external force , the force of friction between the ball and the ground stops it.
Solution 3
The greater is the MASS , the greater is the inertia of the object.
Solution 4
An object possess two kind of inertia, inertia of rest and inertia of motion.A book lying on a table will remain placed at table unless it is displaced by some external force. This is an example of inertia of rest.
A ball rolling on the ground  will continue to roll unless the external force , the force of friction between the ball and the ground stops it.
Solution 5
1 Newton is the force which when applied to a body of unit mass   produces a unit acceleration in it. 1 newton would produce acceleration of   1 ms-2 in a mass of 1 kg.
Solution 6
The acceleration produced bya force in an object is directly proportional to the applied FORCE and inversely proportional to the MASS of the object.
Solution 7
SI unit of force is Newton (N).
Solution 8
Acceleration is the physical quantity associated with N kg-1.
Solution 9
1 N = 105 Dyne.
Solution 10
As mass of loaded van is greater than sports car so it would require more force to stop.
Solution 11
We know force = mass X acceleration.
a= f/m = 12 N / 4 kg. = 3 ms-2
so acceleration of the body would be 3 ms-2.
Solution 12
SI unit of force is Newton whereas CGS unit of force is dyne.
1 newton / 1dyne = 105.
Solution 13
SI unit of momentum is kgms-1.
Solution 14
Momentum is defined as the amount of motion contained in the body. It is given by the product of the mass of the body and its velocity.
Solution 15
Momentum is the physical quantity associated with the motion of the body.
Solution 16
Momentum is possessed by bodies in MOTION.
Solution 17
A fast pitched soft ball has more momentum.
Solution 18
SI unit of momentum is kgms-1 and CGS unit of momentum is g cms-1.
And their ratio is = 1000 X 100 g ms-1= 1:10.
Solution 19
A body at rest has zero momentum as its velocity is zero.
Solution 20
According to Newton's third law, for every action there is always an equal and opposite reaction.
Solution 21
When a force acts on a body then this is called an action.
Solution 22
No, action and reaction never act on a same body they always act simultaneously on two different bodies.
Solution 23
2nd law of motion gives the definition of force.
Solution 24
Newton's third law explains this statement.
Solution 25
Force is a vector quantity.
Solution 26
This means these forces are balanced forces.
Solution 27
Passengers tend to fall sideways when the bus takes a sharp turn due to the inertiaof direction.
Solution 28
Passengers are thrown in the forward direction as the  running bus stops suddenly because due to their inertia of motion, their upper body continues to be in the state of motion even though the lowerbody comes to rest when the bus stops.
Solution 29
Passengers tends to fall in backward direction when bus starts suddenly because due to their inertia of rest, as soon as the bus starts, their lower body comes in motion but the upper body continues to be in the state of rest.
Solution 30
No, internal forces cannot change the velocity of a body.
Solution 31
When a hanging carpet is beaten using a stick, the dust particles will start coming out of the carpet because the part of the carpet where the stick strikes, immediately comes in motion while the dust particle sticking to the carpet remains at rest . Hence a part of the carpet moves ahead alongwith the stick, and the dust particles fall down due to the earth's pull.
Solution 32
When we shake the branches of a tree, the fruits and leaves remain in state of rest while branches comes in rest so fruits and leaves are detached from the tree.
Solution 33
We know force = mass X acceleration
F1 = 10 X 5 = 50 dyne.
F2 = 20 X 2 = 40 dyne.
So first body require more force.
Solution 34 ## Chapter 3 - Laws of Motion Exercise 114

Solution 35
initial velocity of the object = 0 ms-1
Acceleration of the object = 8 ms-2.
Time = 5 s.
Distance covered would be S = ut + 1/2 at2.
S = 1/2 X 8 X 5 X5 = 100 m.
Solution 36
Initial velocity of the truck = 0 ms-1
Distance covered by truck = 100 m
Time taken to cover this distance = 10 s.
We know Distance covered would be S = ut + 1/2 at2.
100 =1/2 Xa X100
a= 2 ms-2.
Mass of truck = 5 metric tons = 5000 kg.
Force acted on truck = mass X acceleration
Force = 5000 X 2 = 10000 N.
Solution 37
Momentum is used for quantifying the motion of body.
Solution 38
When we fire a gun, a force is exerted in the forward in the forward direction as the bullets comes out; in reaction to which an equal and opposite force is act in the backward direction and hence, we feel a backward jerk on the shoulder.
Solution 39
A person applies force on water in backward direction and water according to third law of motion water apply an equal and opposite force in forward direction which helps a person to swim.
Solution 40
Newton's third law of motion is involved in the working of a jet plane.
Solution 41
Yes, a rocket can propel itself in a vacuum once it is given initial velocity.
Solution 42
Action is equal and opposite to reaction but they act on different bodies and object moves as movement requires an unbalanced force and these are provided once inertia is overcome.

## Chapter 3 - Laws of Motion Exercise 125

Solution 1
Sir Isaac Newton stated the law of gravitation.
Solution 2
Every object in the universe attracts every other object with a force directly proportional to the product of their masses  and inversely proportional to the square of distance between them.
Solution 3
Gravity is the force of attraction between the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.
Solution 4
Acceleration due to gravity is the acceleration experienced by a body during free fall.
Solution 5
g = GM/R2.
Solution 6
We know that law of gravitation.
F = G ( m1 X m2)/R2.
Here G is universal constant and is called constant of gravitation. It doesnot depend upon on the value of m1,m2 or R.
Its value is same between any two objects in the universe.
Solution 7
SI unit of constant of gravitation is Nm2kg-2.
Solution 8
we know that law of gravitation.
F =  G ( m1 X m2)/R2.
(a) If distance between them is halved then put R = R/2.
F = 4 X G( m1 X m2)/ R2.
F1 = 4 F.
(b) If distance between them is doubled then put R = 2R.
F = G( m1 X m2)/ 4R2.
F1 = F/4.
(c) If distance between them is made four times then put R = 4R.
F = G( m1 X m2)/16 R2.
F1 = F/16.
(d) If distance between them is infinite then put R = infinite.
F = G( m1 X m2)/ R2.
F1 = 0.
(e) If distance between them is almost zero then put R = 0.
F = G( m1 X m2)/ 0.
F1 = infinite.
Solution 9
All objects in the universe attract each other along the line joining their CENTRES.
Solution 10
The force of attraction between any two material objects is called FORCE OF GRAVITATION.
Solution 11
The gravitational force of the earth is called earth's  GRAVITY.
Solution 12
The Gravity  is a particular case of GRAVITATIONAL FORCE OF EARTH.
Solution 13
The  value of G is extremely  SMALL.
Solution 14
Yes the law of gravitation is also applicable in case of the sun and moon.
Solution 15
we know that law of gravitation.
F =  G ( m1Xm2)/R2.
Mass of earth = 6X1024kg.
Mass of the person = 100 kg.
G = 6.7 X10-11 Nm2kg-2.
Radius of earth = 6.4 X 1014.
F =  (6.7 X10-11X 100 X 6 X1014 )/ (6.4 X6.4 X1012) = 981.4N
Force of gravity due to earth acting on a 100 kg person is 981.4 N.
Solution 16
Objects fall towards the earth due to force of gravitation.
Solution 17
Because the masses of persons are not large enough to overcome the value of small constant of gravitation so the force of gravitation is very small and negligible to feel.
Solution 18
Initial speed of ball is = 4.9 ms-1.
Acceleration due to gravity = -9.8 ms-2.
(a) We know v2 - u2 =2as
At highest point final velocity is zero so
0 - 4.9 X 4.9 = 2 X (-9.8) S
S = 1.125 m
(b) We know v = u + at
0 = 4.9 - 9.8 t
T = 0.5 sec.
(c) for highest point initial velocity is zero
Acceleration due to gravity is = 9.8 ms-2.
Final velocity at ground is v
V2 - 0 = 2 X9.8 X 1.125
V = 4.9 ms-1.
Time taken to reach ground from highest point
V = u + at
4.9 = 0 + 9.8 t
T = 4.9/9.8 = 0.5 sec.
So time of ascent is equal to time descent.
Solution 19
g = GM/R2.
Solution 20
Value of the g at the surface of the earth is 9.8ms-2.
Solution 21
Mass of the body is constant at all positions so mass will not  change. But weight will change as gravity on the surface of earth is almost 6 times than on the surface of the moon, so its weight will increase almost 6 times on the surface of earth.
Solution 22
We will weigh more on the surface of the earth.
Solution 23
Beam balance is used to measure the mass of a body.
Solution 24
Spring scale is used to measure the weight of a body.
Solution 25
The weight is greater at the poles than the equator.
Solution 26
Newton 1N  = 9.8 kgwt.
Solution 27
We will weigh more on earth surface as value of g is greater on earth surface.
Solution 28
No, the force of gravitation between two objects does not depend on the medium between them.
Solution 29
we know that law of gravitation.
F = G ( m1Xm2)/R2.
Now m1 = 2 m1
m2 = 2 m2
R = 2 R
F1 = G ( 2m1 X2 m2)/4R2.
F1 = F
So force between them remains same.

## Chapter 3 - Laws of Motion Exercise 126

Solution 30
Yes, in absence of gravity all freely falling body have same force acting on them.
Solution 31
g= GM/R2
it means acceleration due to gravity is directly proportional to the mass of body and inversely proportional to the square of distance between earth and object.
Solution 32
Yes a body falling freely near the earth surface has a constant acceleration.
Solution 33
As we know
g= 1/R2
so value of g is more at poles than equator so value of g is maximum near a camp site in Antarctica as this lie on the pole.
Solution 34 ## Chapter 3 - Laws of Motion Exercise 128

Solution 1
Force is that external agency which tends to change the state of rest or the state of motion of a body.
Solution 2
1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it.
Solution 3
Newton is the SI unit of force whereas  dyne is the CGS unit of force.
1 N = 105 dyne.
Solution 4
No, force is a vector quantity.
Solution 5
A force can produce MOTION in an objectat rest. It can ACCELERATE an object and can change its DIRECTION of motion.
Solution 6
(a) force changes  the shape of skin.
(b) force produces  stretching in the rubber.
(c) force provides retardation to the car and finally stops the car.
(d) force decreases  the momentum of ball and finally stops the ball.
Solution 7
No, every force does not produce motion in every type of body.
Solution 8
The amount of inertia of a body depends on its MASS.
Solution 9
You can change the direction in which an object is moving  by APPLYING FORCE ON IT.
Solution 10
A man riding on a car has INERTIA of motion.
Solution 11
When a body is at rest , it will continue to remain at rest unless some external force is applied to change its state of rest. This property of body is called inertia of rest.
Solution 12
(i) Weight of the book is action and normal force applied by table on book is reaction.
(ii) Force applied by man on ground is action and force of friction is the reaction.
(iii) Force applied by hammer on nail is action and normal force applied by nail on hammer is reaction to this force.
(iv) Firing of bullet is the action and recoiling of gun is the reaction.
(v) Force applied by us on wall is action and opposite force applied by wall on us or we can say that resistance of wall to our force is reaction.
Solution 13
A book lying on a table will remain placed on  table unless it is displaced by some external force. This is an example of inertia of rest.
A ball rolling on the ground  will continue to roll unless the external force , the force of friction between the ball and the ground stops it.This is an example of inertia of motion.
Solution 14
Unbalance external force causes motion in the body.
Solution 15
Linear Momentum is defined as the amount of motion contained in a body. It is given by the product of the mass of the body and its velocity.
Solution 16
SI unit of momentum is kgms-1.
Solution 17
According to Newton's first law force is that external agency which tends to change the state of rest or the state of motion of a body.
Solution 18
According to Newton's first law, everybody continues in its state of rest or in uniform motion in a straight line unless compelled by some external force to act otherwise.

## Chapter 3 - Laws of Motion Exercise 129

Solution 19
Out of all these, as mass of truck is greatest and mass is measure of inertia so a truck has maximum inertia.
Solution 20
It is advantageous to run before taking a long jump because after running we get motion of inertia which helps in long jumping.
Solution 21
Ball moving on a table top stops eventually due to force of friction between the ball surface and table surface.
Solution 22
Force is equal to the rate of change of linear momentum.
Solution 23
According to newton second law of motion, when a force acts on a body, the rate of change in momentum of a body is equal to the product of mass of the body and acceleration produced in it.
Yes, Newton's first law is contained in the second law as if force is zero then acceleration would be zero which means body would remain in its state of rest or in state of constant motion.
Solution 24
1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it.
1 newton / 1dyne = 105.
Solution 25
1 newton = 1 kg X 1 ms-1 = 1000 g X 100 cms-1 = 105 cms-1.
1 dyne 1 g X 1 cms-1 = 1cms-1.
So 1 newton = 105 dyne.
Solution 26
No, the body will not move as the two forces are equal and opposite and they constitute balanced forces.
Solution 27
As these forces are balanced so they will not affect the motion and motion of the body will remain unaffected.
Solution 28
According to Newton's third law, for every action there is always an equal and opposite reaction. Rocket works on the same principle. The exhaust gases produced as the result of the combustion of the fuel are forced out at one end of the rocket. As a reaction , the main rocket moves in the opposite direction.
Solution 29
According to Newton's third law, every action has equal and opposite reaction so force exerted by the wall on the boy is 30 N.
Solution 30
Newton stated the law of inertia.
Solution 31
Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.
Law of gravitation is called universal because it applies to all bodies of universe.
Solution 32
Gravity is the force of attraction betwen the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.
Solution 33
Person will weigh more at Delhi as we know that gravity decreases with increase in height. Now as Shimla is at a height from Delhi so weight is less in Shimla and more in Delhi.
Solution 34
Spring scale is used to measure the weight of a body.
Solution 35
Gravity is another kind of FORCE. It exerts all through the UNIVERSE. The sun's gravity keeps  the PLANETS in their orbits. Gravity can only be felt with very large MASS.
Solution 36
(i) Objects fall on the earth due to gravitational force between the earth and object.
(ii) Atmosphere doesnot escape because molecules of atmosphere are attracted by earth due to gravitational force of earth.
(iii) A moon rocket needs to reach a certain velocity because during its motion earth attracts the rocket towards it by its gravitational force.
Solution 37
'g' is acceleration due to earth's gravity and 'G' is universal gravitational constant.
Solution 38
Free fall means motion of a body under the gravity of earth only.
Solution 39
Yes, we have a gravitational force of attraction between us and a book. But our mass is very small so the force between us and book is very small almost negligible.
Solution 40
Yes, the force of gravitation of earth affects the motion of moon, because moon is revolving around earth and centripetal force for this revolution is provided by earth's gravitation.
Solution 41
Inertial mass is measure of inertia of the object. According to second law of motion F = m X a
m= F/a and this mass is called as inertial mass.
Newton law of gravitation gives another definition of mass.
F = (G m1m2)/R2
Thus m2 is the mass of the body by which another body of mass m1 attracts it towards it by law of gravitation. This mass is called gravitational mass.
Solution 42
Newton law of gravitation is that Every object in the universe attracts every other object with a force directly proportional to the product of their masses  and inversely proportional to the square of distance between them.
(i) Gravity is the force of attraction between the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.

(ii) 'g' is acceleration due to earth's gravity and 'G' is universal gravitational constant.
Solution 43
Yes, it is true that apple attracts the earth towards it with same force but the mass of earth is so huge that acceleration produced in it due to this force is very much small and negligible to notice.
Solution 45
(i) The force exerted by the block on is the weight of box and that is equal to 20N.
(ii) The force exerted by string on block is equal to the tension in the string and this is also equal to the 20N.
Solution 46
we know F = m X a
m= F/a
so we can calculate mass of each body
Mass of body 1 m1 = 4/8 = 0.5 kg.
Mass of body 2  m2  = 4/20  = 0.2 kg.
Total mass when two masses are tied together  M = 0.5 + 0.2 = 0.7 kg.
Now as force is acting on total mass so acceleration produced is
a=  4/0.7 = 5.71 ms-2.

Solution
Solution
Solution

## Chapter 3 - Laws of Motion Exercise 130

Solution 47 Solution 48
Initial speed of body = 5 ms-1
Final speed of body = 8 ms-1
Time taken to acquire this speed = 2 s.
Acceleration of body = ( v- u)/t
a= (8- 5)/2 = 1.5 ms-2.
Force applied on body = 0.9 N.
we know F = m X a.
m = f/a = 0.9/1.5 = 0.6 kg
mass of the body is 600 gm.
Solution 49 Solution 50
The force that acts on a body for a very short time but produces a large change in its momentum, is known as impulsive force.
Solution 51
initial velocity of body = 0 ms-1.
Final velocity of body = 100 ms-1.
Mass of body = 20 kg.
Force applied = 100N.
We know that
F X t = m (v - u)
100 t = 20 (100 -0)
T = 2000/100 = 20 s.
Solution 52
SI unit of retardation is ms-2.
Solution 53
Force applied is equal to the product of mass and acceleration produced in the body.
F = mass X acceleration.
Solution 54
According to Newton's second law of motion, when a force acts on a body, the rate of change in momentum  of a body equals the product of mass of the body and acceleration produced in it due to that force, provided the mass remains constant.
Mass of  body = 400 g = 0.4kg
Force applied on body = 0.02 N
Acceleration = force/mass = 0.02/0.4 = 0.05 ms-2.
Solution 55
Linear Momentum is defined as the physical quantity which is associated with bodies in linear motion. It is given by the product of the mass of the body and its velocity.
Mass of body = 1 kg
Acceleration produced = 10 ms-2.
Force applied would be = 1 X 10 N = 10 N.
Mass of second body = 4 kg.
As same force has to be applied on second body so force = 10N.
Acceleration produced is = F/M =10/4 = 2.5 ms-2.
Solution 56
Mass of P is m1= m.
Velocity of P is v1 =2 v
Mass of Q is m2 = 2m
Velocity of Q is v2 = v.
(i) inertia of P/inertia of Q = m1/m2 = 1/2.
So ratio of inertia of two bodies is 1:2.
(ii) Momentum of P/momentum of Q = m1v1/m2v2 =  1
So ratio of momentum of two bodies is 1:1.
(iii) As force required to stop them is equal to change in their momentum from moving to rest.
So ratio would be same as the ratio of their momentum i.e 1: 1.
Solution 57
According to newton second law
F = m X a
a= (v - u)/t.
F = m(v -u)/t
F = (mv - mu)/t
As F= m X a
ma = (mv - mu)/t
so rate of change of momentum = mass X acceleration.
This relation holds good when mass remains constant during motion.
Solution 58 Solution 59
According to newton third law, for every action there is always an equal and opposite reaction.
To demonstrate newton third law blow a balloon and hold its neck tightly facing downwards. When we release the balloon, the balloon will moves up instead of falling  to the ground. As air is releasing from bottom of balloon and this air apply equal and opposite force to the balloon and this force helps balloon to move upwards.
Solution 60
time for which force is applied = 0.1 s.
Mass of body = 2 kg
Initial velocity of body = 0 ms-1
Final velocity of body = 2 ms-1.
We know FX t = m (v - u)
F X 0.1 = 2 (2 - 0)
F = 4 /0.1 = 40 N.
Solution 61
mass of ball = 500g = 0.5 kg.
Initial speed of the ball = 30 ms-1
Final speed of ball = 0 ms-1
Time taken by player to stop the ball = 0.03 s.
We know FX t = m (v - u)
F X 0.03 = 0.5 (0 - 30)
F = - 1.5 / 0.03 = - 500 N
(- ) sign shows that player has to apply force in opposite direction of the motion of the ball.
Solution 62
Time for which force is applied =0.1 s.
Mass of the body = 3.2 kg.
Initial speed of body = 0 ms-1
After removal of forces body covers a distance of 3m in 1 second so final speed of body = 3/1 = 3ms-1.
We know FX t = m (v - u)
F X 0.1 = 3.2 (3 -0)
F = 9.6/0.1 = 96 N.
So applied force is 96 N.
Solution 63 ## Chapter 3 - Laws of Motion Exercise 131

Solution 64
Time for which force is applied =3 s.
Mass of the body = 2 kg.
Initial speed of body = 0 ms-1
Force applied = 10 N.
(i) We know F X t = m (v - u)
10 X3 = 2 (v- 0)
v = 15 ms-1.
Final velocity is 15 ms-1.
(ii) As m(v - u) is change in momentum and this is equal to the F X t so change in momentum is equal to the 30 kgms-1.
Solution 65
(i) We always prefer to land on sand instead of hard floor while taking a high jump because sand increases the time of contact.As FX t = m ( v - u ) and our change in momentum is constant so if time increases then force experienced would decrease.
(ii) Again while catching a fast moving ball, we always pull our hands backwards to increase reaction time so force experienced would decrease.
Solution 66
Height of cliff = 98 m.
Initial velocity of stone = 0 ms-1.
Acceleration due to gravity = 9.8 ms-2.

(i) We know H = ut + 1/2 gt2.
98 = 1/2 X 9.8X t2.
t2 = 98X2/9.8 = 20
t= 4.47 sec.
(ii) Final velocity when it strikes the ground
V2 - u2 = 2 g H
V2 = 2X9.8X98
V2= 1920
V= 44.6 ms-1.
Solution 67
Initial speed of ball is = 9.8 ms-1.
Acceleration due to gravity = -9.8 ms-2.
Final speed at maximum height = 0 ms-1.
We know v = u + at
0 = 9.8 - 9.8 t
T = 1 sec.
We know v2 - u2 =2as
At highest point final velocity is zero so
0 - 9.8 X 9.8 = 2 X (-9.8) S
S = 4.9 m.
for highest point initial velocity is zero
Acceleration due to gravity is = 9.8 ms-2.
Final velocity at ground is v
V2 - 0 = 2 X9.8 X 4.9
V = 9.8 ms-1.
Time taken to reach ground from highest point
V = u + at
9.8 = 0 + 9.8 t
T = 9.8/9.8 = 1 sec.
Total time = time of ascent + time of descent.
Total of flight = 1+ 1 = 2 seconds.
Solution 68
Initial speed of ball = 10 ms-1.
Acceleration due to gravity on ball = - 9.8 ms-2
We know that from first equation of motion
v= u + gt.
After 1 sec
v= 10 - 9.8 X1
v= 0.2 ms-1
so velocity after 1 sec would be 0.2 ms-1.

Velocity after 2 seconds
v= 10 - 9.8X2 = 10 - 19.6 = -9.6 ms-1.
Here negative sign shows that velocity is in downward direction and magnitude is 9.6 ms-1.
Solution 69
Maximum Height attained by ball = 19.6 m
Let initial speed of ball = u ms-1.
Acceleration applied on ball due to gravity = -9.8 ms-2.
Final speed of ball at maximum height = 0 ms-1.
We know that from second equation of motion
V2 - u2 = 2as
0 -u2 = 2 X(-9.8)X19.6
u2 = 19.6 X 19.6
u= 19.6 ms-1
so initial speed of ball to attain maximum height of 19.6 m should be 19.6ms-1.
Solution 70
Height of tower = 98 m
Acceleration due to gravity on stone = 9.8 ms-2.
Initial speed of ball= 0 ms-1.
Let initial speed of second stone is v ms-1.
We know from second equation of motion
S = ut + 1/2 a Xt2.
98 = 0 + 1/2 X9.8Xt2.
t2 = 20
t= 4.47 sec.
As second stone is thrown 1 sec later so time taken by second body to cover distance of 98 m is = 4.47 - 1 = 3.47sec.
So again put t= 3.47 sec and S = 98 m in second equation of motion we get
98 = vX3.47 + 1/2 X9.8X3.47X3.47.
98 = 3.47Xv + 59
3.47X v = 98 - 59
v= 39/3.47 = 11.23 ms-1.
Initial speed of second stone should be 11.23 ms-1.

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