# 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.

## Chapter 3 - Laws of Motion Exercise 113

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.

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.

^{-2}in a mass of 1 kg.

^{-1}.

^{5}Dyne.

a= f/m = 12 N / 4 kg. = 3 ms

^{-2}

so acceleration of the body would be 3 ms

^{-2}.

1 newton / 1dyne = 10

^{5}.

^{-1}.

^{-1}and CGS unit of momentum is g cms

^{-1}.

And their ratio is = 1000 X 100 g ms

^{-1}=

^{ }1:10.

F

_{1}= 10 X 5 = 50 dyne.

F

_{2}= 20 X 2 = 40 dyne.

So first body require more force.

## Chapter 3 - Laws of Motion Exercise 114

^{-1}

Acceleration of the object = 8 ms

^{-2}.

Time = 5 s.

Distance covered would be S = ut + 1/2 at

^{2}.

S = 1/2 X 8 X 5 X5 = 100 m.

^{-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 at

^{2}.

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.

## Chapter 3 - Laws of Motion Exercise 125

^{2}.

F = G ( m

_{1 }X

_{ }m

_{2})/R

^{2}.

Here G is universal constant and is called constant of gravitation. It doesnot depend upon on the value of m

_{1},m

_{2}or R.

Its value is same between any two objects in the universe.

^{2}kg

^{-2}.

F = G ( m

_{1}X m

_{2})/R

^{2}.

(a) If distance between them is halved then put R = R/2.

F = 4 X G( m

_{1}X m

_{2})/ R

^{2}.

F

_{1}= 4 F.

(b) If distance between them is doubled then put R = 2R.

F = G( m

_{1}X m

_{2})/ 4R

^{2}.

F

_{1}= F/4.

(c) If distance between them is made four times then put R = 4R.

F = G( m1 X m2)/16 R

^{2}.

F

_{1}= F/16.

(d) If distance between them is infinite then put R = infinite.

F = G( m

_{1}X m

_{2})/ R

^{2}.

F

_{1}= 0.

(e) If distance between them is almost zero then put R = 0.

F = G( m

_{1}X m

_{2})/ 0.

F

_{1}= infinite.

F = G ( m

_{1}Xm

_{2})/R

^{2}.

Mass of earth = 6X10

^{24}kg.

Mass of the person = 100 kg.

G = 6.7 X10

^{-11}Nm

^{2}kg

^{-2}.

Radius of earth = 6.4 X 10

^{14}.

F = (6.7 X10

^{-11}X 100 X 6 X10

^{14})/ (6.4 X6.4 X10

^{12}) = 981.4N

Force of gravity due to earth acting on a 100 kg person is 981.4 N.

^{-1}.

Acceleration due to gravity = -9.8 ms

^{-2}.

(a) We know v

^{2}- u

^{2}=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

V

_{2}- 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.

^{2}.

^{-2}.

F = G ( m

_{1}Xm

_{2})/R

^{2}.

Now m

_{1}= 2 m

_{1}

m

_{2}= 2 m

_{2}

R = 2 R

F

_{1}= G ( 2m

_{1}X2 m

_{2})/4R

^{2}.

F

_{1}= F

So force between them remains same.

## Chapter 3 - Laws of Motion Exercise 126

^{2}

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.

g= 1/R

^{2}

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.

## Chapter 3 - Laws of Motion Exercise 128

1 N = 10

^{5}dyne.

(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.

(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.

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.

^{-1}.

## Chapter 3 - Laws of Motion Exercise 129

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.

1 newton / 1dyne = 10

^{5}.

^{-1}= 1000 g X 100 cms

^{-1}= 105 cms

^{-1}.

1 dyne 1 g X 1 cms

^{-1}= 1cms

^{-1}.

So 1 newton = 10

^{5}dyne.

Law of gravitation is called universal because it applies to all bodies of universe.

(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.

m= F/a and this mass is called as inertial mass.

Newton law of gravitation gives another definition of mass.

F = (G m

_{1}m

_{2})/R

^{2}

Thus m

_{2}is the mass of the body by which another body of mass m

_{1}attracts it towards it by law of gravitation. This mass is called gravitational mass.

(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.

(ii) The force exerted by string on block is equal to the tension in the string and this is also equal to the 20N.

m= F/a

so we can calculate mass of each body

Mass of body 1 m

_{1}= 4/8 = 0.5 kg.

Mass of body 2 m

_{2}= 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}.

## Chapter - Exercise

## Chapter 3 - Laws of Motion Exercise 130

^{-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.

^{-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.

^{-2}.

F = mass X acceleration.

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}.

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}.

_{1}= m.

Velocity of P is v

_{1}=2 v

Mass of Q is m

_{2}= 2m

Velocity of Q is v

_{2}= v.

(i) inertia of P/inertia of Q = m

_{1}/m

_{2}= 1/2.

So ratio of inertia of two bodies is 1:2.

(ii) Momentum of P/momentum of Q = m

_{1}v

_{1}/m

_{2}v

_{2}= 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.

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.

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.

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.

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.

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.

## Chapter 3 - Laws of Motion Exercise 131

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}.

(ii) Again while catching a fast moving ball, we always pull our hands backwards to increase reaction time so force experienced would decrease.

Initial velocity of stone = 0 ms

^{-1}.

Acceleration due to gravity = 9.8 ms

^{-2}.

(i) We know H = ut + 1/2 gt

^{2}.

98 = 1/2 X 9.8X t

_{2}.

t

^{2}= 98X2/9.8 = 20

t= 4.47 sec.

(ii) Final velocity when it strikes the ground

V

^{2}- u

^{2}= 2 g H

V

^{2}= 2X9.8X98

V

^{2}= 1920

V= 44.6 ms

^{-1}.

^{-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 v

^{2}- u

^{2}=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.

^{-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}.

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

V

^{2}- u

^{2}= 2as

0 -u

^{2}= 2 X(-9.8)X19.6

u

^{2}= 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}.

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 Xt

^{2}.

98 = 0 + 1/2 X9.8Xt

^{2}.

t

^{2}= 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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