force-and-laws-of-motionWHERE cd.courseId=8 AND cd.subId=16 AND chapterSlug='force-and-laws-of-motion' and status=1SELECT ex_no,page_number,question,question_no,id,chapter,solution FROM question_mgmt as q WHERE courseId='8' AND subId='16' AND chapterId='189' AND ex_no!=0 AND status=1 ORDER BY ex_no,CAST(question_no AS UNSIGNED) CBSE Class 9 Free NCERT Book Solution for Science

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Chapter 9 : Force and Laws of Motion


We have all heard about the great scientist Newton, he’s proposed different laws to explain motion. In this chapter we will discuss momentum and its effect with example. Conservation of momentum that means momentum can neither be created nor be destroyed.

Exercise 1
Q:
A:

Inertia is the tendency of a body to oppose the changes in their state of rest or motion unless acted upon by an external force. Inertia depends on the mass of the body, higher the mass higher will be the inertia.

a. Mass of a stone is greater than the mass of a rubber ball of the same size. Hence, the inertia of the stone is greater than that of a rubber ball.
b. Mass of a train is greater than the mass of a bicycle. Hence, the inertia of the train is greater than that of the bicycle.
c. Mass of a five rupee coin is greater than that of a one-rupee coin. Hence, inertia of the five rupee coin is greater than that of the one-rupee coin.


Exercise 1
Q:
A:

The velocity of the football changes four times .

(i) When the first player kicks the football, its speed changes from zero to a certain value. 

(ii) When another player kicks the ball towards the goal post , the direction of the ball gets changed. Therefore, its velocity also changes.

(iii) When the goalkeeper collects the ball , the speed of the ball reduces to zero from a certain value . Hence , the velocity of the ball has changed.

(iv) The goalkeeper kicks the ball towards his team players. Hence, the speed of the ball increases from zero to a certain value. Hence, its velocity changes once again. 

Agent supplying the force

In first case - first player
In second case - second player
In third case - goalkeeper
In fourth case - goalkeeper


Exercise 1
Q:
A:

Some leaves of the tree may get detached from a tree, if we vigorously shake its branch due to inertia of rest. Leaves are in the state of rest with respect to the tree so they possess inertia of rest. Due to this reason, the leaves fall down from the tree when shaken vigorously.


Exercise 1
Q:
A:

When a moving bus stops suddenly, the passengers are jerked forward because of inertia the passengers tend to remain in their state of motion even though the bus has come to rest and we fall backwards when bus starts suddenly from rest because of inertia, passengers tend to remain in state of rest though bus starts moving. Hence, the passenger tends to fall backwards when the bus accelerates forward.


Exercise 2
Q:
A:

To pull the cart horse pushes the ground with its foot in the backward direction by pressing the ground. As a reaction of this force, the ground pushes the horse in forward direction. Due to which the horse pulls the cart.


Exercise 2
Q:
A:

When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then equal and opposite reaction forces act on the fireman according to Newton’s third law. Due to this reaction force ,stability of the fireman decreases and fireman faces difficulty to hold the hose.


Exercise 2
Q:
A:

Mass of the rifle, m1 = 4 kg

Mass of the bullet, m2 = 50g = 0.05 kg

Recoil velocity of the rifle = v1

Bullet is fired with an initial velocity, v2 = 35 m/s

Total mass = m1 + m2 = 4 + 0.05 = 4.05 kg

Since the system was at rest , velocity (v) = 0

Momentum ( before firing ) = mass x velocity = ( m1 + m2 )v = ( 4.05 ) 0 = 0 kg m/s

Total momentum = momentum of bullet + momentum of rifle = m1 v1 + m2 v2 

= 0.05 x 35 + 4 x v2 = 7/4 + 4 v2

Since the total momentum remain conserved 

Total momentum before firing = Total momentum after firing

0 = 7/4 + 4 v2

4 v2 = - 7/4

v2 = - 7/16

v2 = -04375 m/s

Thus , the recoil velocity of the pistol is -0.4375 m/s .

The negative sign of velocity means that the rifle will recoil in the opposite direction of the bullet.


Exercise 2
Q:
A:

Mass of one of the objects, m1 = 100 g = 100/1000 = 0.1 kg 

Mass of the other object, m2 = 200 g = 200/1000 = 0.2 kg

Velocity of m1 before collision, v1 = 2 m/s

Velocity of m2 before collision, v2 = 1 m/s

Velocity of m1 after collision, v3 = 1.67 m/s

Velocity of m2 after collision = v4 = ?

By law of conservation of momentum :

Total momentum before collision = Total momentum after collision

 m1v1 + m2v2 = m1v3 + m2v4

(0.1)2 + (0.2)1 = (0.1)1.67 + (0.2)v4

0.4 = 0.167 + 0.2v4

 v4 = 1.165 m/s

The velocity of the second object after collision 1.165 m/s.


Exercise 3
Q:
A:

Yes. It is possible.

(i) The object should already be moving at uniform speed in a straight line path.
(ii) There is no change in speed.
(iii) No direction change.
(iv) Friction force must be zero.
(v) Air resistance must be zero.


Exercise 3
Q:
A:

When a carpet is beaten with a stick, the dust comes out of it because of the law of inertia. Initially the dust particles are at rest along with the carpet but, the dust particles try to resist their state of rest. According to Newton’s first law of motion, the dust particles stay in a state of rest, while the carpet moves. Hence, the dust particles come out of the carpet.


Exercise 3
Q:
A:

When a moving bus suddenly stops, the luggage on the roof tends to continue its state of motion and may fall. Also, when the bus suddenly starts from rest, luggage maintains its rest position and may fall backward. So, it is advised to tie any luggage kept on the roof of a bus with a rope.


Exercise 3

Exercise 3
Q:
A:

Initial velocity, u = 0

Distance travelled, s = 400 m

Time taken, t = 20 s

According to equation of motion:


a = 2 m/s2

1 metric tonne = 1000 kg (Given)

∴ 7 metric tonnes = 7000 kg

Mass of truck, m = 7000 kg

Force, F = Mass × Acceleration

F = ma = 7000 × 2 = 14000 N

The acceleration of the truck is 2 m/s2 and the force acting on the truck is 14000 N.


Exercise 3
Q:
A:

Mass of the stone = 1 kg

Initial velocity of the stone, u = 20 m/s

Final velocity of the stone, v = 0

Distance travelled by the stone, s = 50 m

Using the third equation of motion:

v2 = u2 + 2as

Where,

Acceleration, a

(0)2 = (20)2 + 2 × a × 50

a = −4 m/s2

Force, F = Mass × Acceleration

F = ma

F = 1 × (− 4) = −4 N

The negative sign shows that force of friction is in the opposite direction of motion , the force of friction between the stone and the ice is −4 N.


Exercise 3
Q:
A:

Mass of engine = 8000 kg

Mass of 1 wagon = 2000 kg

Mass of 5 wagons = 2000 x 5 = 10000 kg

Force exerted by the engine, F = 40000 N

Frictional force offered by the track, Ff = 5000 N

(a) Net force, Fa = Force of engine - Force of friction = F − Ff

      = 40000 − 5000 = 35000 N

Hence, the net accelerating force is 35000 N.

(b) Total mass of the train, M = mass of 5 wagons + mass of engine

      = 10000 + 8000 = 18000 kg

From Newton’s second law of motion:

Force = Mass x acceleration

35000 = 18000 x a

a = 35000 / 18000 = 1.94 m/s2

Hence, the acceleration of the train is 1.94 m/s2.


Exercise 3
Q:
A:

Mass of the automobile vehicle, m = 1500 kg

Acceleration of the automobile, a = −1.7 ms−2

We know that

Force = Mass × Acceleration = 1500 × (−1.7) = −2550 N

The negative sign shows that force is in the opposite direction of motion, the force between the automobile and the road is −2550 N.


Exercise 3

Exercise 3
Q:
A:

Since, a horizontal force of 200 N is used to move a wooden cabinet. Thus, from Newton’s third law of motion, an equal amount of force will act in the opposite direction. This opposite force is the frictional force exerted on the cabinet. Hence, a frictional force of 200 N is exerted on the cabinet.


Exercise 3
Q:
A:

Mass of first objects, m1 = 1.5 kg

Mass of second object, m2 = 1.5 kg

Velocity of m1 before collision, v1 = 2.5 m/s

Velocity of m2, moving in opposite direction before collision, v2 = −2.5 m/s

(Negative sign arises because mass m2 is moving in an opposite direction)

We know that

Total momentum before collision = Total momentum after collision

m1v1 + m2 v2 = (m1 + m2) v

1.5(2.5) + 1.5 (−2.5) = (1.5 + 1.5) v

3.75 − 3.75 = 3 v

v = 0 / 3 = 0

Hence, the velocity of the combined object after collision is 0 m/s.


Exercise 3
Q:
A:

Because of the huge mass of the truck, the force of static friction is very high. To move the car / truck, one has to apply a force more than the static friction. Therefore, when someone pushes the truck and the truck does not move, then it can be said that the applied force in one direction is cancelled out by the frictional force of equal amount acting in the opposite direction. Hence , the rationale given by the students is correct.


Exercise 3
Q:
A:

Mass of the hockey ball, m = 200 g = 200 / 1000 = 0.2 kg

Initial velocity, v1 = 10 m/s

Initial momentum = mv1

Final velocity, v2 = −5 m/s

Final momentum = mv2

Change in momentum = mv1 − mv2 = m(v1 - v2)

= 0.2 [10 − (−5)] = 0.2 (15) = 3.0 kg m/s

Hence, the change in momentum of the hockey ball is 3 kg m/s.


Exercise 3
Q:
A:

Mass of bullet, m = 10 g = 10 / 1000 = 0.01 kg

Initial velocity, u = 150 m/s

Final velocity, v = 0

Time taken, t = 0.03 s

Using the first equation of motion

 v = u + at

0 = 150 + (a ×0.03 s)

(Negative sign indicates that the velocity of the bullet is decreasing.)

Now, using the third equation of motion:

v2 = u2 + 2as

0 = (150)2 + 2 (−5000) s

Hence, the distance of penetration of the bullet is 2.25 m.

We know that

Force, F = Mass × Acceleration

F = ma = 0.01 × 5000 = 50 N

Hence, the magnitude of force is 50 N.


Exercise 3
Q:
A:

Mass of the object, m1 = 1 kg

Mass of the wooden block, m2 = 5 kg

Velocity of the object before collision, v1 = 10 m/s

Velocity of the wooden block before collision, v2 = 0 m/s

Total mass of the combined system = m1 + m2 = 1 + 5 = 6 kg

∴ Total momentum before collision = m1 v1 + m2 v2

= 1 (10) + 5 (0) = 10 kg m s−1

Velocity of the combined object = v

Momentum = total mass x velocity of combined object

= (m1 + m2) v = 6v

According to the law of conservation of momentum:

Total momentum before collision = Total momentum after collision

m1 v1 + m2 v2 = (m1 + m2) v

10 = 6 v

The total momentum after collision is also 10 kg m/s.

Hence, velocity of the combined object after collision = 


Exercise 3
Q:
A:

Mass of the object, m = 100 kg

Initial velocity of the object, u = 5 m/s

Final velocity of the object, v = 8 m/s

Time take by the object to accelerate, t = 6 s

Initial momentum = mu = 100 × 5 = 500 kg m/s

Final momentum = mv = 100 × 8 = 800 kg m/s

By using Newton’s second law of motion, F = change in momentum / time taken = 

Force exerted on the object is 50 N.


Exercise 3
Q:
A:

According to the law of conservation of momentum, the momentum of a system before collision is equal to the momentum of the system after collision.

Hence, the change in momentum of the car and insect system is zero.

Thus, Kiran is wrong with the logic as momentum is always conserved, i.e change in momentum of insect must be equal to that of a motorcar.

Akhtar’s logic is incorrect because the mass of the car is very large as compared to the mass of the insect.

Rahul gave the correct logic as insects and motorcar experience the same force and change in momentum.


Exercise 3
Q:
A:

Mass of the dumbbell, m = 10 kg

Distance, s = 80 cm = 80 / 100 = 0.8 m

Acceleration, a = 10 m/s2

Initial velocity of the dumbbell, u = 0

Final velocity of the dumbbell = v

Using the third equation of motion:

v2 = u2 + 2as

v2 = 0 + 2 (10) 0.8

v = 4 m/s

Hence, the momentum with which the dumbbell hits the floor is

= mv = 10 × 4 = 40 kg m/s


Exercise 4

Exercise 4
Q:
A:

Mass of the motor car = 1200 kg

Only two persons are able to move the car with constant velocity. So, the acceleration acquired by the car is given by the third person. Thus the third person is responsible for the acceleration generated.

Acceleration, a = 0.2 m/s2

From Newton’s second law of motion:

Force = Mass × Acceleration

F = 1200 × 0.2 = 240 N

Hence, each person applies a force of 240 N to push the motor car.


Exercise 4
Q:
A:

Mass of the hammer, m = 500 g = 500 / 1000 = 0.5 kg

Initial velocity of the hammer, u = 50 m/s

Time taken, t = 0.01

Since the hammer comes to rest,

Velocity of the hammer, v = 0

Using Newton’s second law of motion:

The hammer strikes the nail with a force of −2500 N. By Newton’s third law of motion, nail exerts an equal force in the opposite direction on the hammer, thus  the force of the nail on the hammer is equal and opposite, i.e., +2500 N.


Exercise 4
Q:
A:

Mass of the motor car, m = 1200 kg

Initial velocity of the motor car, u = 90 km/h = 90 x 5/18 = 25 m/s

Final velocity of the motor car, v = 18 km/h = 18 x 5/18 = 5 m/s

Time taken, t = 4 s

Using the first equation of motion:

v = u + at

5 = 25 + a (4)

a = -5 m/s2

Change in momentum = mv − mu = m (v−u)

= 1200 (5 − 25) = -24000 kg m s−1

Force = Mass × Acceleration

= 1200 × (-5) = -6000 N

Acceleration of the motor car = -5 m/s2

Change in momentum of the motor car = -24000 kg m s−1

Hence, the force required to decrease the velocity is -6000 N.

(Negative sign indicates the retardation, decrease in momentum and retarding force respectively)


Exercise 4
Q:
A:

(a)  Both vehicles experience equal forces of action and reaction because both of the objects are moving.

(b)  Since both truck and car are moving with the same magnitude of velocity v, momentum change for both vehicles is the same.

(c)  Mass and acceleration are inversely proportional to each other. Therefore, car experience greater acceleration due to their smaller mass.

(d)  Due to its smaller mass or opposition to the force exerted on it, the car is likely to suffer more damage than the truck.