Lakhmir Singh Manjit Kaur Class 9 Physics 2nd Chapter “Force and Laws of Motion” solution

Lakhmir Singh Manjit Kaur Class 9 Physics 2nd Chapter “Force and Laws of Motion” solution

Lakhmir Singh and Manjit Kaur Physics solution: “Force and Laws of Motion” Chapter 2. Here you get easy solutions of Lakhmir Singh and Manjit Kaur Physics solution Chapter 2 . Here we have given Chapter 2 all solution of Class 9. Its help you to complete your homework.

  • Board – CBSE
  • Text Book Physics
  • Class – 9
  • Chapter – 02

Lakhmir Singh Manjit Kaur Class 9 Physics 2nd Chapter Solution (First Part)

Very short answer type questions

1.) What name is given to the product of mass and velocity of a body?

Answer: Momentum is the product of mass and velocity of a body.

2.) Name the physical quantity which is considered to be a measure of the quantity of motion of the body.

Answer: Momentum is the physical quantity which is considered to be a measure of the quantity of motion of the body.

3.) What is the SI unit of momentum?

Answer: we know the formula of momentum,

Momentum=.Mass × velocity.

So SI unit of mass is Kg and velocity is m/s.

Therefore SI unit of momentum is Kg m/s

4.) State whether momentum is scalar or vector.

Answer: Momentum is the product of mass and velocity, and velocity is only one vector quantity in momentum. Therefore momentum is also vector quantity.

5.) What is the total momentum of the bullet and the gun before firing ?

Answer: we know the law of conservation of momentum, so total momentum id always conserved. Hence total momentum of bullet and gun is zero.

6.) Name the physical quantity whose unit is kgm/s.

Answer: Momentum is a physical quantity whose unit is kgm/s.

7.) What will be the momentum of a body of mass ‘m’ which is moving with a velocity ‘v’ ?

Answer: If a body having mass moves with velocity v then it’s momentum = p = mv

8.) What is the usual name of the forces which cannot produce motion in a body but only change its shape ?

Answer:  Balanced force cannot produce motion in a body but only change its shape.

9.) Name the unbalanced force which slows down a moving bicycle when we stop pedalling it.

Answer: Frictional force slows down a moving bicycle when we stop pedalling it.

10.) State whether the following statement is true or false :

Unbalanced forces acting on a body change its shape.

Answer: False, balanced force changes shape of body.

11.) When a ball is dropped from a height, its speed increases gradually. Name the force which causes this change in speed.

Answer: A ball is dropped from height then its speed increases gradually because of gravitationalforce exerted by earth.

12.) Name the property of bodies (or objects) to resist a change in their state of rest or of motion.

Answer: The property of bodies (or objects) to resist a change in their state of rest or of motion is inertia.

13.) What is the other name of Newton’s first law of motion?

Answer:  Newton’s First law is discovered by a famous physicist Galileo Galilei

So it is called as Galileo’s law of inertia.

14.) The mass of object A is 6 kg whereas that of another object B is 34 kg. Which of the two objects, A or B, has more inertia?

Answer: Inertia is a property of material associate with mass so if a body having greater mass then it’s inertia also more. In above example mass of body B is greater than body A therefore inertia of body B is also more that body A.

15.) Name the scientist who gave the laws of motion.

Answer: Sir Isaac Newton gave the laws of motion.

16.) State whether force is a scalar or a vector quantity.

Answer: Force has magnitude as well as direction so it is vector quantity.

17.) With which physical quantity should the speed of a running bull be multiplied so as to obtain its momentum?

Answer: Mass is a physical quantity should the speed of a running bull be multiplied so as to obtain its momentum.

Short Answer Type Questions

19.) Explain why, it is easier to stop a tennis ball than a cricket ball moving with the same speed.

Answer: According to second law of motion, force is directly proportional to change in momentum. So magnitude of force is depend on momentum and momentum is depend on mass and velocity. Velocity of both ball is same but mass is different. Cricket ball has greater mass than tennis ball therefore  it is easier to stop a tennis ball than a cricket ball moving with the same speed.

20.) Explain the meaning  of the following equation :

p = m × v

where symbols have their usual meanings.

Answer:

Given formula

p = m × v

We can calculate momentum of the particle by using this formula.

The terms in above formula have usual meaning.

p- Momentum

m – mass of an object.

v- velocity of an object.

21.) Explain how, a karate player can break a pile of tiles with a single blow of his hand.

Answer: Karate player can break a pile of tiles with blow of his hand. There are two reason behind this

I) They hit hand with greater velocity on tiles so magnitude of momentum is large and it produce more effect.

II) They apply such a large force in a very short interval of time, due to this impulse tiles easily breaks.

22.) Calculate the momentum of a toy car of mass 200 g moving with a speed of 5 m/s.

Answer: Given, mass of toy is 200g = 0.2 kg.

Velocity = 5 m/s.

As we know the formula of momentum,

Momentum = p = m × v

p = 0.2 × 5

p = 1 kg m/s.

23.) What is the change in momentum of a car weighing 1500 kg when its speed increases from 36 km/h to 72 km/h uniformly?

Answer: Given, initial speed = u = 36 km/hr

u= 36 km/hr = 36× 5/18 m/s= 10 m/s.

Final velocity = v = 72 km/hr = 72× 5/18 m/s.

v= 20 m/s.

Mass of the car is 1500 kg.

As we know, Change in momentum = final momentum- initial momentum.

Change in momentum = mv – mu

Change in momentum = 1500 × 20 – 1500 ×10

Change in momentum = 1500 ×(20-10)

Change in momentum = 1500× 10

Change in momentum = 15000 kg m/s.

Thus, change in momentum is 15000 kg m/s

24.) A body of mass 25 kg has a momentum of 125 kg.m/s. Calculate the velocity of the body.

Answer: Given, Mass of body is 25kg

Momentum = 125 kg m/s.

As we know the formula of momentum,

Momentum = p = m × v

125 = 25× v

125/25= v

V= 5 m/s.

The velocity of that body is 5 m/s.

25.) Calculate the momentum of the following: 

(a) an elephant of mass 2000 kg moving at 5 m/s

(b) a bullet of mass 0.02 kg moving at 400 m/s.

Answer: (a) Given, mass of elephant is 2000 kg

Velocity = 5 m/s.

As we know the formula of momentum,

Momentum = p = m × v

p = 2000 × 5

p = 10000 kg m/s.

Momentum of the elephant is 10000 kg m/s.

(b) Given, mass of bullet is 0.02 kg.

Velocity = 400 m/s.

As we know the formula of momentum,

Momentum = p = m × v

p = 0.02 × 400

p = 8 kg m/s.

The momentum of bullet is 8 kg m/s.

26.) Which of the two, balanced forces or unbalanced forces, can change the shape of an object ? Give an example to illustrate your answer.

Answer: Balanced forces change the shape of an object. When equal forces applied on the object but the direction of forces are opposite such forces can change the shape of object.

Example-

a) Stretch the spring by two hands. Two hands exerts equal force on the spring but direction is opposite so the shape of spring changes due to tension.

b) Press the balloon by hands. Fingers exerts equal force and change the shape of balloon.

27.) Describe the term ‘inertia’ with respect to motion.

Answer: Inertia is a property associated with mass. The property resist to change in it’s  state of rest or state of uniform motion is called inertia.We can simplify this by a simple example.

a) When you exert force on a rest car then the car does not allow to change in its state of rest. The oppose exert by car is called inertia.

b) When you catch the ball which moving towards you with greater speed then it can injure your hands. This moving ball does not allow to come in rest easily. This property is an example of inertia.

28.) State Newton’s first law of motion. Give two examples to illustrate Newton’s first law of motion.

Answer: Newton’s first law of motion is associated with inertia. It gives explanation about origin of motion.

Statement of  Newton’s first law:

“Everybody continues to be in its state of rest or state of uniform motion until we act an external unbalance force on it.”

This law states that the state of rest and state of uniform motion are equivalent. When we apply external unbalanced force on it then it changes it’s state.

Examples:

i.) Place a book on table. The book remains on same state of rest until we act external unbalanced force on it.

ii.) You throw a ball in space. If we ignore the air resistance then that ball does stop until someone stop it.

29.) On what factor does the inertia of a body depend ? Which has more inertia, a cricket ball or a rubber ball of the same size?

Answer: Inertia is introduced in first law of motion. It is a intrinsic property of material and associated with mass.

Cricket ball has greater mass than rubber ball therefore it has cricket ball has greater inertia than rubber ball.

30.) Why do the passengers in a bus tend to fall backward when it starts suddenly?

Answer: When bus is in state of rest then passenger also in state of rest. But suddenly bus moves forward then passenger does not moves immediately due to inertia. The human body is muscular so lower side moves with  floor of bus but upper side does not movesso passengers fall back. But after few seconds human body accept this change and moves with bus.

The passengers in a bus tend to fall backward when it starts suddenlyfueyoinertia.

31.) Explain why, a person travelling in a bus falls forward when the bus stops suddenly.

Answer: Passengers traveling in bus moves with speed of bus. But suddenly bus drivers breaks up then human body does not accept this change and moves forward with bus. But there is friction between steps and floor of bus so this friction does not allow to move forward . The upper part moves with same speed so passengers falls forward.

The person travelling in a bus falls forward when the bus stops suddenly due to inertia.

32.) Give reason for the following:

When a hanging carpet is beaten with a stick, the dust particles start coming out of it.

Answer:  The dust particle lies on the surface of carpet. When we beat carpet by stick then surface of carpet accelerate but due to inertia dust particle does not gain this speed. These dust particles trying to remains in stationary state. So dust particles coming out from carpet.

When a hanging carpet is beaten with a stick, the dust particles start coming out of it because of inertia.

33.) When a tree is shaken, its fruits and leaves fall down. Why ?

Answer: Earth exerts gravitational force on fruits and leaves of tree. Tree provides sufficient force to leave and fruits so they remains stationary. But when tree shaken, the fruits and leaves are separated due to inertia so force exerted by tree vanishes. The fruits and leaves fall down due to gravitational force of earth.

34.) Explain why, it is dangerous to jump out of a moving bus.

Answer: When we travels in bus that time we moves with the speed of bus. But suddenly we jump out of moving bus then our speed vanishes suddenly and we stop immediately.  Due to inertia our body does not allow sudden change immediately so we fall down.

35.) What is the momentum in kg m/s of a 10 kg car travelling at (a) 5 m/s (b) 20 cm/s, and (c) 36 km/h ?

Answer: (a) Given, mass of car is 10 kg

Velocity = 5 m/s.

As we know the formula of momentum,

Momentum = p = m × v

p = 10× 5

p = 50 kg m/s.

Momentum of the car is 50 kg m/s.

(b) Given, mass of car is 10 kg.

Velocity = 20 cm/s= 0.2 m/s

As we know the formula of momentum,

Momentum = p = m × v

p = 10 × 0.2

p = 2 kg m/s.

The momentum of car is 2 kg m/s.

.(c) Given, mass of car is 10 kg.

Velocity = 36 km/hr= 36 × 5/18 m/s

Velocity = 10 m/s

As we know the formula of momentum,

Momentum = p = m × v

p = 10 × 10

p = 100kg m/s.

The momentum of car is 100 kg m/s.

Long Answer type Questions

36.) (a) Define momentum of a body. On what factors does the momentum of a body depend ?

(b) Calculate the change in momentum of a body weighing 5 kg when its velocity decreases from 20 m/s to 0.20 m/s.

Answer:

Momentum: The concept of momentum is introduced in second law of motion. It is the combination of mass and velocity.

Definition: If a body having mass m

Product of mass and velocity is called moment of the particle.

Momentum is denoted by p.

If a body having mass ‘ m’ moving with velocity ‘v’ then momentum is

p= m × v

Momentum is depend on two factors

(1) Mass of the particle

(2) Velocity of particle.

(b) Given, mass = 5 kg,

initial velocity = u = 20 m/s

Final velocity = v =0.20 m/s.

As we know, Change in momentum = final momentum- initial momentum.

Change in momentum = mv – mu

Change in momentum = (5 × 0.20)– (5 × 20)

Change in momentum = 5  ×(0.20 -20)

Change in momentum = 5 × (- 19.80)

Change in momentum = 99 kg m/s.

Thus, change in momentum is 99 kg m/s

37.) (a) Define the term ‘force’.

(b) State the various effects of force.

Answer: Force is a basic concept in mechanics. It is responsible to change in state.

Definition: Force is nothing but simply pull or push on a body.

Effects of forces:

a.) Force changes the state of object. If the object is in state of rest then force move it.

When object is in motion then force stop it.

b.) Second effect of force is it can accelerate the body. Mechanical force changes speed of vehicle.

c.) Force can change the direction of motion. Batsman can changes the direction of ball moving towards him.

d.) We can change the shape of object by applying force. Blacksmith changes the shape of iron by applying force. Potter make different objects from mud by applying force.

38.) Give one example each where:

(a) a force moves a stationary body.

(b) a force stops a moving body.

(c) a force changes the speed of a moving body.

(d) a force changes the direction of a moving body.

(e) a force changes the shape (and size) of a body.

Answer:

a.) Example of “a force moves a stationary body”- Push a stationary car, magnet pull stationary magnetic material toward it.

b.) Example of “a force moves a stationary body”- player catch the ball and stop it, frictional force stop rolling ball.

c.) Example of “a force changes the speed of a moving body”- mechanical force of motorcycle increases the speed by accelerating and frictional force decreases the speed of motorcycle.

d.) Example of“a force changes the direction of a moving body”- Batsman can changes the direction of ball moving towards him. Sun’s gravitational force changes the direction of earths motion continuously.

e.) Example of “a force changes the shape (and size) of a body”- Blacksmith changes the shape of iron by applying force. Potter make different objects from mud by applying force.

39.) (a) What do you understand by the terms “balanced forces” and “unbalanced forces” ? Explain with examples.

(b) What type of forces — balanced or unbalanced — act on a rubber ball when we press it between our hands? What effect is produced in the ball?

Answer: Force is basic concept in mechanics. Force is nothing but simply push or pull on a body. We can not bind force in a simple definition so we will study effect of this.

There are two types of forces on the basis of effect.

  • Balanced forces
  • Unbalanced forces

Balanced forces: When we apply multiple of forces having equal in magnitude on a body such that resultant of the forces must be zero, these type of forces are called balanced forces. Balanced forces does not change the state of object but changes the shape. We can understand this with a simple example.

Example: (1) Stretch the spring by applying force in two opposite direction. In this example applied force is balanced force. It does not change the state of object but changes it’s shape.

(2) 4 men pushing a stationery car at 4 different sides such that the car remains stationary. In this example, these men are applying force but body does not changes it’s state because of balanced force.

Unbalanced force: when the resultant sum of multiple forces does not vanish then such forces are called as unbalanced forces.Unbalanced forces changes the state of object. We can understand this with a simple example.

(b) When we press tuner ball between our hands then the applied forces are balanced forces. It does not changes the stationary state of ball but changes shape of the ball. Effect of applied balanced forces in this example is change in shape.

Example:

40.) (a) What happens to the passengers travelling in a bus when the bus takes a sharp turn ? Give reasons for your answer.

(b) Why are road accidents at high speeds very much worse than road accidents at low speeds?

Answer:

(a) When bus takes sharp turn then two forces acting on the passengers. We can understand this with the help of ray diagram,

Firstly, due to circular motion, passengers experience  outward force. But this force is not real. It is a type of pseudo force called centrifugal force. It is due to inertia of the passenger.Due to this force passengers trying to throw outward but another type of force called centripetal force balanced this outward force and passengers do not throw out from bus. This inward force is real force and arises due to friction

(b) When bus takes sharp turn then an imaginary force is applied on passengers and passengers trying to throws outwards. This force is called centrifugal force.At same time frictional forces apply in inward direction and neutralise that imaginary force.

(c) According to second law of motion, change in momentum is depend on applied force. If the speed is greater then value momentum also greater hence effect will very much worse. But at lower speed,  change in momentum is minimum and that accident damage less.

2nd Part Solution

Very Short Answer Type Questions

1.) Which physical quantity corresponds to the rate of change of momentum?

Answer: According to second law of motion, force is a physical quantity corresponds to the rate of change of momentum.

 2.) State the relation between the momentum of a body and the force acting on it.

Answer: According to second law of motion, force is directly proportional to change in momentum.

F = dp/dt

Where F – force,

d/dt – rate of change

p- momentum.

In simple way we can write,

F = (mv-mu)/t

Where, mv- final momentum.

mu – initial momentum.

3.) What is the unit of force ?

Answer: Unit of force is Kg m/s2 or Newton.

 4.) Define one newton force.

Answer : one Newton means 1 kg-1 metre/ 1 s2

5.) What is the relationship between force and acceleration ?

Answer: According to second law of motion,

F = m × a

Where

F- force

m- mass

a- Acceleration

6.) If the mass of a body and the force acting on it are both doubled, what happens to the acceleration ?

Answer: As we know the second law of motion,

F = m × a

Therefore acceleration = a F/m

If we doubled force and mass then

2F = 2m ×a

Acceleration = a = F/ m

If we doubled force and mass the there is know change in acceleration.

7.) Name the physical quantity whose unit is ‘newton’.

Answer: Unit of force is Newton.

8) Which physical principle is involved in the working of a jet aeroplane ?

Answer: The principle of conservation of momentum  is involved in the working of a jet aeroplane.

Initial momentum = final momentum

9.) Name the principle on which a rocket works.

Answer: Rockets works on principle of conservation of momentum. It moves forward and realise smoke backward, so total momentum of rocket before and after are equal.

10.) Is the following statement true or false • A rocket can propel itself in a vacuum.

Answer: True, A rocket can propel itself in a vacuum.

11.) What is the force which produces an acceleration of 1 m/s2 in a body of mass 1 kg ?

Answer: Given, Mass = m = 1kg.

Acceleration = 1 m/s2

We know the second law of motion.

F = 1 × 1

F = 1 newton

The force which produces an acceleration of 1 m/s2 in a body of mass 1 kg is 1 Newton.

12.) Find the acceleration produced by a force of 5 N acting on a mass of 10 kg.

Answer: Given, Mass = m = 10 kg.

Force  = 5 Newton.

As we know second law of motion,

F = m × a

a= F/m

a= 5/10

a= 0.5 m/s2

The  acceleration produced by a force of 5 N acting on a mass of 10 kg is a= 0.5 m/s2

13.) A girl weighing 25 kg stands on the floor. She exerts a downward force of 250 N on the floor. What force does the floor exert on her ?

Answer: A girl exert force on floor is 250 N. According to third law of motion, force in action and reaction is equal. So floor exert 250 N force on her.

14.) Name the physical quantity which makes it easier to accelerate a small car than a large car.

Answer: Acceleration is depend on force and mass. The mass of small car is smaller than large car so mass is a quantity which makes it easier to accelerate a small car than a large car.

Short Answer Type Questions

16.) Explain the meaning of the following equation :

F = m × a

where Symbols have their usual meanings.

Answer: The mathematical statement of second law of motion is,

F = m × a

Where,

F – force exerted on a body

m- mass of the body

a- Acceleration produced to applied force.

17.) To take the boat, away from the bank of a river, the boatman pushes the bank with an oar. Why ?

Answer:  We know that according to third law of motion when first body apply a force on second body then second body also exert same force on first. So boatman apply force on a bank of river with oar then bank of river  also exert force on boat.

18.) Why does a gunman get a jerk On firing a bullet?

Answer: We know the law of conservation of momentum,

Momentum of bullet = momentum of gun (initially both are in rest so initial momentum is zero)

When bullet produces momentum in forward direction  then equal momentum will produce by gun in backward direction.

When gunman fire a bullet from gun and bullet moves forward then according to conservation of momentum gun moves backward with some velocity and gunman get a jerk.

19.) If action is always equal to reaction, explain why a cart pulled by a horse can be moved.

Answer: Newton’s third law states that, with every action there is equal and opposite reaction. Horse push the surface so cart moves forward, this is a pair of action and reaction.

20.) Explain how a rocket works.

Answer: Rocket works on the principle of Newton’s third law. Rocket throws smoke downward and moves upward. Moving upward is a action and throwing smoke downward is a reaction.

21.) Do action and reaction act on the same body or different bodies ? How are they related in magnitude and direction ? Are they simultaneous or not ?

Answer:  We have to understand statement of third law of motion. The third law states that, “ With every action there is equal and opposite reaction and both occurs simultaneously.

Action and reaction act on different bodies. The magnitude of action and reaction is same but direction is different and both occurs simultaneously.

22.) If a man jumps out from a boat, the boat moves backwards. Why ?

Answer: The answer of this question is based on Newton’s third law.When a man jumps out from the boat then he applied force on but in bachelor’s direction. The man jumps out from the boat is action and moving boat in backward is reaction, both occurs on different objects.

23.) Why is it difficult to walk on a slippery road ?

Answer: When we try to walk on slippery road then slippery road does not produced reaction force in exact backward direction. If a road is though road then it produced sufficient reaction force so we can walk easily on it.

24.) Explain why, a runner presses the ground with his feet before he starts his run.

Answer: The faster the runner is pushed to ground, the faster he will bounce in front. So he will gain sufficient initial speed. That is the reason behind this.

25.) A 60 g bullet fired from a 5 kg gun leaves with a speed of 500 m/s. Find the speed (velocity) with which the gun recoils (jerks backwards).

Answer: Given, mass of bullet m1 = 60 g = 0.06 kg,

Mass of gun m2 =5 kg, final speed of bullet = v1 = 500 m/s,

Final Speed of gun = v2

According to conservation of momentum,

Initial momentum of gun+ Initial momentum of bullet = final momentum of gun+ final momentum of bullet

But initially both are in rest so momentum is zero

0 =  final momentum of gun+ final momentum of bullet

0 =  m1 × v1+ m2 × v2

0 = 0.06 × 500 + 5 × v2

0 = 30 + 5 v2

V2 = – 30 /5

V2 = -6 m/s

Minus sign indicates that gun moves backward with speed 6 m/s.

The recoil speed of gun is 6 m/s.

26.) A 10 g bullet travelling at 200 m/s strikes and remains embedded in a 2 kg target which is originally at rest but free to move. At what speed does the target move off ?

Answer: Given, mass of bullet m1 = 10 g = 0.01 kg,

Mass of target m2  =2 kg, initial speed of bullet = u1 = 200 m/s,

Initial Speed of target =u2=0 m/s,

Final velocity of bullet = v1 and. Final velocity of target = v2

We know the law conservation of momentum

Initial momentum of gun+ Initial momentum of target = final momentum of gun+ final momentum of target

m1 × u1 + m2 × u2 = m1 × v1 + m2 × v2

As given in problems, after collision bullet and target moves together so they have same final velocity.

v1 = v2 = V

m1 × u1 + m2 × u2 =( m1  + m2)v

0.01 × 200 + 2 × 0 = (0.01+ 2) v

2 + 0 = 2.01 v

V = 2/ 2.01

V =  0.995 m/s.

The target and bullet moves together with speed of 0.995 m/s

27.) A body of mass 2 kg is at rest. What should be the magnitude of force which will make the body move with a speed of 30 m/s at the end of 1 s ?

Answer: Given, mass of body = 2 kg. It is at rest so initial speed = 0 m/s.

Final speed = 30 m/s

Time = 1 second

According to second law of motion

Force =  mass × (final speed – initial speed)/time

Force = 2 × ( 30 -0)/1

Force = 60 N.

The magnitude of force needed to change the speed from zero to 30 m/s in one second is 60 N.

28.) A body of mass 5 kg is moving with a velocity Of 10 m/s. A force is applied to it so that in 25 seconds, it attains a velocity of 35 m/s. Calculate the value of the force applied.

Answer: Given, mass of body = 5 kg.

Initial speed = 10 m/s.

Final speed = 35 m/s

Time = 25 second

According to second law of motion

Force =  mass × (final speed – initial speed)/time

Force = 5 × ( 35 – 10)/25

Force = 5× (25)/25

Force = 5 N

The magnitude of force needed to change the speed from 10 m/s to 35 m/s in 25 second is 5 N.

29.) A car of mass 2400 kg moving with a velocity of 20 m/s is stopped in 10 seconds on applying brakes. Calculate the retardation and the retarding force.

Answer: Given, mass = 2400 kg,

Initial speed = 20 m/s

Finally car stop so,  final speed = 0 m/s.

Time = 10 seconds.

Acceleration = (final speed – initial seed)/time

Acceleration = (0-20) / 10

Acceleration= -2 m/s2

Where minus sign indicates retardation.

Now we calculate force,

According to second law,

Force = mass × acceleration

Force = 2400 ×(-2)

Force = -4800.

The retardation of car -2 m/s2is and – 4800 N force is required to stop the car.

30.) For how long should a force of 100 N act on a body of 20 kg so that it acquires a velocity of 100 m/s ?

Answer: Given, mass of body = 20 kg.

Initial speed = 0 m/s.

Final speed = 100 m/s

Force = 100 N

As we know second law of motion,

Force =  mass × (final speed – initial speed)/time

100 = 20 × ( 100-0)/T

100= 20× 100/T

T = 20 × 100/100

T = 20 seconds.

A force of 100 N act on a body of 20 kg so that it acquires a velocity of 100 m/s is 20 seconds.

31.) How long will it take a force of 10 N to stop a mass of 2.5 kg which is moving at 20 m/s ?

Answer: Given, mass of body = 2.5 kg.

Initial speed = 20 m/s.

Final speed = 0 m/s

To stop the object we must apply force in opposite direction so Force = -10 N

As we know second law of motion,

Force =  mass × (final speed – initial speed)/time

-10 = 2.5 × ( 0-20)/T

T = 2.5 × (-20)/-10

T = 5 seconds

The object moving with velocity 20 m/s is stopped by 10 N in 5 seconds.

32.) The velocity of a body of mass 10 kg increases from 4 m/s to 8 m/s when a force acts on it for 2 s.

(a) What is the momentum before the force acts ?

(b) What is the momentum after the force acts ?

(c) What is the gain in momentum per second ?

(d) What is the value of the force ?

Answer: Given, mass = 10 kg, initial speed 4 m/s, final speed = 8 m/s. Time = 2 seconds.

a) Initial Momentum =mass ×initial velocity

Initial momentum= 10 × 4

Initial momentum = 40 kg m/s

b) Final Momentum =mass × final velocity

Final momentum= 10 × 8

Final momentum = 80 kg m/s

c) Gain in momentum per seconds = (Final momentum- initial momentum)/time

Gain = (80-40)/2

Gain = 40/2

Gain = 20

The gain in momentum per seconds is 20 kg m/s2

(Note: Gain in momentum per second is nothing force)

d) As we know the second law of motion,

Force =  mass × (final speed – initial speed)/time

Force = 10 × ( 8-4)/2

Force = 10 × (4)/2

Force = 20 N

The value of force is 20 N.

33.) A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 s to move through the barrel of the gun and acquires a velocity of 100 m/s. Calculate :

(i) the velocity with which the gun recoils.

(ii) the force exerted on gunman due to recoil of the gun

Answer: Given, mass of bullet m1 = 30 g = 0.03 kg,

Mass of gun m2  =3 kg, time = 0.003 seconds, final speed of bullet = v1 = 100 m/s.

Final Speed of gun = v2

i) Now we calculate recoil velocity of gun using law of conservation of momentum.

According to conservation of momentum,

Initial momentum of gun+ Initial momentum of bullet = final momentum of gun+ final momentum of bullet

But initially both are in rest so momentum  is zero

0 =  final momentum of gun+ final momentum of bullet

0 =  m1 × v1 + m2 × v2

0 = 0.03 × 100 + 3 × v2

0 = 3 + 3 v2

V2 = – 3/3

V2 = -1 m/s

Minus sign indicates that gun moves backward with speed 1 m/s.

The recoil speed of gun is 1 m/s.

  • Force exerted on gunman due to recoil of the gun is

Force = change in momentum of gun/ time

Force = 3 × 1/ 0.003

Force = 1000 N

The force exerted on gunman due to recoil of the gun is 1000 N.

34.) Draw a diagram to show how a rocket engine provides a force to move the rocket upwards. Label the diagram appropriately.

Answer: the following diagram shows that how a rocket engine provides a force to move the rocket upwards.

35.) Name the laws involved in the following situations :

(a) the sum of products of masses and velocities of two moving bodies before and after their collision remains the same.

(b) a body of mass 5 kg can be accelerated more easily by a force than another body of mass 50 kg under similar conditions

(c) when person A standing on roller skates pushes another person B (also standing on roller skates) and makes him move to the right side, then the person A himself gets moved to the left side by an equal distance.

(d)  If there were no friction and no air resistance, then a moving bicycle would go on moving for ever.

Answer: The laws involving in the situations is,

  • The sum of products of masses and velocities of two moving bodies before and after their collision remains the same- Law of conservation of momentum.
  • A body of mass 5 kg can be accelerated more easily by a force than another body of mass 50 kg under similar conditions- Newton’s second law of motion.
  • When person A standing on roller skates pushes another person B (also standing on roller skates) and makes him move to the right side, then the person A himself gets moved to the left side by an equal distance-Newton’s third law of motion.
  • If there were no friction and no air resistance, then a moving bicycle would go on moving for ever- Newton’s first law of motion.

Long Answer Type Questions

36) (a) State and explain Newton’s second law of motion.

(b) A 1000 kg vehicle moving with a speed of 20 m/ s is brought to rest in a distance of 50 metres :

(i) Find the acceleration.

(ii) Calculate the unbalanced force acting on the vehicle.

Answer:

(a) Newton second law of motion:

Statement: Rate of change  momentum is directly proportional to applied force on it.

A body having mass m is moving with constant velocity u. Due to external force F, the body increases velocity and moves with final velocity v then according to second law of motion,

F = (mv – mu)/t

F = m(v-u)/t———1

But we know definition of acceleration

Acceleration = a = (v-u)/t

Equation 1 becomes

F = m × a

This is the another form of second law of motion.

Applied force is the product of mass and acceleration.

(b)

Given , mass of vehicle = 1000kg. Initial speed = u = 20 m/s. Distance = s = 50 m. Finally it will be in rest so final velocity =u =0 m/s.

.(i) we have to calculate acceleration

As, we know third equation of kinematics,

V2 = u2 + 2× a× s

02 = 202 + 2×a× 50

0= 400 +100a

a= -400/100

a= -4 m/s2

The acceleration produced in the vehicle is due to external force is -4 m/s2

(ii) Given, a = -4 m/s2

Mass = 1000 kg

From second law of motion,

Force = m × a

Force = 1000 ×(-4 )

Force = -4000 N.

The force needed to stop that vehicle is -4000 N.

37.) (a) Explain why, a cricket player moves his hands backwards while catching a fast cricket ball.

(b) A 150 g ball, travelling at 30 m/ s, strikes the palm of a player’s hand and is stopped in 0.05 second. Find the force exerted by the ball on the hand.

Answer:

Answer: Second law state that date of change of momentum is equal to applied force.

F = (mv-mu)/t

The cricket player increases the time for changing velocity from certain value to zero so he require minimum force.

But if he stop the ball immediately then his hand hearts due to greater value of force.

(b) Given, mass of ball = 150 g = 0.15 kg,

Initial velocity = u = 30 m/s, time = 0.05 s.

Finally ball stop so final velocity is zero.

From second law of motion,

Force =  mass × (final speed – initial speed)/time

F = 0.15 × ( 0-30)/0.05

F = 0.15× (-30/0.05

F = 0.15× (-600)

F = -90 N

Minus sign indicate that force is applied to decorate.

The force required to stop the ball is 90 N.

38.) (a) State Newton’s third law of motion and give two examples to illustrate the law.

(b) Explain why, the fireman direct a power stream of water on fire from hose pipe, the hose pipe bends to go back word.

Answer: (a) Statement of Newton’s third law of motion-

“ With every action, there is equal and opposite traction and both occurs simultaneously”.

We can calculate force in complex situation easily using this law.

Let’s understand the law with a example.

Examples:

When you throw a ball on wall then ball return back.

  • Ball exerts a force on wall is an action and wall exert force of same magnitude on ball is reaction. Action is happened on wall but reaction is on ball.There is no time lack between action and reaction.
  • When you trying to high jump then u push the earth. Pushing earth is an action but u will bounce is a reaction. Action is on earth but reaction is on you. There is no time lack between action and reaction.

(b) according to third law of motion, when fireman direct a power stream of water on fire from hose pipe then with this action there is exact reaction happened in opposite direction so hose pipe bends to go back.

39.) (a) State the law of conservation of momentum.

(b) Discuss the conservation of momentum in each of the following cases :

(i) a rocket taking off from ground.

(ii) flying of a jet Aeroplane.

Answer: (a)

Law of conservation of momentum:

The law of conservation of momentum states that total momentum is conserved.  In simple way we say that in absence of external force initial momentum is equal to final momentum.

The mathematical form of law of conservation of momentum is

Initial momentum = final momentum.

(b) (i) When rocket taking off it release the smoke in downward  direction it effect rocket moves upward. Hence total momentum is zero.

(ii) When jet Aeroplane movies forward then it release gas in backward direction. The momentum of gas in forward direction is equal to momentum of jet Aeroplane in forward direction.

40.) (a) If a balloon filled with air and its mouth untied, is released with its mouth in the downward direction, it moves upwards. Why ?

(b) An unloaded truck weighing 2000 kg has a maximum acceleration of 0.5 m/s2. What is the maximum acceleration when it is carrying a load of 2000 kg ?

Answer: When we release air from mouth of balloon indownward direction then it moves upward. The air in balloon movies with greater velocity in downward, so with same momentum balloon moves upward.

According to law of conservation of momentum if a balloon filled with air and its mouth untied, is released with its mouth in the downward direction, it moves upwards.

(b) Given, Mass of unloaded truck = m = 2000 kg.

Acceleration of unloaded truck = a = 0.5 m/s2.

Mass of load is=M= 2000 kg. And acceleration of truck with load is = A.

The force required to accelerate empty truck is,

Force = m × a

Force = 2000× 0.5

Force = 1000 N.

The acceleration of truck with load is calculate by using second law,

Acceleration  with load is = a =F/(m+M)

a= 1000/(2000+2000)

a= 1000/4000

a= 0.25 m/s2

The maximum acceleration of truck with load is 0.25 m/s2

 

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Updated: May 28, 2022 — 3:41 pm

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