Force and Laws of Motion Extra Questions and Answers Class 9 Science Chapter 9

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Force and Laws of Motion Extra Questions and Answers Class 9 Science Chapter 9

Force and Laws of Motion Extra Questions for Very short

1.) Why does bicycle slow down when you stop pedalling?

Answer: Frictional force exerts in opposite direction so it slows down bicycle when we stop pedaling.

2.) Why does the ball accelerate when we release it from certain height?

Answer: The ball released from certain height get accelerate because of gravitational force.

3.) If net force acting on the object is zero then calculate its acceleration.

Answer: According to 1st law of motion, if net force is zero then acceleration of that object also zero.

4.) State the direction of Frictional force.

Answer: Frictional force opposes the motion therefore it’s direction is always opposite to displacement.

5.) How do we minimize the effect of frictional force in machines?

Answer: We minimize the effect of frictional force using lubricants.

6.) On what factor does the inertia depends?

Answer: Inertia is depend on mass of that object.

7.) Give the example of third law of motion.

Answer: Example of third law of motion-

A ball hits on wall and returned back.

8.) State the mathematical form of second law of motion.

Answer: Mathematical form of second law of motion-

F = m × a

Where

F- Net force

m – mass of the object

a – acceleration.

9.) If a book of mass 0.7 kg is put on table then calculate gravitational force acting on that book. (g = 10 m/s²)

Answer: Given, Mass = 2 kg, g = 10 m/s²,

According to second law of motion,

F = m × a

F = 0.7 × 10

F = 7 N.

10.) If a book of mass 0.7 kg is put on table then calculate net force acting on that book.

Answer: The book is at rest so according to 1st law of motion, Net force on book is zero.

11.) What is the sign of acceleration of ball which thrown vertically upward.

Answer: The speed of ball decreases so it’s acceleration becomes negative.

12.) Which concept is introduced in second law of motion?

Answer: Momentum is introduced in second law of motion.

13.) How will you calculate momentum of the object of mass ‘m’ moving with velocity ‘v’.

Answer: Momentum of the object is calculated by,

Momentum = mass × velocity

P = m × v.

14.) If the object of mass 4 kg is moving with velocity 12 m/s then calculate it’s momentum.

Answer: Given, Mass = 4 kg,  velocity = 12 m/s.

As we know,

Momentum = mass × velocity

Momentum = 4 × 12

Momentum = 48 kg m/s.

15.) Is momentum depend on direction?

Answer: yes, momentum is depend on direction of velocity.

Force and Laws of Motion Extra Questions for Short Type

1.) Explain balanced force with example.

Answer:  If sum of forces acting on a body is zero then such forces is called as balanced forces.

Example-Two boyspull a pen in opposite direction and the pen does not move either side. In this case the forces acting on the pen are balanced forces.

2.) Why does external unbalanced force not require for uniform motion?

Answer: According to first law of motion, external force is required to change the state of object. If the object is at rest then external unbalanced force changes that state. Also if the object is in motion then external unbalanced force changes it’s speed or direction. So according to second law of motion external unbalanced force is not required to keep a body in uniform motion.

3.) A stone of mass 7 kg is dropped from height of 5 m. Calculate momentum of stone when it reaches at ground. (g = 10 m/s²)

Answer: Given, Mass = m = 7 kg, Height = s = 5 m,

g = 10 m/s², It dropped from certain height so initial velocity = 0 m/s.

We must know the velocity when itreaches at ground forcalculating momentum.

According to third equation of motion,

v² = u² + 2as

v² = 0² +2 × 10 × 5

v² = 100

Taking square root on both side,

v = 10 m/s.

As we know,

Momentum = m× v

Momentum = 7 × 10

Momentum = 70 kg m/s.

Momentum of the stone will be 70 kg m/s.

4.) Explain, how can we jump on the basis of third law of motion.

Answer: Third law of motion states that, with every action there is equal and opposite reaction and both occurs simultaneously.

When we push the earth downward then earth can push us upward. Pushing earth downward is an action and earth pushes us upward is a reaction and both occurs simultaneously. The magnitude of action and reactions must be same.

5.) A car having mass 10000 kg moving initially with speed of 90 km/h. Suddenly driver breaks and it stop after 5 seconds. Calculate force exerted by breaks.

Answer: Given, Mass of the car = 10000 kg,

Initial speed = u = 90 km/h = 90 × 5 / 18 m/s

u = 25 m/s,

Finally it stops so final velocity becomes zero.

v = 0 m/s, time = t = 5 s.

Firstly we have to calculate acceleration of the car.

According to 1st equation of motion,

v = u + at

0 = 25 + a × 5

a = -25/5

a = -5 m/s².

Minus sign indicates the retardation.

Now we calculate force using second law of motion.

Force = mass × acceleration

Force = 10000 × (-5)

Force = – 50000 N

Minus sign indicates that force opposes the motion.

The force exerted by breaks is 50000 N.

6.) Why is it advised to wear seat belts while driving a car?

Answer: We can not drive car with constant velocity. Sometimes we accelerate it when road is clear and we breaksit when road is crowded.  Also we know that our body has inertia because of mass. So body does not accept this change immediately when we breaks or accelerate the car. Therefore we must wear a seat belt while driving.

7.) Is speed of bullet depend on mass of gun from which it fired? Explain the answer.

Answer: Yes,  speed of bullet is depend on mass of gun from which it fired. According to conservation of linear momentum, total momentum of gun and bullet is zero. Hence the product of speed and mass of bullet is equal to product of mass and speed of gun. Gun and bullet are at rest before firing so the sum of initial momentum is zero.

According to law of conservation of momentum,

Initial momentum =final momentum.

0 = final momentum of gun + final momentum of bullet.

0 = mass of gun × speed of gun + mass of bullet + speed of bullet

On basis of above formula we conclude that, speed of bullet is depend mass of gun

We must increase the mass of gun for greater speed of bullet.

8.) If a ball of mass 200 gm strike on a  wall with a speed 20 m/s and return back with same speed then calculate change in momentum.

Answer: Given, Mass of ball = m = 200 gm = 0.2 kg,

Initial speed = u = 20 m/s,

It return back with same speed but direction is opposite , so final speed = v = – 20 m/s.

According to law of conservation of momentum,

Change in momentum = final momentum – initial momentum.

Change in momentum = mv – mu

Change in momentum = m (v-u)

Change in momentum = 0.2 ( -20-20)

Change in momentum = 0.2 ×(-40)

Change in momentum = -8 kg m/s.

Thus change in momentum will be -8 kg m/s.

9.) Calculate the momentum of a ball of mass 300 gm when it fall from height of 800 m. (Take g = 10 m/s²)

Answer: Given, Mass = m = 300g = 0.3 kg,

Height = s = 80 m.

Initial velocity = u=  0 m/s.

Acceleration = 10 m/s².

We calculate final velocity using 3rd law,

v² = u² + 2as

v² = 0² + 2 × 10 ×80

v² = 1600

Taking square root on both side,

v = 40 m/s.

The final velocity of ball is 40 m/s.

Now we calculate momentum,

As,

Momentum = mv – mu

Momentum = 0.3× 40 – 0.3 × 0

Momentum = 12 kg m/s.

The momentum of ball is 12 kg m/s.

10.) A car of mass 9000 kg start from rest and gain velocity 20 m/s in 5 s. Calculate force exerted by engine.

Answer: Given, Mass = 9000 kg, Initial velocity = 0 m/s,

Final velocity = 20 m/s, time = 5 s.

We calculate acceleration using first law of motion,

v = u + at

20 = 0 + 5a

a = 20/5 m/s.

Acceleration = 4 m/s².

The acceleration of the car is 4 m/s².

According to second law of motion,

Force = mass × acceleration

Force = 9000 × 4

Force = 36000 N.

The force exerted by engine is 36000 N.

11.) A bullet of mass 50 gm fired from gun of 10 kg and moves with velocity 200 m/s strike on stationary wooden block.The bullet penetrate the wooden block and stop after 200 m. Calculate time required to stop the bullet and force exerted by wooden block on bullet.

Answer: Given, Mass of bullet = 50 g = 0.05 kg,

Initial velocity = u = 200 m/s, Finally it stop after penetration so final velocity becomes zero.

Final velocity = v = 0 m/s

Displacement = s = 200 m.

We calculate acceleration using 3rd equation of motion

v² = u² + 2as

0²= 200² + 2 a ×200

0 = 40000 + 400a

-40000/400= a

a = -100 m /s².

Minus sign indicates decreasing velocity.

Now we calculate force exerted by wooden block,

According to second law of motion,

Force = mass × acceleration

Force = 0.05 × 100

Force = 5 N.

The wooden block exerts a force of 5 N on bullet.

12.) A bullet of mass 100 gm is fired from gun of mass 4 kg. If the velocity of bullet is 100 m/s them calculate recoil velocity of gun.

Answer: Given, Mass of bullet = m = 100 gm = 0.1 kg,

Mass of gun = M = 4 kg,

Initially both at rest so initial momentum of gun and bullet is zero.

Final velocity of bullet = 100.m/s,

Let Assume, Final velocity of bullet = v m/s,

Final velocity of gun = V m/s .

We can find recoil velocity of gun using law of conservation of momentum,

Momentum before fire = momentum after fire

0 = momentum of bullet + momentum of gun

0 = mv + MV

0 = 0.1 × 100 + 4V

0 = 10 + 4V

V = -10 /4

V = -2.5 m/s.

Minus sign indicates that the velocity of gun and bullet is opposite in direction.

Thus the recoil velocity of gun is 2.5 m/s.

13.) An astronaut jump from spaceship in interstellar space and moves with 20 m/s. Calculate its speed after 2 minutes.

Answer:  An astronaut jumps from spaceship in interstellar space means  there will be no force. The total force acting on the astronaut is zero.

According to first law of motion, if external forces acting on the object is zero then the value of acceleration also zero.

Thus the force in interstellar space is zero, Acceleration also becomes zero.

So there is no change in magnitude as well as direction of velocity.

Thus the astronaut moves with constant velocity of 20 m/s.

14.) (a) State second law of motion.

(b) If a ball of mass 6 kg throws by exerting force of 12 N then calculate acceleration produced in that ball.

Answer: (a) Second law of motion states that,

“Rate of change of momentum is equal to applied forces on it”.

The mathematical form of second law of motion is as,

Force = mass × acceleration.

(b) Given,

Mass of ball = 6 kg, Force = 12 N.

According to second law of motion,

Force = mass × acceleration

Acceleration = force / mass

Acceleration= 12/6

Acceleration = 2 m/s².

The acceleration of ball is 2 m/s².

Force and Laws of Motion Extra Questions for Long Type

1.) (a) What is the SI unit of force?

(b) Converts 15 kg m/s² into gram cm /s².

Answer: (a) SI unit of force is kg m/s² or N.

(b) Given 1 kg m/s².

As we know,

15 kg =15×10³ g,

15 m = 15× 10² cm.

Therefore,

15 kg m/s² = 15 × 10³ × 10² g cm/s².

15 kg m/s² = 15 ×10⁵ g cm /s².

2.) How did 1st law of motion discover? Also state and explain this law.

Answer: Aristotle studied the motion in scientific way. He was trying to resolve question below-

Is an external force required to keep a body in uniform motion?

He put his theory on the basis of common example but felt.

After that Galileo Galilei explained the answer of above question. He corrected theory of Aristotle and put correct explanation.

He introduced the concept of frictional force which opposes the motion.

Galileo concluded that-

  • State of rest and state of uniform motion are equivalent.
  • Force is required to change the state of object.
  • The force responsible for change the state must be unbalanced and external.
  • The body does not changes it state by itself.

1st law of motion states that,

“An object is in state of rest or state of uniform  motion until we act an external unbalanced force on it”.

Example-1)  A book on table will be remains in same state until someone moves it.

2) A ball moving with uniform velocity does not stop until some one exert external unbalanced force on it.

3.) Which thing has greater inertia

i.) Rubber ball

ii.) Stone of 5 kg

iii.) Refrigerator

(b) A stone of mass 1 kg rolls on the floor with initial speed 30 m/s and stop after 5 seconds. Calculate magnitude and direction of Frictional force.

Answer: (a) Inertia is a property of material which depend on mass of the object. Thus mass of refrigerator is greater than others. So momentum of refrigerator is greater.

(b) Given, initial speed =u = 30 m/s,

Finally it stop, so final speed =v = 0 m/s,

Time = 5 s, Mass of stone = 1 kg

Firstly we calculate acceleration produced in stone by Frictional force.

According to 1staw of motion,

v = u + at

0 = 30 + a×5

a = -30/5

a = -6 m/s².

The acceleration of stone is -6 m/s².

As,

Force = mass × acceleration

Force = 1 ×(-6)

Force = -6 N.

The magnitude of force is 6 N and it exert in opposite directions of motion.

4.) Car A having mass 10000 kg moves with velocity 20 m/s and car B having mass 20000 kg moves with velocity 15 m/s. Which car has greater momentum?

Answer: Given, Mass of car A = m = 10000 kg,

Velocity of car A = v = 20 m/s.

Mass of car B= M = 20000 kg,

Velocity of car B = V= 15 m/s.

Firstly we calculate momentum of car A,

Momentum of car A= mass × velocity

= 10000 × 20

= 200000 kg m/s

The momentum of car A is 200000 kg m/s.

Similarly we calculate momentum of car B ,

Momentum of car B = mass × velocity

= 20000 × 15

= 300000 kg m/s

The momentum of car B is 300000 kg m/s.

Thus momentum of car B is greater than car A.

5.) A bullet of mass 40 gm is fired from gun and moves with 160 m/s. It penetrate the wooden block of mass 5 kg and moves together.  Calculate velocity of wooden block and bullet. (Ignore air resistance)

Answer: Given, Mass of bullet = m = 40 gm = 0.04 kg,

Mass of wooden block = M = 5 kg.

Initial velocity of bullet = u = 160 m/s,

The wooden block initially at rest so

Initial velocity of wooden block = U = 0 m/s.

Finally both moves with same speed therefore

Final velocity of bullet = final velocity of wooden block = v m/s.

According to conservation of linear momentum,

Initial momentum of bullet + Initial momentum of wooden block= final momentum of bullet + final momentum of wooden block

mu + MU = mv + Mv

0.04 × 160 + 5 × 0 = 0.04v + 5v

6.4 +0 = v(0.04 + 5)

6.4 = v (5.04)

v = 6.4 / 5.04

v = 1.27 m/s.

The wooden block and bullet moves with 1.27 m/s.


Updated: September 11, 2021 — 11:40 am

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