Force and laws of motion Worksheet-4

Force and laws of motion Worksheet-4

The car A of mass 1500 kg, traveling at 25 m/s collides with another car B of mass 1000 kg traveling at 15 m/s in the same direction. After collision the velocity of car A becomes 20 m/s Calculate the velocity of car B after the collision.

A. 22.5 m/s B. 225 m/s

C. 82.5 m/s D. None of these

A bullet of mass 10 moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900 m/s. what is the velocity acquired by the block ?

A. 4.4 m/s B. 44 m/s C. 20 m/s D. 5.6 m/s

The rockets work on the principle of conservation of :

A. mass B. Energy C. Momentum D. Velocity

An object of mass 2 kg is sliding with a constant velocity of 4 m/s on a frictionless horizontal table. The force required to keep this object moving with the same velocity is :

A. 32 N B. 0 N C. 2 N D. 8 N

According to the third law of motion, action and reaction :

A. Always act on the same body but in opposite direction

B. Always on different bodies in opposite directions

C. Have same magnitudes and directions

D. Act on either body at normal to each other

The unit of measuring momentum of a moving body is :

A. m s^–1 B. kg.ms^–1 C. kg.ms^–2 D. Nm²kg^–2

A boy of mass 50 kg standing on ground exerts a force of 500 N on the ground. The force exerted by the ground on the boy will be :

A. 50 N B. 2500 N C. 10 N D. 500 N

Car A (1150 kg), Car B (720 kg), Car C (600 kg) and Car D(2500 kg), all are running at the same speed of 50 m/s under identical conditions. If all these cars are hit from behind with the same force and they continue to move forward, the maximum acceleration will be produced in :

A. A B. B C. C D. D

The acceleration produced by a force of 5 N acting on a mass of 20 kg in m/s² is :

A. 4 B. 100 C. 0.25 D. 2.5

Which of the following situations involves newton’s second law of motion ?

A. A force can stop a lighter vehicle as a heavier vehicle which are moving

B. A force can accelerate a lighter vehicle more easily than a heavier vehicle which are moving

C. A force exerted by a lighter vehicle on collision with a heavier vehicle results in both the vehicles coming to a standstill

D. A force exerted by the escaping air from a balloon in the downward direction makes the balloon to go upwards

A fielder pulls his hands backwards after catching the cricket ball. This enables the fielder to :

A. Exert larger force on the ball

B. Reduce the force exerted by the ball

C. Increase the rate of change of momentum

D. Keep the ball in hands firmly

Answer:

Explanation: First let’s calculate the total momentum of both the cars, before and after the collision.

(a)

= 1500 × 25

= 37500 kg.m/s

= 1000 × 15

= 15000 kg.m/s

= 37500 + 15000

= 52500 kg.m/s ……….(1)

(b) After collision, the velocity of car A is 20 m/s.

Momentum of car A = 1500 × 20

(after collision) = 30000 kg.m/s

After collision, suppose the velocity of car B is v m/s

Momentum of car B = 1000 × v

(after collision) = 1000 v kg.m/s

Total momentum of car A and car B

= 30000 + 1000 v ……….(2)

(after collision)

According to the law of conservation of momentum :

52500 = 30000 + 1000 v

1000 v= 52500 – 30000

1000 v = 22500

v = 22.5 m/s

hence the velocity of car B after the collision will be 22.5 m/s.

Explanation: Mass of the bullet, m₁ = 10 g

= (10/1000)kg

= 0.01 kg

velocity of the bullet, v₁= 400 m/s

Momentum of the bullet = m₁ × v₁

= 0.01 × 400 kg.m/s ……..(1)

Now, the bullet gets embedded into a wooden block of mass 900 g. Now the mass of wooden block along with the embedded bullet will be 900 + 10 = 910 g.

Mass of wooden block + Bullet, m₂ = 900 + 10

= 0.91 kg

Velocity of wooden block + bullet, v₂ = ?

Momentum of wooden block + bullet = m₂ × v₂

= 0.91 × v₂ kg.m/s ……….(2)

According to the law of conservation of momentum, the two momenta as given by equations (1) and (2) should be equal.

m₁ × v₁ = m₂ × v₂

0.01 × 400 = 0.91 × v₂

= 4.4 m/s

Hence the velocity acquired by the wooden block (having the bullet embedded in it) is 4.4 metres per second.

Explanation: Acceleration is inversely proportional to mass if force is constant.

Explanation: Using relation F = ma where m is mass and a is acceleration.