1 .A car of mass 1000kg and a bus of mass 8000kg are moving with the same velocity of 36

km/h. Find the forces to stop both the car and the bus in 5s.

2. A mechanic strikes a nail with a hammer of mass 500g moving with a velocity of 20m/s The

hammer comes to rest in 0.02s.after striking the nail. Calculate the force exerted by the nail on

the hammer.

3. A bullet of mass 10g traveling with a velocity of 100 m/s penetrates in a wooden plank and is

brought to rest in 0.01s Find (a) the distance through which the bullet penetrates in the wooden

plank and (b) the force exerted on the bullet.

1. Initial velocity = 36 km/h =10 m/s

Final velocity = 0 m/s

Time = 5 sec

According to the first equation of motion, v = u + at

0 =10+a×5

a = -2 m/s²

The forces to stop both the car in 5s = ma= 1000×2 = 2000 N

The forces to stop both the bus in 5s = = ma= 8000×2 = 16000 N

2. Mass of the hammer, m = 500 g = 0.5 kg

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

Time taken by the nail to the stop the hammer, t = 0.02 s

Velocity of the hammer, v = 0 (since the hammer finally comes to rest)

From Newton’s second law of motion:

= 500 N

The hammer strikes the nail with a force of −500 N. Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., +500 N.

3. Now, it is given that the bullet is travelling with a velocity of 100 m/s.

Thus, when the bullet enters the block, its velocity = Initial velocity, u = 100 m/s

Final velocity, v = 0 (since the bullet finally comes to rest)

Time taken to come to rest, t = 0.01 s

According to the first equation of motion, v = u + at

Acceleration of the bullet, a

0 = 100 + (a ×0.01 s)

a = -100/0.01= -10000 m/s²

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

According to the third equation of motion:

v² = u² + 2as

0 = (100)² + 2 (−10000) s

s = -10000/20000= 0.5 m

Hence, the distance of penetration of the bullet into the block is 0.5 m.

From Newton’s second law of motion:

Force, F = Mass × Acceleration

Mass of the bullet, m = 10 g = 0.01 kg

Acceleration of the bullet, a = 10000 m/s²

F = ma = 0.01 × 10000 = 100 N

Hence, the magnitude of force exerted by the wooden block on the bullet is 100 N.