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

- 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/s ^{2}*

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

The forces to stop both the bus in 5s = = m*a=* 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/s^{2}

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

According to the third equation of motion:

*v*^{2} = *u*^{2} + 2*as*

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/s^{2}

*F* = *ma* = 0.01 × 10000 = 100 N

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

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