Now, it is given that the bullet is travelling with a velocity of 150 m/s.

Thus, when the bullet enters the block, its velocity = Initial velocity, *u *= 150 m/s

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

Time taken to come to rest, *t *= 0.03 s

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

Acceleration of the bullet, *a *

0 = 150 + (*a *× 0.03 s)

(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 = (150)^{2} + 2 ( – 5000) s

Hence, the distance of penetration of the bullet into the block is 2.25 m.

From Newton’s second law of motion:

Force, *F *= Mass x Acceleration

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

Acceleration of the bullet, *a *= 5000 m/s^{2}

*F *= *ma = 0.01×5000 = 50 N*