Given, initial velocity, u = 150 ms−1, final velocity, v=0 (since the bullet finally comes to rest), mass of the bullet, m=10 g=0.01 kg and time taken to come to rest, t=0.03 s.
Let the acceleration of a bullet be a.
We use first equation of motion, v=u+at,
⇒0=150+a×0.03
⇒a = −1500.03=−5000 ms−2
(Negative sign indicates that the velocity of the bullet is decreasing.)
Now we use third equation of motion:
v2=u2+2as,
⇒0=1502+2×(−5000)s
⇒s=2250010000=2.25 m
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton's second law of motion:
Force, F=m×a=0.01×5000=50 N
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.