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)
A = -150/0.03 = – 5000 m/s²
(Negative sign indicates that the velocity of the bullet is decreasing.)
According to the third equation of motion:
v²= u² + 2as
0 = (150)² + 2 ( – 5000) s
S = – (150)²/-2(5000) = 22500/10000 = 2.25m
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/s²
F = ma = 0.01×5000 = 50 N

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