A bullet of mass 10g travelling horizontally with a velocity of 150m/ s strikes a statutory wooden block and comes to rest in 0.03s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet ?

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

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

where , a is the acceleration of the bullet

0 = 150 + (a × 0.03)

Concept Note:- 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

0 = 22500 + 2×(-5000)×s

0=22500 – 10000 s

10000 s = 22500

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

From Newton’s second law of motion:

Concept Note:- Force, F = Mass Acceleration

Mass of the bullet, m = 0.01 kg

Acceleration of the bullet, a = -5000 m/s2

F = ma = 0.01×(-5000) = -50 N

Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.