Question

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

Answer 1

Explanation:

Initial velocity of the bullet, u=150  ms−1

Final velocity of the bullet, v=0

Mass of the bullet, m=10 g=0.01 kg

Time taken by the bullet to come to rest, t=0.03 s

Let distance of penetration be s

v=u+at

0=150+a(0.03)  

⟹a=−5000 ms−2    (- sign shows retardation)

v2=u2+2as

0=1502+2×(−5000)×s

s=2.25 m

Magnitude of the force exerted by the wooden block, ∣F∣=m∣a∣

∣F∣=0.01×5000=50 N

Answer 2

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

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Answer refers in attachment.

It helps you.