Question

A bullet of mass 10g travelling horizontally with a velocity of 160m/s strikes a stationary wooden block and comes to rest in 0.02s. The distance of penetration of the bullet in to the block will be

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)

v

2

=u

2

+2as

0=150

2

+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

Answer:

1.60m

Explanation:

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