Question

A bullet of mass 10g traveling horizontally with a velocity of 150ms-1 strikes a stationary wooden block and comes to rest in0.03s. Cal.the distance of penetration of the bullet into the block.Also.Cal.the magnitude of the force exerted by the wooden block on the bullet..

Answer 1

Answer:

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

Explanation:

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