Question

A bullet of mass 10g travelling horizontally with the velocity 150m/s-1 strikes a stationary 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 force exerted on the wooden block by the bullet. ​

Answer 1

Answer:

50n

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

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