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Question

A rifle bullet of mass 5 g shot into a still target with a velocity of 150 m s1, comes to rest in the target after penetrating about 30 cm. What is the resistive force (in N) offered by the target to the bullet?
  1. -187.5

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Solution

The correct option is A -187.5
Given
The mass of the bullet, m=5 g=5×103 kg
The initial velocity of the bullet, u=150 m s1
As the bullet came to rest in the target, its final velocity, v=0
The penetrated length in the target, s=30 cm=30×102 m
According to the third equation of motion,
v2u2=2as, as the body decelerates once it hits the target.
0(150)2=2a×30×102

a=150×15060×102

a=37,500 m s2
The opposing force exerted by the target is given by Newton's second law of motion: Force acting on a body accelerates or decelerates the body. In this case, the opposing force decelerates the bullet, and forces it to come to rest.
Force F=-ma
F=5×103×37500
F=187.5 N

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