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Question

A gun weighing 4 kg fires a bullet of 40 g with a velocity of 500 ms−1. The velocity with which gun recoils is

A
5 ms1
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B
10 ms1
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C
20 ms1
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D
4 ms1
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Solution

The correct option is A 5 ms1
Mass of gun, m1=4 kg.
Let recoil velocity of gun be v1, initial velocity of gun be u1, initial velocity of bullet be u2
Mass of bullet, m2=40 g=0.04 kg
Velocity of bullet after firing be v2=500 ms1.

Using law conservation of momentum :
m1u1+m2u2=m1v1+m2v2 ...(1)
Total momentum of gun and bullet initially is zero, as the bullet is in rest before firing, i.e initial velocities of bullet and gun are 0 ms1
(m1+m2)×0 = m1v1+m2v2
v1=m2v2m1
Here the negative sign suggests that this velocity is in the opposite direction to the motion of bullet.
Recoil velocity, v1=m2v2m1=0.04×5004=5 ms1

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