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Question

A robber fires 100 bullets of mass 100 gm each from a machine gun, each at a speed of 500 m/s. If the first 40 bullets are fired in 10 s, what is the average force exerted on the machine gun by the robber during this period?


A
250 N
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B
50 N
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C
2000 N
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D
200 N
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Solution

The correct option is D 200 N
Mass of each bullet is m=100 gm=0.1 kg
Considering bullets were initially at rest inside machine gun,
Change in momentum per bullet ΔP=(mvfmvi)=mvf0
ΔP=0.1×500=50 N-s

Total number of bullets fired in t=10 s is n=40
Hence average force imparted by the machine gun to the bullets is:
Favg=n×ΔPΔt
Favg=40×5010=200 N

By Newton's 3rd law, the gun will also experience the same amount of force in a direction opposite to the firing direction.
i.e the robber will have to apply a force 200 N in order to hold the machine gun.

Physics

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