A robber fires 100 bullets of mass 100gm each from a machine gun, each at a speed of 500m/s. If the first 40 bullets are fired in 10s, what is the average force exerted on the machine gun by the robber during this period?
A
250N
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B
50N
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C
2000N
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D
200N
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Solution
The correct option is D200N Mass of each bullet is m=100gm=0.1kg
Considering bullets were initially at rest inside machine gun,
Change in momentum per bullet ΔP=(mvf−mvi)=mvf−0 ∴ΔP=0.1×500=50N-s
Total number of bullets fired in t=10s is n=40
Hence average force imparted by the machine gun to the bullets is: Favg=n×ΔPΔt ⇒Favg=40×5010=200N
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 200N in order to hold the machine gun.