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Question

A 10 g bullet moving at 100 m/s strikes a log. Assume that bullet undergoes uniform deacceleration and stops in 6.0 cm, find the magnitude of average force experienced by the log.

A
600 N
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B
1233.33 N
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C
833.3 N
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D
1266.66 N
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Solution

The correct option is C 833.3 NGiven: Mass of bullet; m=10×10−3 kg Initial velocity of bullet; vi=100 m/s As bullet is coming to rest. ∴Final velocity of bullet; vf=0 m/s Using Impulse |J|=|ΔP| =|m(vf−vi)|=mvi =10×10−3×100 =1 N-s We know, |Average force| =|Impulse|time Given that, bullet stops after travelling S=6.0 cm with initial velocity of vi=100 m/s From : v2f=v2i−2aS .∴a=v2i2S=(100)22×0.06=8.3×104 m/s2 From : vf=vi−at t=via=1008.3×104=1.2×10−3 s ∴|Average force| experienced by bullet =|Impulse|time =11.2×10−3=833.3 N By Newton's 3rd law of motion, we can say log will also experience average force of same magnitude but in opposite direction. Average force experienced by log = 833.3 N

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