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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 N
Given:
Mass of bullet; m=10×103 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(vfvi)|=mvi
=10×103×100
=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=v2i2aS
.a=v2i2S=(100)22×0.06=8.3×104 m/s2

From : vf=viat
t=via=1008.3×104=1.2×103 s

|Average force| experienced by bullet =|Impulse|time =11.2×103=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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