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Question

A lead bullet of mass 2 kg, travelling with a velocity of 20 ms1, comes to rest after penetrating 20 m in a still target. Find the average acceleration (a) of the bullet and resistive force (F) applied by the target.

A
a=20 ms2​​
F=40 N
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B
a=40 ms2
F=80 N
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C
a=10 ms2
F=20 N
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D
a=0 ms2
F=0 N
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Solution

The correct option is C a=10 ms2
F=20 N
Given:
Initial velocity, u=20 ms1
Final velocity, v=0
Distance travelled, s=20 m
Mass, m=2 kg

From third equation of motion, we have
v2=u2+2as
02=202+(2×a×20)
a=10 ms2

Using Newton's second law of motion,
F=ma=2×(10)
F = 20 N
So, resistive force is 20 N.

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