A lead bullet of mass 2kg, travelling with a velocity of 20ms−1, comes to rest after penetrating 20m in a still target. Find the average acceleration (a) of the bullet and resistive force (F) applied by the target.
A
a=20ms−2 F=40N
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B
a=40ms−2 F=80N
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C
a=−10ms−2 F=20N
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D
a=0ms−2 F=0N
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Solution
The correct option is Ca=−10ms−2 F=20N Given:
Initial velocity, u=20ms−1
Final velocity, v=0
Distance travelled, s=20m
Mass, m=2kg
From third equation of motion, we have v2=u2+2as 02=202+(2×a×20) ⇒a=−10ms−2
Using Newton's second law of motion, F=ma=2×(−10) F=−20N
So, resistive force is 20N.