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Q. A bullet of mass $0.04\, kg$ moving with a speed of $90\, ms ^{-1}$ enters a heavy wooden block and is stopped after a distance of $60\, cm$. The average resistive force exerted by the block on the bullet is

Laws of Motion

Solution:

$a=\frac{-u^{2}}{2 s}=\frac{-90^{2} ms ^{-2}}{2(0.6)}=-6750\, ms ^{-2}$
$F=m a=(0.04 kg )\left(-6750 ms ^{2}\right)=-270\, N$
$-ve$ sign indicates retarding force.