Average force necessary to stop a bullet of mass 20, g moving with a speed of 250 ,m/s , as it penet

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4-1.Newton's Laws of Motion

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The average force necessary to stop a bullet of mass $20\, g$ moving with a speed of $250 \,m/s$, as it penetrates into the wood for a distance of $12\, cm$ is

A

$2.2 \times {10^3}N$

B

$3.2 \times {10^3}N$

C

$4.2 \times {10^3}N$

D

$5.2 \times {10^3}N$

Solution

(d) $u = 250\,m/s$, $v = 0$, $s = 0.12\,metre$

$F = ma = m\left( {\frac{{{u^2} – {v^2}}}{{2s}}} \right) = \frac{{20 \times {{10}^{ – 3}} \times {{(250)}^2}}}{{2 \times 0.12}}$

$\therefore F = 5.2 \times {10^3}\,N$

Standard 11

Physics

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