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4-1.Newton's Laws of Motion
easy
A bullet of mass $50$ gram is fired from a $5 \,kg$ gun with a velocity of $1km/s$. the speed of recoil of the gun is .......... $m/s$
A
$5$
B
$1$
C
$0.5$
D
$10$
Solution
(d)According to conservation of momentum
${m_B}{v_B} + {m_G}{v_G} = 0$ ==> ${v_G} = – \frac{{{m_B}{v_B}}}{{{m_G}}}$
${v_G} = \frac{{ – 50 \times {{10}^{ – 3}} \times {{10}^3}}}{5} = – 10\,m/s$
Standard 11
Physics
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