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Q.
A man fires a bullet of mass $200\,gm$ at a speed of $5\,m/s$. The gun is of one $kg$ mass. By what velocity the gun rebounds backward?
AIPMTAIPMT 1996Laws of Motion
Solution:
Mass of bullet $(m_1)=200\, gm=0.2\,kg;$
Speed of bullet $(v_1) = 5 \,m/sec$. and mass of gun $(m_2) = 1 \,kg.$
Before firing, total momentum is zero.
$\therefore $ After firing total momentum is $m_1 v_1 + m_2 v_2$
From the law of conservation of momentum $m_1 v_1 + m_2 v_2=0$
or $v_2=\frac{-m_1v_1}{m_2}$
$=\frac{-0.2\times5}{1}$
$-1 \,m/\sec.$