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Question

Physics Question on laws of motion

A man fires a bullet of mass 200gm200\,gm at a speed of 5m/s5\,m/s. The gun is of one kgkg mass. By what velocity the gun rebounds backward?

A

1 m/s

B

0.01 m/s

C

0.1 m/s

D

10 m/s

Answer

1 m/s

Explanation

Solution

Mass of bullet (m1)=200gm=0.2kg;(m_1)=200\, gm=0.2\,kg;
Speed of bullet (v1)=5(v_1) = 5 m/sec. and mass of gun (m2)=1kg.(m_2) = 1 \,kg. Before firing, total momentum is zero.
\therefore After firing total momentum is m1v1+m2v2m_1 v_1 + m_2 v_2
From the law of conservation of momentum m1v1+m2v2=0m_1 v_1 + m_2 v_2=0
or v2=m1v1m2v_2=\frac{-m_1v_1}{m_2}
=0.2×51=\frac{-0.2\times5}{1}
1m/sec.-1 \,m/sec.