Q: A gun of mass M fires a bullet of mass m with maximum speed v. Given that m < M .The kinetic energy of the gun will be
(a) 1/2 mv2
(b) 1/2 Mv2
(c) more than 1/2 mv2
(d) less than 1/2 mv2
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Ans: (d)
Sol: $\displaystyle 0 = m \vec{v} + M \vec{v_1}$
$ v_1 = – \frac{m}{M} v $
Kinetic energy of the gun will be
$ K.E = \frac{1}{2} M v_1^2 $
$ K.E = \frac{1}{2} M (\frac{m}{M} v)^2 $
$ K.E = \frac{1}{2 M} m^2 v^2 $
$ K.E = \frac{m}{M} ( \frac{1}{2 } m v^2 ) < \frac{1}{2 } m v^2 $