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Q.

A gun of mass M fires a bullet of mass m with a velocity v relative to the gun. The average force required to bring the gun to rest in 0.5 sec. is

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a

$\frac{2 Mmv}{M + m}$

b

$\frac{Mmv}{2 (M + m)}$

c

$\frac{3 Mmv}{2 (M + m)}$

d

$\frac{Mmv}{(M + m)}$

answer is A.

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Detailed Solution

$Velocity of gun = v_{1} Velocity of bullet = v - v_{1}$

$m (v - v_{1}) = {Mv}_{1} \Rightarrow v_{1} = \frac{mv}{m + M}$

The momentum imparted on the gun: P=Mv₁

$F = (\frac{\frac{Mmv}{M + m}}{0.5}) = \frac{2 Mmv}{(M + m)}$

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