Question

A rifle with a weight of 25 N fires a 4.0 g bullet with a speed of 290 m/s.

Answer 1

Answer:

Explanation:

Given

Weight of rifle $W=25\ N$

mass of rifle $M=\frac{25}{10}=2.5\ kg$

mass of bullet $m=4\ gm$

speed of bullet $u=290\ m/s$

As there is no net force on the bullet-rifle system therefore momentum is conserved

$0=Mv+mu$

$v=-\frac{mu}{M}$

$v=-\frac{4\times 10^{-3}\times 290}{2.5}$

$v=-0.464\ m/s$

i.e. opposite to the direction of bullet speed

(b)Weight of Man $=750\ N$

Combined weight of man and rifle$=750+25=775\ N$

mass of man-rifle system $M'=\frac{775}{10}=77.5\ kg$

Now man and rifle combinedly recoil

therefore

$0=(M')v '+mu$

$v'=-\frac{mu}{M'}$

$v'=-\frac{4\times 10^{-3}\times 290}{77.5}$

$v'=0.0149\ m/s$