Question

A 12.70 g bullet has a muzzle velocity (at the moment it leaves the end of a firearm) of 430 m/s when rifle with a weight of 25.0 N. Calculate the recoil speed (in m/s) of the rifle.

Answer 1

Answer:

2.1844 m/s

Explanation:

The principle of conservation of momentum can be applied here.

when two objects interact, the total momentum remains the same  provided no external forces are acting.

Consider the whole system , gun and bullet. as an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum,

$0=m_{bullet}*v_{bullet} + m_{rifle}*v_{rifle}$

assume the bullet goes to right side and the gravitational acceleration =10 $ms^{-2}$

so now the weight of the rifle=$\frac{25}{10}$  

$0=m_{bullet}*v_{bullet} + m_{rifle}*v_{rifle} \\\\0=(12.70*10^{-3}) *430ms^{-1} +(\frac{25}{10} )*v_{rifle} \\v_{rifle} =-2.1844ms^{-1}$

this is a negative velocity to the right side. that means the rifle recoils to the left side