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

2.1844 m/s

Step-by-step 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=
`(25)/(10)`

`0=m_(bullet)*v_(bullet)  + m_(rifle)*v_(rifle)  \\\\0=(12.70*10^(-3)) *430ms^(-1) +((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