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,
[tex]0=m_{bullet}*v_{bullet} + m_{rifle}*v_{rifle} \\[/tex]
assume the bullet goes to right side and the gravitational acceleration =10 [tex]ms^{-2}[/tex]
so now the weight of the rifle=[tex]\frac{25}{10}[/tex]
[tex]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}[/tex]
this is a negative velocity to the right side. that means the rifle recoils to the left side
