A bullet of mass m, flying horizontally at a velocity Vo, hits a bar lying on a smooth floor and pierces it through.

A bullet of mass m, flying horizontally at a velocity Vo, hits a bar lying on a smooth floor and pierces it through. The mass of the bar is M, the speed of the bullet after the flight is V. Determine the speed of the bar.

Given:
m is the mass of the flying bullet;
v0 – initial bullet velocity;
M is the mass of the bar;
v is the speed of the bullet after leaving the bar.
It is required to determine the speed of movement of the bar v1 after being pierced by a bullet.
According to the problem, the bar was at rest before interacting with the bullet. Then, according to the law of conservation of momentum (momentum):
m * v0 = m * v + M * v1;
M * v1 = m * v0 – m * v;
M * v1 = m * (v0 – v1);
v1 = m * (v0 – v1) / M.
Answer: the speed of movement of the bar after interaction with the bullet will be equal to v1 = m * (v0 – v1) / M.