posted by Joan .
A physics teacher shoots a .30 caliber rifle at a 0.47 kg block of wood. The rifle and wood are mounted on separate carts that sit atop an air track (like a linear air hockey table--ie. frictionless). The 5.7 kg rifle fires a 27 gram bullet at 227 m/s in the positive direction.
What would be the velocity of the rifle after the bullet is fired?
And, what would be the velocity of the block of wood with the bullet lodged inside?
The total momentum of gun and bullet after firing remains zero. That means the momentum of the bullet and the gun are equal and opposite in sign.
Let V be the recoil velocity of the gun
5.7*V = 0.027*227
Solve for V.
Momemtum of bullet and block are the same before and after impact.
227*0.027 = (0.47 +0.027)*Vfinal
Solve for Vfinal/
V = 1.20 m/s
Vfinal = 12.33 m/s