spring-loaded ballistic cart measuring 490 g is in contact with a second 880 g cart. The carts are initially at rest on a level surface. The spring is released and the lighter cart is observed to move at +0.72 m/s afterward. What is the velocity of the other cart?

Answer 1

According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

The initial momentum is zero since both carts are at rest. Therefore, the final momentum is also zero.

Let v2 be the velocity of the heavier cart after the collision.

The momentum of the lighter cart (490 g) is given by:

(0.490 kg)(0.72 m/s) = 0.353 kg·m/s

The momentum of the heavier cart (880 g) is given by:

(0.880 kg)(v2) = 0.880v2 kg·m/s

Since the total momentum after the collision is zero, the equation becomes:

0.353 kg·m/s + 0.880v2 kg·m/s = 0

Simplifying the equation:

0.880v2 kg·m/s = -0.353 kg·m/s

Solving for v2:

v2 = -0.401 m/s

Therefore, the velocity of the other cart is -0.401 m/s.