A spring-loaded ballistic cart measuring 100 g is in contact with a second 540 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 +0.96 m/s afterward. What is the velocity of the other cart?
Since the carts are initially at rest and in contact, the total momentum of the system before the spring is released is zero.
After the spring is released, the lighter cart moves with a velocity of +0.96 m/s. Let's denote the velocity of the heavier cart as v.
Using the principle of conservation of momentum, the total momentum of the system after the spring is released is given by the sum of the momenta of the two carts:
(0.1 kg)(0.96 m/s) + (0.54 kg)(v) = 0
0.096 + 0.54v = 0
0.54v = -0.096
v = -0.096 / 0.54
v ≈ -0.178 m/s
Therefore, the velocity of the other cart is approximately -0.178 m/s.