A spring-loaded ballistic cart measuring 430 g is in contact with a second 710 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 +0.14 m/s afterward. What is the velocity of the other cart?
Let's assume the velocity of the other cart is v.
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 0 because both carts are initially at rest.
The final momentum is given by the sum of the momentum of the lighter cart (430 g) and the momentum of the other cart (710 g).
The momentum of an object is calculated by multiplying its mass by its velocity.
Therefore, the final momentum is (430 g)(0.14 m/s) + (710 g)(v).
Since the carts are initially at rest, the total momentum before the collision is 0.
Therefore, the equation becomes:
0 = (430 g)(0.14 m/s) + (710 g)(v)
To find v, we can solve this equation:
0 = (430 g)(0.14 m/s) + (710 g)(v)
0 = 60.2 g·m/s + (710 g)(v)
Rearranging the equation:
-60.2 g·m/s = (710 g)(v)
Dividing both sides by 710 g:
-0.0847676056 g/s = v
So, the velocity of the other cart is approximately -0.0848 m/s.