Question

A 4.00 kg metal cart is sitting at rest on a frictionless ice surface. Another metal cart whose mass is 1.00 kg is fired at the cart and strikes it in a one-dimensional elastic collision. If the original velocity of the second cart was 2.00 m/s, what are the velocities of the two carts after the collision?

Answer 1

The final velocity of the combined carts after the collision is 0.40 m/s while the final velocity of the 1.00 kg cart is 2.00 m/s

To solve this problem, we can apply the principles of conservation of momentum and kinetic energy. Since the collision is elastic, both momentum and kinetic energy should be conserved.

Let's denote the initial velocity of the 4.00 kg cart as v1 and the initial velocity of the 1.00 kg cart as v2. The momentum before the collision is given by: mv1 + mv2 = (4.00 kg)(0 m/s) + (1.00 kg)(2.00 m/s) = 2.00 kg*m/s.

After the collision, the two carts will move together as one system. Let's denote the final velocity of the combined carts as vf. Since we have an elastic collision, the total momentum after the collision should also be 2.00 kg*m/s: (4.00 kg + 1.00 kg)vf = 5.00 kg*vf = 2.00 kg*m/s. Therefore, the velocity of the combined carts after the collision is vf = 0.40 m/s.

Since kinetic energy is conserved in an elastic collision, we can also calculate the final velocity of the 1.00 kg cart using the kinetic energy equation: (1/2)(1.00 kg)(v2^2) = (1/2)(1.00 kg)(vf^2). Solving for vf, we find vf = 2.00 m/s.