Two air carts of mass = 0.85kg and = 0.49kg are placed on a frictionless track.?

Question

Cart 1 is at rest initially, and has a spring bumper with a force constant of 610 N/m . Cart 2 has a flat metal surface for a bumper, and moves toward the bumper of the stationary cart with an initial speed = 0.68m/s . Assume that positive -axis is directed toward the direction of motion of cart 2.
a.) How much energy is stored in the spring bumper when the carts have the same speed?
b.) What is the final speed of the cart 1 after the collision?
c.) What is the final speed of the cart 2 after the collision?

Answer 1

This is a conservation of problem - no friction so the total energy is constant throughout as is the total momentum.

The three energies are the kinetic energies of the two carts and the potential energy of the spring. The spring starts out relaxed so it has no potential energy.

Kinetic energy = (1/2)MV^2

Cart 1 starts out at rest so it has no kinetic energy

You know the initial velocity and mass of cart 2 so you can compute its kinetic initial energy energy which is the total energy in the system Et.

When everything is done, the spring is again relaxed, but now both carts are moving. Let's denote their velocities by V1 and V2. So the sum of the two kinetic energies must equal the total energy.

Also, we have conservation of momentum, so the momentum in must equal the momentum out.

momentum = MV

For cart 1 the initial velocity was 0 so its momentum was too. We have the mass and initial velocity of cart 2 so we can compute its initial momentum, which is the momentum of the entire system, Pt.

So we have Pt = M1V1 + M2V2

Two equations in two unknowns give us enough data to compute V1 and V2, the final velocities of the carts.

Now comes the interesting part, what happens as the two carts collide and the spring is being deflected.

Consider what happens as cart 2 hits the bumper of cart 1. The spring starts deflecting, and in doing so, imparts a force to both cart 1 and cart 2. The force is in a direction so as to reduce the speed of cart while increasing the (initially 0) speed of cart 1. In the process, the spring is deflected.

Eventually the two carts have the same speed. This is not their final speeds because the spring, still being deflected, continues to exert force on each.

Because the spring is deflected, it is hard to use conservation of energy directly. But we can still use conservation of momentum.

We know the total momentum and we know the two masses. When they carts have the same velocity, there is only one unknown - that velocity - so we can compute it.

Knowing that velocity, we can compute the two kinetic energies. That tells us the energy left over that has to be in the spring.