1. Jun 23, 2011

### AJKing

1. The problem statement, all variables and given/known data

An empty ore cart coasts at a constant horizontal speed under an ore chute. Ore falls vertically into the car. What effect does the addition of the ore have on the kinetic energy of the loaded car?

2. Relevant equations

$KE = \frac{1}{2}mv_{o}^2 - \frac{1}{2}mv^2$

3. The attempt at a solution

The solution given:

I don't understand this. Shouldn't the added ore increase the carts mass, thus altering its Kinetic Energy?

Not to mention the fact that dropping a mass onto a moving object would surely increase its frictional force, altering its velocity.

Am I misunderstanding something? Is this an isolated scenario and the question is just hiding that from me?

2. Jun 23, 2011

### AJKing

I think I understand now.

Let me know if this is wrong:

$KE = \frac{1}{2}m_{o}v_{o}^2 = \frac{1}{2}mv^2$

Thus, because the mass increases, the velocity must decrease equally if we only consider kinetic energy as a factor in this situation.

But why isn't momentum conserved?

3. Jun 23, 2011

### Staff: Mentor

The ore falling vertically into the cart initially has no horizontal momentum. But each bit of mass of the ore is effectively undergoing an inelastic collision with the cart and its then current contents.

Kinetic energy is not conserved in an inelastic collision. But momentum is. The total mass M is increasing. What must be happening to V in order to keep M*V constant?