# Elastic Collision of two wood blocks

1. Mar 12, 2006

### Jacob87411

Two blocks are free to slide along a frictionless wooden track ABC as shown in Figure P9.20. The block of mass m1 = 5.07 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2 = 11.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

I put the figure attached. I just want to know if Im missing something here. Theres no friction, and its an elastic collision. Isn't energy conserved and the block will just rise back to 5m?

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2. Mar 12, 2006

### Chi Meson

That would happen only if the two blocks had the same mass. Since they don't the first block will not come to rest after the collision, but rebound backwards. The energy will now be split (not evenly) between the two masses.
What you got to do is set up two equations, one using the conservation of momentum, and one for conservation of kinetic energy (which only happens in perfectly elastic collisions). There are two unknown variables, the speeds of 1 and 2 after the collision.

Simultaneous equations will take you to a quadratic solution. That means there will be two sets of answers. One of these will be the speeds before the collision, the other set will be the correct answer.

3. Mar 12, 2006

### lightgrav

The relative velocity after an elastic collision is the opposite (negative) of
the relative velocity before the collision. This, and momentum equation,
are not as tough to solve as momentum and KE.

In the center-of-mass reference frame, the initial velocities (inward)
both become opposite directions during an elastic collision. This is easy.

4. Mar 13, 2006

### Chi Meson

OK, so that way is easier, but...

OK it's a lot easier.

5. Oct 18, 2006

### j_salazar87

ok i have the same problem and cant arrive at a solution can someone help out

thanks