physics question

Question

Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m1 = 4.93 kg is released from the position shown, at height h = 5.00 m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass m2 = 9.70 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

Answer 1

Without the actual figure, I could be off a little.  That said, the technique to approach a problem like this is to break it into phases based on what is conserved.  

 

Initially, energy is maintained (while converted from potential to kinetic) as the block slides down from the slope onto the flat portion.  Equating initial potential energy with the kinetic energy of the block once it reaches the flat portion allows you to calculate the velocity of the block before the collision.

 

mgh = 1/2 m v²

 

In the second phase, assess the elastic collision.  Since it's elastic, both kinetic energy and momentum are maintained.  This gives us a pair of equations, equating kinetic energy before and after and equating momentum before and after.

 

1/2 m₁ v_1i² = 1/2 m₁v_1f² + 1/2 m₂v_2f²

m1v1i =  m₁v_1f + m₂v_2f

 

Third, we assess the movement of the first block as it exchanges kinetic energy for potential energy.  Since we know its velocity from phase 2, we can then use:

 

mgh = 1/2 m v_1f²

 

in order to find the final height that the block will reach.