Two blocks are free to slide along the frictionless wooden track ABC shown in th

Question

Two blocks are free to slide along the frictionless wooden track ABC shown in the figure below. A block of mass m1=5.00kg is released from rest at point A. The front end of the block has 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=10.0 kg, initially at rest. This means that the two blocks never touch and the collision is perfectly elastic. Calculate the maximum height to which m1 rises up the slope after the collision.

Explanation / Answer

Answer: 0.112m Work: Well, with an elastic collision both momentum and energy are conserved. So, first thing's first - find the potential energy of the block at its starting point: PE = mgh PE = (5.65)(9.81)(5) = 277.13 From there, we can find the velocity of block of mass m1: KE = 277.13 = (1/2)mv^2 277.13 = (1/2)(5)v^2 v^2 = 110.85 v = 10.53 Now, we set up a system of equations that will allow us to solve for the velocities of the blocks. We know that momentum is conserved, so we write that the momentums of the blocks will have to add up to 10.53. (However, one of the blocks' individual momentums may be greater than the starting momentum because one may have negative momentum, due to directions of the blocks.) (m1)(v1) + (m2)(v2) = (10.53)(5.65) (5.65)(v1) + (8.5)(v2) = 59.5 Also, energy of the blocks is conserved, so we find that: (1/2)(m1)(v1)^2 + (1/2)(m2)(v2)^2 = 277.13 (

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