Kinetic energy and elastic collision physics problem?
Problem: Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m1 = 4.98 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 = 9.00 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.
figure: http://www.webassign.net/serpse8/9-p-027.gif
Steps on how the solution is reached are appreciated. Thanks!
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As M1 moves downward 5.00 meters, its potential energy is converted into kinetic energy. Since the collision is elastic, all this kinetic energy is split between the two blocks, as they exert the force of repulsion on each other.
Initial PE = 4.98 * 9.8 * 5
KE of M1 before the interaction = Initial PE = 4.98 * 9.8 * 5 = 244.02
KE = ½ * M1 * v^2 = 244.02
v^2 = 2 * 244.02 = 488.04
The velocity of M1 before the interaction = √488.04
Momentum is always conserved.
Momentum after interaction = Momentum before interaction
Momentum before interaction = M1 * v = 4.98 * √488.04
Since, M1 rises after the elastic collision, M1 is going back up the hill after the interaction. So, the velocity of M1, after the interaction is negative.
Eq. #1 4.98 * -v1 + 9 * v2 = 4.98 * √488.04
Since the interaction is elastic, the kinetic energy after the interaction is equal to the kinetic energy before the interaction.
KE before = ½ * 4.98 * 488.04
KE after = ½ * 4.98 * (-v1)^2 + ½ * 9 * (v2)^2
½ * 4.98 * (v1)^2 + ½ * 9 * (v2)^2 = ½ * 4.98 * 488.04
Divide both sides by ½
Eq. #2 4.98 * (v1)^2 + 9 * (v2)^2 = 4.98 * 488.04
You have 2 equations in2 variables. Solve Eq #1 for v1 and substitute into Eq. #