Consider a frictionless track as in figure. A block of mass m1=5 kg is released

Question

Consider a frictionless track as in figure. A block of mass m1=5 kg is released from a height of 5.0 m from point A. It makes a head on collision elastically at point B with a block of mass m2=12 kg that is initially at rest. http://cs-suse-4.ok.ubc.ca/autoedu/loadFile.jsp?user=bulent&tag;=TrackMomentum A) What is the speed of m1 at point B just before the collision? B) What is the speed of m1 at point B just after the collision? C) What is the maximum height to which m1 rises after the collision?

Explanation / Answer

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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