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Momentum Unit

PROGRESSIVE SCIENCE INITIATIVE® (PSI®)

2 Comments

Amy Calhoun • 4 months, 2 weeks agologin to reply

On #33 of the Momentum MC questions, my students and I are getting letter A for the answer, but on the answer key, it says letter E. Is this a typo? If not can someone explain how to get to answer E? Thanks!

John Ennis • 4 months, 2 weeks agologin to reply

Amy - Actually, the answer to #32 is B, but the answer key says A, so I'm fixing that. The answer to #33 is E. I answered this 2 years ago on the thread for the MC, but here it is again! This is a tricky one. We treat the collision as a perfectly elastic collision because of the spring. The spring does compress - even though it is a frictionless surface, the spring has mass, so it has inertia and the force of the block hitting it compresses the spring. It's a little different from two blocks hitting each other, where we assume the collision is instantaneous and there is no deformation as in the spring case. For the spring, the collision takes a little longer. So, when the block first contacts the spring, it begins slowing down until it reaches v' = 0 and the spring starts speeding up until it reaches v' = v. But in between, the spring and the block reach the same velocity, and then they separate as the block continues slowing down and the spring is speeding up. The block is slowing down due to the equal and opposite spring force exerted on it. The maximum compression is released right before the block and spring separate. At this point the collision resembles a perfect inelastic collision, so v(spring/block system) is found by conservation of momentum: Mv = (M + M)v(spring/block) v(spring/block) = v/2. Using conservation of energy: 1/2 Mv^2 = 1/2 (M + M)(v/2)^2 + 1/2 Kx^2. Solve for x, the maximum compression, and the answer is v(M/2K)^1/2. John

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