A 1.1 kg block is placed at rest on a ramp at some height above the bottom of th

Question

A 1.1 kg block is placed at rest on a ramp at some height above the bottom of the track. It collides elastically with a 0.1 kg block at rest at the bottom of the track.(Figure 1) After colliding elastically both blocks make it safely around a loop-the-loop which has a radius 0.6 m. The track is frictionless. What is the minimum height the 1.1 kg block must be released above the bottom of the track so that both blocks safely traverse the loop-the-loop?

I have mgh = .5mv^2 and mg=mv^2/r (for the circular loop) and I get h=.299 but this is incorrect

Explanation / Answer

m1gh = 0.5(m1)(v1)^2, for collision, use conservation of momentum: (m1)(v1) = (m1 + m2)(v2), then (m1 + m2)g = (m1 + m2)(v2)^2/R

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