A block of mass m is released and slides without friction down an initially stat

Question

A block of mass m is released and slides without friction down an initially stationary ramp of mass m (where is some number). The block starts at a height h above the horizontal surface on which the ramp rests, and the ramp itself is free to slide on this surface without friction. The ramp is curved such that the block can slide smoothly off the ramp and onto the horizontal surface. Assume that the physical size of the block is negligible compared to h and compared to the curvature of the ramp. (a) Find the velocity of the block (vb) and the velocity of the ramp (vr) after the two separate, in terms of some or all of g, h, and m. (b) The block, now sliding freely to the right along the horizontal surface, bounces elastically off an immovable bumper, thereby changing its direction but maintaining its speed. For what values of will the block eventually catch up to the ramp? (c) Now take = 4. The block catches up to, and begins sliding up on, the ramp. At the moment that the block comes to rest (instantaneously) relative to the ramp, what is the velocity vf of the ramp/block system?

Explanation / Answer

A) momentum conservation in horizontal direction

=> mV_b = mV_r

=> V_b = V_r

and potential energy = kinetic energy

=> 0.5*m*V_b^2 + 0.5*m*V_r^2 = mgh

=> V_r = sqrt(2gh/(^2 + ))

V_b = *sqrt(2gh/(^2 + ))

B) V_b > V_r

=> *sqrt(2gh/(^2 + )) > sqrt(2gh/(^2 + ))

=> > 1 Answer

C) momentum conservation in horizontal direction

m*V_b + m*V_r = (1+)m*V

=> v = 2*V_r/(1+) = 8*sqrt(2gh/20)/5 Answer

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