A cart of mass m1=10 kg. slides down a frictionless ramp and is made to collide

Question

A cart of mass m1=10 kg. slides down a frictionless ramp and is made to collide with a second cart of mass m2=20 kg. at the bottom of the ramp on a horizotal portion of the track. Cart m2 then heads along the horizonal track and into a vertical loop of radius 0.25 m.
a) Determine the height (h) from which from which cart 1 would need to start to make sure that cart 2 completes the loop without leaving the track? (Assume an elastic condition).
b) Find the height needed if instead cart 2 (more massive) is allowed to slide down the ramp on the track into the smaller cart 1?

Explanation / Answer

The height of the cart of mass m1 = 10kg the initial height be h the mass of the second cart of mass m2 = 20kg the radius of the vertical loop r = 0.25m (a) From law of cosnervation of energy m1gh = 1/2 m1v1^2 then v1^2 = 2gh Now from law of conservation of momentum m1v1 = m2v2 then v2 = m2/m1 v1 = (20/10) (2gh) = 4gh again from law of conservation of energy 1/2m2v2^2 = m2g(2r) 0.5 (4gh) = 2gr therefore h = r = 0.25 m (b) If instead of m1 we use m2 then v1 = gh now law of cosnervation of energy 1/2m1v1^2 = m1gh 1/2(gh) = g(2r) therefore h = 4r = 1m

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