A steel ball of mass 0.5kg is fastened to a cord 70cm long and fixed at the far

Question

A steel ball of mass 0.5kg is fastened to a cord 70cm long and fixed at the far end, and is released when the cord is horizontal. At the bottom of its path, the ball strikes a 2.5kg steel block and initially at rest on a frictionless surface.

a) If the collision is elastic find the speeds of the steel ball and the block after the collision.

b) If the collision is perfectly inelastic, what is the height to which this new system will ascend and what angle does the cord make with the vertical at that point?

c) How much energy was lost during the collision?

Explanation / Answer

Apply Law of Conservation of Moentum

0.5*(2*9.8*0.70) = 0.5*v1 + 2.5v2

Apply Conservation of Kinetic Energy

0.5*0.5*(2*9.8*0.70) = 0.5*0.5*v1^2 + 0.5*2.5*v2^2

By Solving we get

v1 = Velocity of Steel Ball = - 2.469 m/sec

v2 =Velocity of Block = 1.234 m/sec

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