A steel ball of mass m 1 = 1.2 kg and a cord of length of L = 2.1 m of negligibl

Question

A steel ball of mass m1 = 1.2 kg and a cord of length of L = 2.1 m of negligible mass make up a simple pendulum that can pivot without friction about the point O, as in the figure below. This pendulum is released from rest in a horizontal position, and when the ball is at its lowest point it strikes a block of mass m2 = 1.2 kg sitting at rest on a shelf. Assume that the collision is perfectly elastic and that the coefficient of kinetic friction between the block and shelf is 0.10. What is the velocity of the block just after impact?

Explanation / Answer

At its highest point, the ball will have potential energy mgh whereh=1.8m

therefore PE=1.2*9.8*2.1=24.696 J

at lowest point PE=0 , and KE=1/2mv^2

Equating the two:

24.696=1/2*1.2*v^2

v = sqrt(24.696/0.6)

v = 6.4156 m/s

since collision is elastic, all momentum is transfered to 2ndobject.

m1u1=m2v2 (both massesare the same, so velocity gets interchanged)

v2 = 6.4156 m/s

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