A 1.0-kg steel ball and a 2.0-m cord of negligible mass makeup a simple pendulum

Question

A 1.0-kg steel ball and a 2.0-m cord of negligible mass makeup a simple pendulum that can pivot without friction about thepoint O, as in the picture below. This pendulum isreleased from rest in a horizontal position, and when the ball isat its lowest point it strikes a 1.0-kg block sitting at rest on ashelf. Assume that the collision is perfectly elastic andthat the coefficient of kinetic friction between the block andshelf is 0.10. (a) What is the velocity of the lbock just after impact? (b) How far does the block slide before coming to rest(assuming that the shelf is long enough)?

Explanation / Answer

m = 1.0-kg, L = 2.0-m, = 0.10. (a) the velocity of the block just after impact = v
use energy conservation: mv2/2 = mgL
v = (2gL) = 6.26 m/s

(b) the distance the block slide before coming to rest = d
note the masses of the ball and the block are equal, and thecollision is elastic, so they exchange velocities during thecollision. The velocity of the block after collision is v = 6.26m/s
work done by the friction = mg*d*cos(180) = -mgd =change in kinetic energy = 0 - mv2/2 = -mgL
d = L/ = 20 m

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