A steel ball of mass m 1 = 1.1 kg and a cord of length of L =1.8 m of negligible

Question

A steel ball of mass m1 = 1.1 kg and a cord of length of L =1.8 m of negligible mass make up asimple pendulum that can pivot without friction about the pointO, as in the figure below. This pendulum is released fromrest in a horizontal position, and when the ball is at its lowestpoint it strikes a block of mass m2 =1.1 kg sitting at rest on a shelf.Assume that the collision is perfectly elastic and that thecoefficient of kinetic friction between the block and shelf is0.10.
(a) What is the velocity of the block just after impact?
1       m/s

(b) How far does the block slide before coming to rest (assumingthat the shelf is long enough)?
2 m

Explanation / Answer

At its highest point, the ball will have potential energy mgh whereh=1.8m therefore PE=1.1*9.8*1.8=19.404 J at lowest point PE=0 , and KE=1/2mv2 Equating the two: 19.404=1/2*1.1*v2 v=6 m/s since collision is elastic, all momentum is transfered to 2ndobject. m1u1=m2v2 (both massesare the same, so velocity gets interchanged) v2=6 m/s now the 2nd block's KE=1/2*1.1*6*6 =19.8 J Work done against friction till it stops = Ff* S(where S is thedistance where it stops) and since Ff=N (where N=mg=normal reaction), 0.1*1.1*9.8*S=19.8 S=18.37m

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