The velocity of both suitcases after the collision, vA1 and vB1, may be found from the conservation of momentum and the restitution coefficient, Cr. Cr=(vB1-vA1)/vA0=0.3 so vB1-vA1=0.3*vA0 or vA1=vB1-0.3*vA0. From the conservation of momentum we get 20/32.2*vA0=20/32.2*vA1+10/32.2*vB1. This gives us 20/32.2*vA0=20/32.2*(vB1-.03*vA0)+10/32.2*vB1. Solving this for vB1, the after collision velocity of the 10 lb suitcase, we get 0.62*vA0=0.62*vB1-0.0189vA0+0.31*vB1, or 0.7289*vA0=0.93vB1, or vB1=0.784*vA0.
The distance covered by the sliding suitcase will be such that the kinetic energy imparted to the suitcase B by the collision equals the work done by friction in stopping B. The work of friction Wf is the normal force, 10 lb, times the coefficient of friction, 0.4, times the distance slid, 11.3 ft. Wf=10*0.4*11.3 = 45.2 ft-lb. The kinetic energy given B in the collision was 1/2*20/32.2*vB12 = 0.31*(0.784*vA0)2 = 0.19*vA02. But vA02 is 2*32.2*h, so 0.19*64.4*h=45.2 or h=3.68 ft.
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