Baggage Handling Collision


Two baggage handlers get into an argument about the height of the top end of a practically frictionless baggage ramp. They let a 20 pound suitcase slide down and strike a stationary 10 pound suitcase at the bottom of the ramp at floor level. They look up the coefficient of friction between the 10 lb bag and the floor and find it to be 0.4. They find the coefficient of restitution in a collision between these suitcases to be 0.3. If the 10 lb suitcase slides 11.3 feet along the floor after the collision, what was the height above the floor from which the 20 lb suitcase was released?


We know that the velocity of the 20 lb suitcase, A, just before it strikes the 10 lb suitcase, B, at floor level is the square root of 2*g*h where g is 32.2f/s2 and h is the height we are looking for, let's call it vA0.

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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