Baggage Handling Collision

## Question:

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?
## Answer:

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/s^{2} 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*vB1^{2} = 0.31*(0.784*vA0)^{2} =
0.19*vA0^{2}. But vA0^{2} is 2*32.2*h, so
0.19*64.4*h=45.2 or h=3.68 ft.

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