Collision on a Rough Surface

## Question:

Two cars, A and B, have equal mass, m. Car A is traveling with a
speed of 12 fps when it collides head on with car B. If the
driver of car B locks up his brakes at the instant of the
collision, how far does car B slide. Take the coefficient of
kinetic friction between car B and the road to be 0.5. The
coefficient of restitution in the collision between the cars is
0.6, a very bouncy pair of bumpers.
## Answer:

The coefficient of restitution, Cr, is the ratio of the velocity
of separation between the two objects after the collision to the
velocity of approach before the collision. Bear in mind that the
words separation and approach imply opposite algebraic signs so
Cr=(vB1-vA1)/(vA0-vB0) where 1 identifies
the after collision velocities and the 0 identifies the before
collision velocities. In this case then since
Cr=0.6, vB0=0 and vA0=12 f/s,
vB1-vA1=0.6*12=7.2f/s, so
vA1=vB1-7.2. We know that in a collision momentum is
conserved so m*vA0=m*vA1+m*vB1, or 12=vA1+vB1. But
vA1=vB1-7.2 so 12=2*vB1-7.2 or
vB1=19.2/2 = 9.6f/s.
Car B will stop sliding when its kinetic energy has all been
converted to heat in the friction of the tires against the road.
This is the case when the work done by the braking force equals
the kinetic energy that car B had immediately after the
collision. Let's call the distance slid x so the work of the
braking force is x times the force of friction. The force of
friction is the normal force holding the tires against the road
times the coefficient of kinetic friction. The normal force is
the mass of car B times the acceleration of gravity. Putting this
all together we get the work of friction,
Wf=m*32.2*0.5*x.

The kinetic energy imparted to car B by the collision is 1/2*m*vB1^{2} = 46.08*m ft-lb. The car
will stop sliding when this is equal to Wf. So
16.1*m*x=46.08*m at the end of the slide. This gives us
x=2.86ft.

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