Collision on a Rough Surface


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.


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*vB12 = 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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