Car A weighs 1515 lb and was traveling eastward. Car B weighs 1125 lb and was traveling westward at 41.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 18.5 ft...

Question

You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 1515 lb and was traveling eastward. Car B weighs 1125 lb and was traveling westward at 41.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 18.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750

What speed 𝑣 (in miles per hour) was car A traveling just before the collision?

Answer 1

Momentum before collision is the sum of the momentum vectors for each car.

Car A momentum = 1515 lb /g x Va Car B momentum = -1125 /g x Vb.

Total momentum before collision = 1515 lb /g x Va – 1125lb/g x Vb g is the gravitation acceleration (here 32 ft/s²) that is necessary to convert weight to mass (the minus sign in front of 1125 is due to the fact that the momenta are in opposite directions.

By conservation of momentum, the final momentum equals initial total momentum.  (1515 + 1125) /g x Vf  = 1515 lb /g x Va - 1125 lb /g x Vb. In this case g cancels.

Since Vb is given as 41 mph, we have two unknowns Va and Vf; so we will need two equations to solve.

The initial kinetic energy of the combined cars is ½ x M x Vf². As the cars come to a stop, the work done by the frictional force must equal the initial kinetic energy. µ x Normal force x distance = ½ x M x Vf² = 0.750 x (1515 + 1125 ) x 18.5 ft = ½ x (1515 + 1125) / g x Vf². Note that the normal force is equal to the combined weight of the cars; whereas mass is weight divided by g.

We can solve the last equation for Vf as there is only one unknown, Vf.

Knowing Vf, we can calculate Va from the momentum equation.