Question

You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light when it was hit from behind by car A, of mass 1500 kg. The cars locked bumpers during the collision and slid to a stop. Measurements of the skid marks left by the tires showed them to be 7.30 m long, and inspection of the tire tread revealed that the coefficient of kinetic friction between the tires and the road was 0.65.
(a) What was the speed of car A just before the collision?
(b) If the speed limit was 35 mph, was car A speeding, and if so, by how many miles per hour was it exceeding the speed limit?

Answer 1

Answer:

Explanation:

Force of friction at car B ( break was applied by car B ) =μ mg = .65 x  2100 X 9.8  = 13377 N .

work done by friction = 13377 x 7.30 = 97652.1 J

If v be the common velocity of both the cars after collision

kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²

= 1800 v²

so , applying work - energy theory ,

1800 v² = 97652.1

v² = 54.25

v = 7.365 m /s

This is the common velocity of both the cars .

To know the speed of car A , we shall apply law of conservation of momentum  .Let the speed of car A before collision be v₁ .

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 7.365

v₁ = 17.676 m /s

= 63.63 mph .

( b )

yes Car A was crossing speed limit by a difference of

63.63 - 35 = 28.63 mph.