Question

A 920 kg sports car collides into the rear end of a 2300 kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, knowing that the coefficient of kinetic friction between tires and road is 0.31, calculates the speed of the sports car at impact. What was that speed?

Answer 1

After the collision, Using energy conservation

Work done by friction = Kinetic energy of both cars

W = KE)total

Ff*d = 0.5*M*V^2

Ff = uk*M*g

uk*M*g*d = 0.5*M*V^2

V = sqrt (2*uk*g*d)

Using given values:

V = sqrt (2*0.31*9.81*2.8) = 4.13 m/sec

Speed of both cars at impact = 4.13 m/sec

Now using momentum conservation

Pi = Pf

m1*v1 + m2*v2 = M*V

v1= speed of sport car = ?

v2 = speed of SUV = 0 m/sec

920*v1 + 2300*0 = (920 + 2300)*4.13

v1 = 3220*4.13/920

v1 = 14.45 m/sec = speed of sports car at impact