Question

A 990-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 2.5 m before stopping. The police officer, knowing that the coefficient of kinetic friction between tires and road is 0.80, calculates the speed of the sports car at impact. What was that speed? (You will need to split the problem into two parts. In one part you will use conservation of linear momentum. In the other part you will use the work-energy principle.)

Answer 1

Answer:

$v_1 = 20.8 m/s$

Explanation:

Since we know that coefficient of the friction on the road is given as

$\mu = 0.80$

here the distance that the two cars will move is given as

$d = 2.5 m$

now we can find the speed just after two cars are locked together

$v_f^2 - v_i^2 = 2 ad$

here we know that acceleration is due to friction

$ma = -\mu mg$

$a = -\mu g$

$0 - v_i^2 = 2(-\mu g)d$

$v_i = \sqrt{2\mu g d}$

$v_i = \sqrt{2(0.80)(9.81)(2.5)}$

$v_i = 6.26 m/s$

now by momentum conservation of two cars

$m_1v_1 = (m_1 + m_2) v$

so we have

$(990)v_1 = (2300 + 990) (6.26)$

$v_1 = 20.8 m/s$