Question

A 930-kg sports car collides into the rear end of a 3000-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, estimating the coefficient of kinetic friction between tires and road to be 0.80, calculates the speed of the sports car at impact.

Answer 1

The speed of the sports car at impact is 28 m/s

Step-by-step explanation:

Given;

mass of sport car, m₁ = 930-kg

mass of SUV, m₂ = 3000-kg

Coefficient of kinetic friction = 0.80

distance moved by the cars before stopping, s = 2.8 m

let the speed of the sport car at impact be u₁

Also, let the speed of the SUV at impact be u₂

The deceleration of the cars after the impact can be calculated using frictional force formula;

`F_k = \mu N\\\\(m_1 +m_2)(-a) = \mu g(m_1 +m_2)\\\\a = - (\mu g(m_1 + m_2))/((m_1 + m_2)) \\\\a = -\mu g\\\\a = -(0.8 *9.8)\\a = -7.84\ m/s^2`

The common velocity of the cars after the impact can be calculate using kinematic equation

v² = u² + 2as

0 = u² + (2 x -7.84 x 2.8)

0 = u² - 43.904

u² = 43.904

u = √43.904

u = 6.626 m/s

Finally, we determine the speed of the sports car at impact by applying principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = u(m₁ + m₂)

Since the SUV stopped at a red light, u₂ = 0

930u₁ + 0 = 6.626(930 + 3000)

930u₁ = 26040.18

u₁ = 26040.18 / 930

u₁ = 28 m/s

Thus, the speed of the sports car at impact is 28 m/s