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A 960-kg sports car collides into the rear end of a 2200-kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.9 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 :

The speed of the sports car at impact is;

u₁ = 22.2 m/s

  • We are given;

Mass of sports car; m₁ = 960 kg

Mass of SUV; m₂ = 2200 kg

Coefficient of kinetic friction; μ_k = 0.8

skid distance; d = 2.9 m

  • Using newton's equation of motion to get;

v² = u² + 2ad

where a = μ_k * g = 0.8 * 9.8 = 7.84 m/s²

v is final speed of cars after collision

u is initial speed before collision = 0 m/s

d is skid distance = 2.9 m

Thus;

v² = 0² + 2(7.84 * 2.9)

v² = 45.472

v = √45.472

v = 6.743 m/s

Using conservation of momentum, we have;

m₁u₁ + m₂u₂ = (m₁ + m₂)v

where;

u₁ is the speed of the sports car at impact.

u₂ is speed of the suv at impact = 0 m/s

Thus;

960u₁ + 2200(0) = (960 + 2200)6.743

960u₁ = 21307.88

u₁ = 21307.88/960

u₁ = 22.2 m/s

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