Question

A 930-kg sports car collides into the rear end of a 2000-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

Answer:2.103 m/s

Explanation:

Given

mass of sports car $m=930\ kg$

mass of SUV $M=2000\ kg$

Suppose u is the velocity if sports car before collision

Conserving momentum we get

$mu=(M+m)v$

$v=\dfrac{2000+930}{930}\times u$

$v=3.15\cdot u$

After collision the combined mass drag 2.8 m and finally stops

From work energy theorem work done by friction is equal to change in kinetic energy of the  combined mass system

$\frac{1}{2}(M+m)v^2=\mu (M+m)gx$

where $\mu =\text{coefficient of friction}$

$x=\text{drag distance}$

$v=\sqrt{2\mu gx}$

$v=\sqrt{2\times 0.8\times 9.8\times 2.8}$

$v=6.626\ m/s$

Initial velocity $u=\frac{6.626}{3.15}$

$u=2.103\ m/s$