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

Step-by-step 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=(2000+930)/(930)* 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

`(1)/(2)(M+m)v^2=\mu (M+m)gx`

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

`x=\text{drag distance}`

`v=√(2\mu gx)`

`v=√(2* 0.8* 9.8* 2.8)`

`v=6.626\ m/s`

Initial velocity
`u=(6.626)/(3.15)`

`u=2.103\ m/s`