1 answer

A 2000 [Ib) sports car collides into the rear end of a 5000 [ib] SUV stopped...

Question:

A 2000 [Ib) sports car collides into the rear end of a 5000 [ib] SUV stopped at a red light. The bumpers lock, the brakes are
A 2000 [Ib) sports car collides into the rear end of a 5000 [ib] SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 15.0 [t] before stopping. The police officer, knowing that the coefficient of kinetic friction between tires and road is 0.85 calculates the speed of the sports car at impact. What was that speed? cooh/ -m12.72

Answers

First, find the velocity of the locked cars after the collision

Given that the cars skid forward 15ft before stopping

Kinetic energy after collision is lost due to friction

KE, = W friction

KE; = Ffriction *

KE;=uN*d

Mvi = u Mg*d

bu; = 4g d

Vj = 2 * ug*d

v_{f}=\sqrt{2*0.85*32.2ft/s^{2}*15ft}

Uf = 28.655ft/s

=========================

During collision momentum is conserved

Initial momentum = Final momentum

mcarcar + mtruck Utruck = Muf

mcarucar + mtruck Utruck = (mcar + mtruck)UF

2000lb*u_{car}+5000lb*0ft/s=(2000lb+5000lb)*28.655ft/s

2000*u_{car}+5000*0=(2000+5000)*28.655

2000*u_{car}=7000*28.655

u_{car}=\frac{7000*28.655}{2000}

ANSWER: {\color{Red} u_{car}=100.2925ft/s}

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