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 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

Answer 1

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

$KE_{f}=W_{friction}$

$KE_{f}=F_{friction}*d$

$KE_{f}=\mu N*d$

$\frac{1}{2}Mv_{f}^{2}=\mu Mg*d$

$\frac{1}{2}v_{f}^{2}=\mu g*d$

$v_{f}=\sqrt{2*\mu g*d}$

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

${\color{Red} v_{f}=28.655ft/s}$

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During collision momentum is conserved

Initial momentum = Final momentum

$m_{car}u_{car}+m_{truck}u_{truck}=Mv_{f}$

$m_{car}u_{car}+m_{truck}u_{truck}=(m_{car}+m_{truck})v_{f}$

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

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

$2000*u_{car}=7000*28.655$

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

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

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