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**NOTE: The subject should say Inelastic Collision**

1. Homework Statement

1. Homework Statement

A

*980kg*sports car collides into the rear end of a

*2300kg*SUV stopped at a red light. The bumpers lock, the breaks are locked, and the two cards skid forward

*2.6m*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. What was that speed?

I believe the relevant givens to be as follows:

__Givens__

m

_{A}=980kg

m

_{B}=2300kg

v

_{B}=0

ΔX=2.6m

μk=0.80

## Homework Equations

I believe the relevant equations to be as follows:

__Equations__

*f*=μF

_{N}

K=½mv

^{2}

½m

_{A}v

_{A}

^{2}+½m

_{B}v

_{B}

^{2}=½m

_{A}v'

_{A}

^{2}+½m

_{B}v'

_{B}

^{2}

## The Attempt at a Solution

Before the cars collide, the SUV isn’t moving so all the kinetic energy is in the sports car. The equation simplifies to:

m

_{A}v

_{A}

^{2}=m

_{A}v'

_{A}

^{2}+m

_{A}v'

_{A}

^{2}

I know that v'

_{A}=v'

_{B}because the cars stick to each other, so I can simplify the equation even further:

m

_{A}v

_{A}

^{2}=v'

^{2}(m

_{A}+m

_{B})

Now to find v'

^{2}, I'm going to need force of friction.

F

_{N}=(980kg+2300kg)(9.8m/s

^{2})=32144N

f

_{FR}=(0.80)(32144N)=25715.2N

Now I will use F=ma to find the acceleration of the two masses as they slide the 2.6m.

a=(25715.2N)/(980kg+2300kg)=7.84m/s

^{2}

Since I know the acceleration, I should now be able to calculate v' using a kinematic equation. At the moment of impact, the initial velocity would be v

_{0}=0 since the SUV is not moving.

v

^{2}-v

_{0}

^{2}=2aΔX

v=sqrt(2aΔX)=sqrt[2(7.84m/s

^{2})(2.6m)]=v'

Now I should be able to find v

_{A}using v

_{A}=sqrt([v'

^{2}(m

_{A}+m

_{B})]/(m

_{A}))

But when I run that calculation, I get v

_{A}≈11.68 m/s. But I should be getting 21 m/s according to the book.

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