Calculating initial velocity in elastics collision

[blixel]

NOTE: The subject should say Inelastic Collision

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=½mv²
½m_Av_A²+½m_Bv_B²=½m_Av'_A²+½m_Bv'_B²

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_Av_A²=m_Av'_A²+m_Av'_A²

I know that v'_A=v'_B because the cars stick to each other, so I can simplify the equation even further:
m_Av_A²=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/s²)=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/s²

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 v₀=0 since the SUV is not moving.
v²-v₀²=2aΔX
v=sqrt(2aΔX)=sqrt[2(7.84m/s²)(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.

 

[TSny]

You assumed the collision is elastic (KE is conserved). Is that an appropriate assumption for this problem?

 

[BvU]

Your expression for conservation of kinetic energy is only valid for an elastic collision. Not here ('bumpers lock')