# Calculating initial velocity in elastics collision

1. Dec 11, 2016

### blixel

NOTE: The subject should say Inelastic Collision

1. The problem statement, all variables and given/known data

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
mA=980kg
mB=2300kg
vB=0
ΔX=2.6m
μk=0.80

2. Relevant equations
I believe the relevant equations to be as follows:

Equations
f=μFN
K=½mv2
½mAvA2+½mBvB2=½mAv'A2+½mBv'B2

3. 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:
mAvA2=mAv'A2+mAv'A2

I know that v'A=v'B because the cars stick to each other, so I can simplify the equation even further:
mAvA2=v'2(mA+mB)

Now to find v'2, I'm going to need force of friction.
FN=(980kg+2300kg)(9.8m/s2)=32144N
fFR=(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 vA using vA=sqrt([v'2(mA+mB)]/(mA))

But when I run that calculation, I get vA≈11.68 m/s. But I should be getting 21 m/s according to the book.

Last edited: Dec 11, 2016
2. Dec 11, 2016

### TSny

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

3. Dec 11, 2016

### BvU

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