# Homework Help: Collision - need big help

1. Mar 8, 2004

### Drakon25th

I need help with this question:
A pair of bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear. One has a mass of 450kg and the other 550kg, owing to differences in passenger mass. if the lighter one approaches at 4.5m/s and the other is moving at 3.7m/s, calculate a)their velocities after the collision and b) the change in momentum of each

ok this is what i know
m1 = 450 kg, v1 = 4.5 m/s
m2 = 550 kg, v2 = 3.7 m/s

i know i'm suppose to use these formulas:
m1v1+m2v2 = m1v1' + m2v2'
and
(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)m1v1^2' + (1/2)m2v2^2'

my teacher gave us this hw without really explaining anything and the example my book gives to me is very confusing

so i know m1v1+m2v2 = total momentum in this system, which is:
450kg(4.5m/s)+550kg(3.7m/s)=4060 Ns
and the total KE after the collision is:
.5(450kg)(4.5m/s)^2+.5(550kg)(3.7m/s)^2 = 8321J

then my book becomes confusing and i get lost

what do i do?

2. Mar 8, 2004

### Drakon25th

basically this is the work my book gives for the example it has provided:

first: the same two equations i gave
next: using the momentum equation (m1v1+m2v2=m1v1'+m2v2') they get:
v1' = v1 - (m2/m1)v2'

(v2 = 0 in this equation in case you're wondering why that might be missing)

then they say they substituted that into the KE equation and got:

(v2)'^2 - (v2')((2m1v1)/(m1+m2) = 0

i dont know how they got that equation

(info bout this example:
m1 = 1.01u, v1 = 3.6*10^4m/s
m2 = 4.0u, v2 = 0 m/s)

3. Mar 8, 2004

4. Mar 9, 2004

bump?

5. Mar 9, 2004

### Drakon25th

if im right, my anwer is v1' = 3.73m/s and v2' = 4.34m/s

6. Mar 9, 2004

### HallsofIvy

You know "450kg(4.5m/s)+550kg(3.7m/s)=4060 Ns
and the total KE after the collision is:
.5(450kg)(4.5m/s)^2+.5(550kg)(3.7m/s)^2 = 8321J"

so, taking v1 and v2 as the velocities after the collision, you know

450v1+ 550v2= 4060 and
.5(450)v12+ .5(550)v22= 8321.

Solve those 2 equations for v1 and v2.