# Cart Collision Problem

1. Nov 25, 2006

### crimsondarkn

A 2.4 kg dynamics cart with a linear elastic spring attached to its front end is moving at 1.5 m/s [W] when it collides head on with a stationary 3.6 kg cart.

What is the velocity of each cart at minimum separation?

The answer is 0.6 m/s [W]

Can anyone show me the steps?

2. Nov 26, 2006

### Andrew Mason

There is a very easy way to do this. Since there are no external forces, the total momentum the system does not change. What is the speed of the centre of mass? (What is the speed of the observer who sees the two cars colliding with equal and opposite momenta?)

What is the relationship between the speed of the centre of mass and the speeds relative to the centre of mass of the two cars at minimum separation?

AM

3. Nov 26, 2006

### crimsondarkn

Oh I got it, thanks!

This is what I did..

m1v1+m2v2=m1v'1+m2v'2

m2v2 becomes zero since its stationary

m1v1=v'(m1+m2)

-3.6=v'(6)

v'=-3.6/6

v'= -0.6 m/s ---> 0.6m/s[W]

I didn't know it was completely elastic collision, that's why I was stuck.

4. Nov 26, 2006

### crimsondarkn

I'm stuck on part c now...

Calculus the change in total kinetic energy of the system at minimum separation. Here's my work

v'=-0.6m/s

1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2+1/2m2v2'^2

cross out all the 1/2s and also v2 becomes zero

m1v1^2 = m1v1'^2 + m2v2'^2

(2.4)(-1.5)^2 = (2.4)(0.6)^2 + (3.6)(0.6)^2

5.4 = 0.864 + 1.296

5.4 = 2.16

5.4-2.16 = 3.24 J

And this is where i'm stuck...

The answer at the back is -1.6 J

5. Nov 26, 2006

### Staff: Mentor

Don't write this as an equation (since they are not equal!) and don't cross anything out. You have the speeds and the masses, so just calculate the initial and final KEs by plugging into the KE formula. Then compare KEi to KEf. (By "final" I mean at the point of minimum separation.)