# Elastic Collision of moving cart

1. Feb 16, 2007

### A_Moose

1. The problem statement, all variables and given/known data
A 6.0 kg cart moving at 6.0 m/s collides with a 2.0 kg cart moving at 2.0 m/s in the same direction on the same track. The collision is cushioned by a perfectly elastic bumper attached to one of the carts. What are the velocities of the carts after the collision? What was the maximum potential energy stored in the bumper during the collision.

2. Relevant equations
1/2m1v12 + 1/2m2v22 = 1/2m1v'12 + 1/2m2v'22

3. The attempt at a solution
I tried using the conservation of momentum and kinetic energy equations found in my textbook, but the answer never turns out properly. (It shows the answer, but not the work, in my book.) I think it's because it isn't a head-on collision, they are both going the same way. Any help would be appreciated.
Thanks.

Last edited: Feb 17, 2007
2. Feb 16, 2007

### Staff: Mentor

That's all you need to solve for the final speeds--show what you did.

3. Feb 16, 2007

### A_Moose

Alright... M1 = 6 V1 = 6 M2 = 2 V2 = 2
Plugging those into the conservation of kinetic energy equation (the one I listed above) I got:
25 = 3v1'2 + v2'2

Plugging them into the conservation of momentum equation:
40 = 6v1' + 2v2'

Here's where I'm stuck... If I solve for one variable, then put it into the other equation, I never seem to get the right answer, and I've tried a few times.
The right answer: 4.0 m/s right, 8.0 m/s right, max PE = 12 J

4. Feb 16, 2007

### turdferguson

For elastic collisions, the following equation can be easily derived using conservation of kinetic energy:
v1 - v2 = v2 -v1

5. Feb 16, 2007

### A_Moose

But with that equation, would I just do what I was doing before, solve for one and plug it into the other equation?
Also, does that work when the objects are going in the same direction?

6. Feb 17, 2007

### Staff: Mentor

How did you get 25? The initial KE should equal:
$$1/2(6)(6^2) + 1/2(2)(2^2)$$

This one's good.

Try again with the corrected KE.
This equation is derived by combining conservation of energy with conservation of momentum. It's fine to use, if he can derive it. But his original approach should work just fine.

The fact that the original velocities are in the same direction doesn't matter.

7. Feb 17, 2007

### A_Moose

Hmm... I have no idea what I was thinking when I put 25... :grumpy:
Well, thanks a ton for the help, I really appreciate it.

Just to make sure... The 2 equations, in their correct forms, are:
112 = 3v1'2 + v2'2
and
40 = 6v1' + 2v2'
Then, with those, I can solve for the correct velocities, right?

Edit:Now I see where I went wrong, thanks to you.
I got 25 because, for some reason when finding the kinetic energy for the first mass, I ADDED m1 and v12 rather than MULTIPLIED.

Last edited: Feb 17, 2007
8. Mar 14, 2007

### linuxux

What would be the potential energy stored in the bumper, would it just be the the difference in initial velocities squared times half the mass of the colliding cart?

9. Mar 15, 2007

### Staff: Mentor

maximum PE stored

To find the maximum PE stored in the bumper, first find the minimum KE that the carts must have when they are squeezed together. (Hint: Momentum is conserved.)

10. Mar 15, 2007

### linuxux

Im still stuck on trying to find the potential energy stored by the bumper! im confused, can someone show me how to do it, my textbook has no examples of elastic collision when the object are traveling in the same direction!

11. Mar 15, 2007

### Staff: Mentor

As I suggested, first find the KE of the carts at the moment the bumper is maximally compressed. (Another hint: At that moment, both carts move as one.)

12. Mar 15, 2007

### linuxux

ah!, now i see, thanks.

13. Mar 16, 2007

### linuxux

hello Doc, in regards to the PE stored in the bumper, is it calculated like this:

since momentum is conserved, and the point at which the bumper is maximally compressed is when the carts move as one, then
$$m_{1}v_{1}+m_{2}v_{2}=\left(m_{1}+m_{2}\right)v^{'}$$

$$v^{'}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}$$

$$v^{'}=\frac{6\cdot6+2\cdot2}{6+2}=5m/s$$

initial KE of cart:
$$\frac{1}{2}v_{1}^{2}m_{1}=\frac{1}{2}\left(6\right)^{2}\cdot6=18J$$

after colission and when stuck to cart 2, therefore maximally compressing bumper, KE of cart:
$$=\frac{1}{2}\left(5\right)^{2}\cdot6=15J$$

since $$\Delta KE_{cart 1 and 2}= PE_{bumper}$$:

$$=\left(\frac{1}{2}\left(5\right)^{2}\cdot6-\frac{1}{2}\left(6\right)^{2}\cdot6\right)+\left(\frac{1}{2}\left(5\right)^{2}\cdot2-\frac{1}{2}\left(2\right)^{2}\cdot2\right)=18J$$

so the PE of the bumper is 18J?...is that right?

Last edited: Mar 17, 2007
14. Mar 17, 2007

### Staff: Mentor

Your calculation of the velocity of the combined carts at max compression is correct, but not your comparison of KE. You must compare the total KE of both carts together, not just the change in KE of one cart.

15. Mar 17, 2007

### vijiraghs

collsion

A 6.0 kg cart moving at 6.0 m/s collides with a 2.0 kg cart moving at 2.0 m/s in the same direction on the same track. The collision is cushioned by a perfectly elastic bumper attached to one of the carts. What are the velocities of the carts after the collision? What was the maximum potential energy stored in the bumper during the collision.

For an elastic collision both kE and momentum are conserved.If m1, m2 are the masses,u1, u2 are the velocities before collision , v1, v2 are the velocities after collision then
v1=u1(m1-m2) / (m1+m2) + 2m2u2 / (m1+m2) and
v2= 2m1u1 / (m1+m2) + u2(m2-m1) / (m1+m2)
but i have no idea to calculate potential energy.

16. Mar 17, 2007

### linuxux

i think the kinetic energy of the 1st cart decreases because it slows down, but the kinetic energy of the second cart increase since it speeds up after the collision, so you you determine the kinetic energies of both cart prior to the collision, then their kinetic energies after the collision using the velocity of $$v^{'}$$ which is calculated above and add everything together in a "final minus initial" manner. It should be a positive number, meaning that energy was supplied, and this is true since the smaller cart speed up. i calculated 18J as the potential energy.

Last edited: Mar 17, 2007
17. Mar 17, 2007

### vijiraghs

but the answer was zero, not a +ve number.

18. Mar 17, 2007

### vijiraghs

initial kinetic energy of cart1 =108(k1)
initial kinetic energy of cart2=4(k2)
final kinetic energy of cart1=48(k3)
final kinetic energy of cart2=64(k4)
so, k1+k2=k3+k4=112

19. Mar 17, 2007

### Staff: Mentor

So far, so good.

Redo this calculation. You have an arithmetic error.

Redo: Another arithmetic error.

Here's what you should be doing. Initially, all the energy is KE. Calculate that! (For both cars, of course.) When the bumper is maximally compressed, the energy will be partly KE and partly spring PE. Calculate the KE part and subtract it from the total energy to find the spring PE.

Your approach is off by a minus sign, and your answer is incorrect due to arithmetic error.

20. Mar 17, 2007

### Staff: Mentor

Your answer is zero because you compared intial KE to final KE. Since energy is conserved, of course they are equal.

To find the PE stored when the bumper is compressed (remember that the bumper only stores energy temporarily) you must calculate the KE of the carts when the bumper is compressed. Then compare that KE to the total energy.