# Elastic Collision of moving cart

• A_Moose
In summary: Jinitial kinetic energy:=\frac{1}{2}\left(6\right)^{2}\cdot6+\frac{1}{2}\left(2\right)^{2}\cdot2=38Jso the change in kinetic energy is 42J, and this is equal to the potential energy stored by the bumper, which is the difference between the final KE and initial KE. So the potential energy stored by the bumper is 42J. Is this correct?In summary, the

## Homework Statement

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.

## Homework Equations

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

## 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.

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A_Moose said:

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

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

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

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?

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

Plugging them into the conservation of momentum equation:
40 = 6v1' + 2v2'
This one's good.

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.
Try again with the corrected KE.
turdferguson said:
For elastic collisions, the following equation can be easily derived using conservation of kinetic energy:
v1 - v2 = v2 -v1
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.

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.

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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?

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.)

Im still stuck on trying to find the potential energy stored by the bumper! I am 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!

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.)

ah!, now i see, thanks.

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?

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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.

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.

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.

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

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

linuxux said:
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
$$m_{1}v_{1}+m_{2}v_{2}=\left(m_{1}+m_{2}\right)v^{'}$$

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

$$v^{'}=\frac{6\cdot6+2\cdot2}{6+2}=5m/s$$
So far, so good.

initial KE of cart:
$$\frac{1}{2}v_{1}^{2}m_{1}=\frac{1}{2}\left(6\right)^{2}\cdot6=18J$$
Redo this calculation. You have an arithmetic error.

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

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

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

vijiraghs said:
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
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.

## 1. What is an elastic collision?

An elastic collision is a type of collision between two objects where there is no loss of kinetic energy. The objects bounce off each other with the same speed and energy before and after the collision.

## 2. What are the conditions for an elastic collision to occur?

The conditions for an elastic collision to occur are that the objects must be in motion, there must be no external forces acting on them, and the objects must be able to deform and then return to their original shape after the collision.

## 3. How is the momentum conserved in an elastic collision?

In an elastic collision, the total momentum of the system before the collision is equal to the total momentum after the collision. This is known as the law of conservation of momentum.

## 4. Can an elastic collision occur between two objects of different masses?

Yes, an elastic collision can occur between two objects of different masses. The mass of the objects does not affect the conservation of kinetic energy in an elastic collision.

## 5. Is kinetic energy conserved in an elastic collision?

Yes, kinetic energy is conserved in an elastic collision. This means that the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision.