# Homework Help: Collision of two carts

1. Mar 22, 2012

### eagles12

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

A cart of mass m moves with a speed v on a frictionless air track and collides with an identical cart that is stationary. If the two carts stick together after the collision, what is the final kinetic of the system in terms of m and v?

2. Relevant equations

pi=pf

3. The attempt at a solution

mivi=mfvf
0=mvf
so m=vf
is that right?

2. Mar 22, 2012

### BruceW

your first line is correct (mivi=mfvf), but I don't understand the rest of your working. One step at a time, what is the initial momentum?

EDIT: remember, the momentum of the system is conserved, so you should work out the total momentum.

3. Mar 22, 2012

### eagles12

the initial momentum is 0 because it says the cart starts at rest

4. Mar 22, 2012

### BruceW

But there are two carts to begin with. You need to find the total momentum, because it is the momentum of the system which is conserved.

5. Mar 22, 2012

### HallsofIvy

The initial momentum of one cart is 0 but it is total momentum that is conserved. BruceW was asking for the total momentum.

6. Mar 22, 2012

### eagles12

so instead of 0=mvf i should do
m1iv1i+m2iv2i=m1fv1f+m2fv2f
which is the same as
mv1i=mv1f+mv2f
mv1i=2mvf
so vf=mv1i/2m
is that right?

7. Mar 22, 2012

### BruceW

Exactly. Well written, too :) The examiner will like it when you set out your steps nicely, as you have done here.

So you've now got vf (and you can further simplify the fraction). And the next step is to get "the final kinetic of the system in terms of m and v" (I'm guessing this means the final kinetic energy).

8. Mar 23, 2012

### eagles12

so if i simplify i get vf=v/2
is that the correct simplification?

9. Mar 23, 2012

### BruceW

Yep, that's right. next step is to find the final kinetic energy.

10. Mar 23, 2012

### eagles12

oh right.
so i can use Ei=Ef
and then say
Ki+Ui=Kf+Uf
do i have to do this twice, once for each cart?

11. Mar 23, 2012

### eagles12

so Kf=1/2mvf
which means
Kf=1/2m (v/2)
kf=mv/4
but it said this is wrong!

12. Mar 23, 2012

### BruceW

energy is not necessarily conserved in this collision. In problems like this, if you're unsure, then a good rule of thumb is to use momentum conservation. You have already used conservation of momentum to get the final velocity of the two carts stuck together, so what is the final kinetic energy?

13. Mar 23, 2012

### eagles12

but i didnt think final kinetic energy was in the momentum conservation formula

14. Mar 23, 2012

### emailanmol

Hello eagles12,

See the question states that cart 1 was moving initially and cart 2 was at rest.After collision, both start moving with a common velocity v.

Using momentum conservation you found out the new velocity as u/2 (where u is initial velocity of cart1).

Now you need to find final kinetic energy of the two carts.

You know their masses m and their velocites.So just apply the formula of kinetic energy.

15. Mar 23, 2012

### eagles12

i applied the formula of kinetic energy Kf=1/2m (v/2)
and then kf=mv/4
but it said that this was incorrect

16. Mar 23, 2012

### emailanmol

Also it is v important to realise that during head on collisions (which you will deal with at school level)
its the TOTAL momentum which is conserved .

That is
m1u1 +m2u2=m1v1+m2v2

Later in higher classes you wil learn about oblique collisions in which individual components are also conserved.(But lets just move past that right now!).

Also during collisions although total energy is conserved,
using conservation of energy formula doesn't help.(except if the collision is elastic)

This is because some of the energy is lost as sound or heat (etc) and you cannot acount how much amount was lost.

K1 +U1=k2 + U2

you never know about the values of U2 and U1 (which include terms like sound energy, heat, potential energy, )
so using conservation of energy gives you no extra information.

However, by using conservation of momentum you can find v2 and v1 and thus obtain the final kinetic energies (as you know m1 and m2).

After this you can use K1-K2 =U1-U2 and thus find how much energy was lost as sound or heat or to elasticity :-)

Last edited: Mar 23, 2012
17. Mar 23, 2012

### emailanmol

Formula of kinetic energy for a body having velocity u and mass is mu^2/2 not mu/2

18. Mar 23, 2012

### eagles12

if i don't know U1 and U2 how do I figure out K1-K2=U1-U2

19. Mar 23, 2012

### BruceW

You must find the final kinetic energy without using the principle of energy conservation. emailanmol was saying that once you do find the kinetic energy, then you can calculate U2-U1. But this isn't in the question, so you don't need to worry about doing this for this question.

You got pretty close here:
As emailanmol was saying, it is this bit which was incorrect. Remember that the two carts get stuck together, so what is their mass? And it looks like you've also forgotten to square the speed.

20. Mar 23, 2012

### emailanmol

First lets focus on finding K2
(We are drifting away from the main question here).

You know velocity of each cart is v/2. Mass is m for each cart

What is the kinetic energy of one cart?
What is the total kinetic energy of both carts?

Remember kinetic emergy is mu^2/2

21. Mar 23, 2012

### eagles12

isnt the velocity of them both together v/2?
so then m(v/2^2)/2

22. Mar 23, 2012

### BruceW

yes, they are both moving at speed v/2. But you shouldn't be using m as the mass. You can either think of them as two carts both moving at the same speed (in which case, each of them will contribute toward the total KE). Or you can say the two carts make a new object which has a new mass. Both are correct, but you need to do one of them.

23. Mar 23, 2012

### eagles12

now i got mv^2/8
but it says that is incorrect also

24. Mar 23, 2012

### BruceW

how did you get that answer? what are you using for the mass?

25. Mar 23, 2012

### eagles12

im using m for mass. should it be 2m?
then i would get mv^2/4