Inelastic collision of two identical carts

[eagles12]

Homework Statement

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?

Homework Equations

pi=pf

The Attempt at a Solution

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

 

[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.

 

[eagles12]

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

 

[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.

 

[HallsofIvy]

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

 

[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?

 

[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).

 

[eagles12]

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

 

[BruceW]

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

 

[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?

 

[eagles12]

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

 

[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?

 

[eagles12]

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

 

[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.

 

[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

 

[emailanmol]

Also it is v important to realize 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 let's 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.


So in your formula,
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 :-)

 

[emailanmol]

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

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

 

[eagles12]

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

 

[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:

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

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.

 

[emailanmol]

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

Read my last post.

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

 

[eagles12]

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

 

[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.

 

[eagles12]

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

 

[BruceW]

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

 

[eagles12]

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

 

[BruceW]

That's right. You've got the answer. Do you fully understand it? You had the final speed of the two carts, and you were calculating the total final kinetic energy, so you see why 2m was used for the mass?

 

[emailanmol]

BruceW has made a v important point.
You need to understand how you got the answer.
I would suggest you read each and every post in this thread again , and understand what the poster wanted to convey.