# Question envolves the centre-of-mass and momentum.

• Philippe007
In summary: KE'sYou can't because you have three velocities and really only two equations. The equation u1x = v2x - v1x was derived by combining the conservation of momentum and conservation of kinetic energy equations.
Philippe007

## Homework Statement

An electron collides elastically with a hydrogen atom that is initially at rest. Assume all the motion occurs along a straight line. What fraction of the electron's initial kinetic energy is transferred to the atom? (Take the mass of the hydrogen atom to be 1840 times the mass of an electron.)

KE = (1/2)mv^2

uapp = vsep

p = mv

## The Attempt at a Solution

I used the formulas above, and came up with 3 equations:

p-cons

u1x = v1x + 1840v2x

u1x = v2x - v1x

KE

u1x^2 = v1x^2 + 1840v2x^2

But i keep trying and can't use them to find anyone velocity.

Philippe007 said:
p-cons

u1x = v1x + 1840v2x

u1x = v2x - v1x
Just combine these to get V1 in terms of U1. That's all you need to compare initial and final KE of the electron.

(Note that the second of these two equations comes from combining conservation of momentum with conservation of energy.)

Philippe007 said:
But i keep trying and can't use them to find anyone velocity.
You can't because you have three velocities and really only two equations. The equation u1x = v2x - v1x was derived by combining the conservation of momentum and conservation of kinetic energy equations.

In any case, it doesn't matter because the problem is asking for the fraction of the electron's initial kinetic energy transferred to the atom. If you write that down mathematically, you'll see what you need is a ratio of two of the velocities, which is a quantity you can solve for.

I have tried combining the two equations, but that only gives me a new equation with two unknown velocities, and I need all three velocities to find the inital and final KE.

How would I use an equation with 2 unknown variables to compare the initial and final KE??

Help??

Philippe007 said:
I have tried combining the two equations, but that only gives me a new equation with two unknown velocities, and I need all three velocities to find the inital and final KE.
As I said earlier, just solve for V1 in terms of U1. Then you can compare the ratio of the final and initial KE of the electron. (You don't need all three velocities.)

Doc Al said:
As I said earlier, just solve for V1 in terms of U1. Then you can compare the ratio of the final and initial KE of the electron. (You don't need all three velocities.)

V1i = 1841v2f

How can this be used to compare the two KE?

since KEi = (1/2)mv^2

KEi = (1/2)(1)(1841v2f)^2

It doesn't give an actual number, and again there ends up with two variables

and then KEf would be:

KEf = (1/2)(1)(v1f)^2 + (1/2)(1840)(v2f)^2

neither of these can really be compared to give a percentage??

Philippe007 said:
I didn't say anything about adding the equations. Solve for V1 in terms of U1. (Hint: Eliminate V2.)

Doc Al said:
Just combine these to get V1 in terms of U1. That's all you need to compare initial and final KE of the electron.

(Note that the second of these two equations comes from combining conservation of momentum with conservation of energy.)

You did say to combine the equations. But, when you do solve for u1, there are still 2 variables, and you can't find the KEf or KEi without at least one of these, and therefore can't compare the two.

Philippe007 said:
You did say to combine the equations. But, when you do solve for u1, there are still 2 variables, and you can't find the KEf or KEi without at least one of these, and therefore can't compare the two.
You have two equations and three unknowns. Eliminate V2. Then express V1 in terms of U1.

Do it.

Ok:

u1x = v1x + 1840v2x

u1x = v2x - v1x

so...

v2x= u1x + v1x

and then...

u1x = v1x + 1840(u1x + v1x)

u1x = v1x + 1840u1x + 1840v1x

-1839u1x = 1841v1x

u1x = -1841v1x/1839

there.

What is this even showing me. I have still have 2 unknowns?

Philippe007 said:
u1x = -1841v1x/1839

there.
Good. (I meant to say, Express V1 in terms of U1.)

What is this even showing me. I have still have 2 unknowns?
Now:
What's the initial KE of the electron? (This will be in terms of U1, of course.)
What's the final KE of the electron? (Express it in terms of U1.)

Compare! (Find the ratio: KEf/KEi)

So, I did what you said, and its starting to look like an answer, thanks for that.

But now I have:

u1x = (u1x)^2 ((1839/1841)^2 + (1840)(2/1840)^2)

with the left side being inital and the right side being the final KE's

I was wondering though, how would I convert that into a percentage?

Last edited:
Philippe007 said:
So, I did what you said, and its starting to look like an answer, thanks for that.

But now I have:

u1x = (u1x)^2 ((1839/1841)^2 + (1840)(2/1840)^2)

with the left side being inital and the right side being the final KE's

I was wondering though, how would I convert that into a percentage?
That equation couldn't possibly be correct. You have a velocity on the LHS, not a KE, and the RHS side is a velocity squared, also not a KE.

Write down KEi in terms of u1x and KEf in terms of v1x. Divide one by the other. You'll end up with a ratio which you can evaluate given your earlier result.

Philippe007 said:
So, I did what you said, and its starting to look like an answer, thanks for that.

But now I have:

u1x = (u1x)^2 ((1839/1841)^2 + (1840)(2/1840)^2)

with the left side being inital and the right side being the final KE's

I was wondering though, how would I convert that into a percentage?
Not sure what you did here.

KEi = 1/2m(U1)^2
KEf = 1/2m(V1)^2

Express V1 in terms of U1, then rewrite KEf in terms of U1.

Doc Al said:
Not sure what you did here.

KEi = 1/2m(U1)^2
KEf = 1/2m(V1)^2

Express V1 in terms of U1, then rewrite KEf in terms of U1.

Oh i think its because i used KEf = 1/2m1(v1f)^2 + 1/2m2(v2f)^2

Wouldnt that be right?

No, that ratio would come out to be 1 because it's an elastic collision, i.e. total KE(initial)=total KE(final). Go back and see what the question was actually asking for.

Philippe007 said:
Oh i think its because i used KEf = 1/2m1(v1f)^2 + 1/2m2(v2f)^2

Wouldnt that be right?
No. Just compare the initial and final KE of the electron, not the total KE.

## 1. What is the center-of-mass and why is it important in science?

The center-of-mass is the point at which the mass of an object or system is evenly distributed and behaves as if all the mass was concentrated at that point. It is important in science because it helps us understand the motion and stability of objects and systems.

## 2. How is the center-of-mass calculated?

The center-of-mass is calculated by finding the average position of all the mass in an object or system. This can be done by multiplying the mass of each individual component by its distance from a chosen reference point, and then dividing the sum of these values by the total mass.

## 3. What is momentum and how is it related to the center-of-mass?

Momentum is a measure of an object's motion, calculated by multiplying its mass by its velocity. The center-of-mass and momentum are related because the center-of-mass of a system will continue to move at a constant velocity unless acted upon by an external force, making it a useful reference point for studying the momentum of a system.

## 4. Can the center-of-mass be outside of an object?

Yes, the center-of-mass can be outside of an object if the mass distribution within the object is uneven. This is often the case for irregularly shaped objects where the mass is concentrated towards one side.

## 5. How does the center-of-mass affect collisions between objects?

The center-of-mass plays a crucial role in determining the outcome of collisions between objects. In a perfectly elastic collision, the center-of-mass of the system will remain unchanged, while in an inelastic collision, the center-of-mass will shift towards the heavier object. Understanding the center-of-mass helps scientists predict and analyze the outcomes of collisions in various scenarios.

• Introductory Physics Homework Help
Replies
16
Views
4K
• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
21
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
863
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
775
• Introductory Physics Homework Help
Replies
12
Views
4K