Understanding Elastic Collisions: Solving for Total Kinetic Energy and Momentum

[nlsherrill]

Homework Statement

I the center of mass reference frame, a particle with mass m1 and momentum p1 makes an elastic head on collision with a second particle of mass m2 and momentum p2=-p1. After the collision the first particles momentum is p1'. Write the total kinetic initial energy in terms of m1, m2, and p1 and the total final kinetic energy in terms of m1, m2, and p1'. Show that p1' = +/- p1. If p1' = -p1, the particle is merely turned around by the collision and leaves with the speed it had initially. What is the situation for p1' = +p1 solution?

Homework Equations

The Attempt at a Solution

The back of the book just gives this solution that I have no idea how they came to the conclusion

ans for a): Ki=Kf=P1^2/2(m1^2 +6m1m2 + m2^2/m1^2m2 + m1m2^2)

ans for b): the particles do not collide

Part b doesn't sound so bad, but as for part a I am not sure how that is the answer.

 

[willem2]

I think formula in the book for the kinetic energy is actually wrong. It's good enough to derive the answer since

$$K_i = K_f = {\frac { 1}{2} {P_1}^2 f(m1,m2)$$

will give the same result for any function f that depends on only m1 and m2.

The correct answer has a much simpler f.

Use the ordinary equation for kinetic energy, and use $$P_1 = m_1 v_1$$
$$P_2 = m_2 v_2$$ and $$P_1 = -P_2$$ to eliminate the other variables

 

[nlsherrill]

I think formula in the book for the kinetic energy is actually wrong. It's good enough to derive the answer since

$$K_i = K_f = {\frac { 1}{2} {P_1}^2 f(m1,m2)$$

will give the same result for any function f that depends on only m1 and m2.

The correct answer has a much simpler f.

Use the ordinary equation for kinetic energy, and use $$P_1 = m_1 v_1$$
$$P_2 = m_2 v_2$$ and $$P_1 = -P_2$$ to eliminate the other variables

The formula in the book is wrong? How do you know what book I'm using?

Kind of confused with the last part of your post. The answer can't have velocity in there, so that's clearly P/m, but I think the answer in the back of the book added the kinetic energy relative to the center of mass reference frame...which would be zero right?

 

[willem2]

The formula in the book is wrong? How do you know what book I'm using?

I meant the expression you gave for the kinetic energy in a). I tought this came from your book.

Kind of confused with the last part of your post. The answer can't have velocity in there, so that's clearly P/m, but I think the answer in the back of the book added the kinetic energy relative to the center of mass reference frame...which would be zero right?[/QUOTE]

I don't know what you mean with "kinetic energy relative to the center of mass frame"
What is asked is the kinetice energy in the center of mass frame, which is just
$$(1/2)m_1 {v_1}^2 + (1/2)m_2 {v_2}^2$$
for each object, where v1 and v2 are the speeds in the center of mass frame.

you need to convert this to an expression only P1, m1 and m2

 

[nlsherrill]

I meant the expression you gave for the kinetic energy in a). I tought this came from your book.

Kind of confused with the last part of your post. The answer can't have velocity in there, so that's clearly P/m, but I think the answer in the back of the book added the kinetic energy relative to the center of mass reference frame...which would be zero right?

I don't know what you mean with "kinetic energy relative to the center of mass frame"
What is asked is the kinetice energy in the center of mass frame, which is just
$$(1/2)m_1 {v_1}^2 + (1/2)m_2 {v_2}^2$$
for each object, where v1 and v2 are the speeds in the center of mass frame.

you need to convert this to an expression only P1, m1 and m2[/QUOTE]

Since v=p/m, i would just substitute that in for the velocities in the kinetic energy equation you listed right? doing this still wouldn't give me the book answer, so there must be something else.

 

[willem2]

Since v=p/m, i would just substitute that in for the velocities in the kinetic energy equation you listed right? doing this still wouldn't give me the book answer, so there must be something else.

It won't give you the book answer because it's wrong.

 

[nlsherrill]

It won't give you the book answer because it's wrong.

My answer would be wrong? You indicated that getting the kinetic energy of the system equation in terms of only m1, m2, p1 was what you do, and to do that you have to get rid of velocity which is easy? Now I am even more confused.

edit: sorry I reread your earlier response. You believe the book answer for part a) is incorrect, and so do other students in my class, so I hope the way you told me was the correct way. thanks for your help.

 

[willem2]

My answer would be wrong? You indicated that getting the kinetic energy of the system equation in terms of only m1, m2, p1 was what you do, and to do that you have to get rid of velocity which is easy? Now I am even more confused.

The book answer, this: Ki=Kf=P1^2/2(m1^2 +6m1m2 + m2^2/m1^2m2 + m1m2^2)

is wrong.

 

[nlsherrill]

The book answer, this: Ki=Kf=P1^2/2(m1^2 +6m1m2 + m2^2/m1^2m2 + m1m2^2)

is wrong.

okay yeah sorry I read your earlier post wrong. thanks for your time/help/patience.