# Electron-Atom 1-D Collision

1. Jan 15, 2012

### R2Zero

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

An electron, mass m, collides head-on with mass M, initially at rest. As a result of the collision, a characteristic amount of energy E is stored internally within the atom. What is the minimum initial speed v0 that the electron must have? (Hint: Conservation principles lead to a quadratic equation for the final electron speed v and a quadratic equation for the final atom speed V. The minimum value v0 follows from the requirement that the radical in the solutions for v and V be real.)

2. Relevant equations

$$P=mv$$
$$KE = mv^2/2$$

3. The attempt at a solution

At first glance, I can tell its a momentum problem since its a collision and I'm given variables for mass and velocity. Since characteristic energy E is stored as a result of the collision, I know that this is an inelastic collision because kinetic energy is not conserved.

Since I know that the conservation of momentum still applies, I can say:

(1) mv0 = mv + MV

And also, since I know that total energy is conserved, I can also say:

(2) (1/2)mv02 = (1/2)mv2 + (1/2)MV2 + E

I've worked through this problem several times, and I know that somewhere I will get a quadratic equation for v or V, depending on which of the two I eliminate, then end up using the quadratic formula to get a solution.

If I eliminate the variable V (via substitution or whatever technique), I'm "supposed" to end up with the equation:

(3) (1 + m/M)v2 - (2m/M)v0v + 2E - (1 - m/M)v02 = 0.

I understand the concepts behind how the result is gotten, the problem is that when I do the substitution, I get the same result, except I end up with a 2E/m instead of a 2E. I'm not sure if I'm right or not, because if I attach units to equation (3) that my book and the TA gave me, it doesn't make sense to add Joules (kg m^2/s^2) to an equation where the rest of the units are m^2/s^2.

Moving along, solving for v yields:

(4) v = (m/M)v0 (+/-) [v02 - 2E(m+M)/mM]^(1/2)

from there you can just say that the minimum value of v0 is the square root of 2E(m+M)/mM.

As I stated before, I know the basic concepts, but I'm not sure how they got (3) from (1) and (2). Did the book and the TAs make a mistake?

2. Jan 16, 2012

### Redbelly98

Staff Emeritus
Welcome to Physics Forums.
An often useful way to spot an error is to look at the units associated with various terms in an equation. Look at the units of each term in the equation you are "supposed to" get:

(1 + m/M)v2: units are _____
(2m/M)v0v: units are _____
2E: units are _____
(1 - m/M)v02: units are _____
If these terms are to be added to and/or subtracted from each other, then the units must all agree. Do they? Finally, what happens if you have 2E/m instead of 2E?

3. Jan 16, 2012

### R2Zero

The units for 2E would be Joules, or kg (m^2/s^2), while the rest of the values are (m^2/s^2) or v^2, as the m/M units of kg/kg cancel out. Thus, 2E/m seems to work out in terms of units. I continued on the problem using 2E/m, and after tedious amounts of algebra I got that v0 must be at least [2E(m+M)/m]^(1/2), which looks similar to the book's answer of [2E(m+M)/(mM)]^(1/2).

4. Jan 16, 2012

### Redbelly98

Staff Emeritus
I agree that 2E/m is better than 2E, which must be wrong based on looking at the units. I haven't work this out myself to confirm your 2E/m.

However, note that your final answer has units of Energy1/2, which is not a velocity, whereas the book's answer does result in velocity units.

Whenever I get a result with the wrong units, I backtrack through my work to find the step where the units first became inconsistent. That's almost certainly where the error happened.

5. Jan 16, 2012

### R2Zero

I reworked the problem and found my mistake somewhere when I was trying to simplify the huge radical that came from using the quadratic formula. I finally got the right answer from the book, |v0| > [2E(M+m)/Mm]1/2.

In short, the book answer was right, but equation (3) provided by the book and my TA was wrong, as I reached the answer using 2E/m in place of 2E. Thanks for everything Redbelly!

6. Jan 16, 2012

### Redbelly98

Staff Emeritus
You're welcome. Hooray for units!