Electron motion after collision in a magnetic field

[jjson775]

Homework Statement

One electron collides elastically with another electron at rest. After their collision, the radii of their trajectories are 1.00 cm and 2.40 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.044T. Determine the energy in KeV of the incident electron.

Relevant Equations

r = mv/qB

Isn‘t this a straightforward problem of conservation of momentum? See attached.

 

[TSny]

Although momentum will be conserved in the collision, the collision is not necessarily a "head on" collision. Immediately after the collision, the two electrons are not necessarily moving along the same line. Since momentum is a vector quantity, the initial momentum of the incoming electron is the vector sum of the two outgoing momentum vectors. There is not enough information given about the directions of the outgoing momenta to use conservation of momentum.

But note that your are given that the collision is elastic.

 

[kuruman]

There is not enough information given about the directions of the outgoing momenta to use conservation of momentum.

When two equal masses collide elastically head-on, they exchange velocities. This cannot be the case here because we are told that both electrons move after the collision. Therefore, the collision is not head-on.

When two equal masses collide elastically not head-on with the target mass at rest, the angle between the final velocities (in the lab frame) is 90^o as required by momentum conservation and the Pythagorean theorem.

To @jjson775 : You need to rethink your approach given that the collision is not head-on.

 

[TSny]

When two equal masses collide elastically not head-on with the target mass at rest, the angle between the final velocities (in the lab frame) is 90^o as required by momentum conservation and the Pythagorean theorem.

Yes, using these facts about elastic collisions of equal-mass particles, you can deduce the directions of the individual momenta of the two electrons immediately after the collision. Nice.

But it seems to me that it's easier to solve the problem using only conservation of energy.

 

[jjson775]

I added the kinetic energy (a scalar quantity) of the 2 electrons and immediately got the right answer. The problem statement gave no information about the momenta after the collision so no way to add the vectors. Thanks

 

[kuruman]

But it seems to me that it's easier to solve the problem using only conservation of energy.

Not really, both are equally easy because the energy conservation solution morphs into the momentum conservation solution. One doesn't need the angles of the final momenta relative to the direction of the initial momentum, the Pythagorean theorem takes care of that. Of course, both approaches require finding the final speeds from the trajectories.
Energy conservation $$\frac{P_0^2}{2m}=\frac{P_1^2}{2m}+\frac{P_2^2}{2m}$$ Momentum conservation $$\frac{P_0^2}{\cancel{2m}}=\frac{P_1^2}{\cancel{2m}}+\frac{P_2^2}{\cancel{2m}}.$$

 

[TSny]

Not really, both are equally easy because the energy conservation solution morphs into the momentum conservation solution. One doesn't need the angles of the final momenta relative to the direction of the initial momentum, the Pythagorean theorem takes care of that. Of course, both approaches require finding the final speeds from the trajectories.
Energy conservation $$\frac{P_0^2}{2m}=\frac{P_1^2}{2m}+\frac{P_2^2}{2m}$$ Momentum conservation $$\frac{P_0^2}{\cancel{2m}}=\frac{P_1^2}{\cancel{2m}}+\frac{P_2^2}{\cancel{2m}}.$$

Yes, I see. Writing the momentum conservation as $P_0^2 = P_1^2 + P_2^2$ does require knowing that the final directions are perpendicular. But that's fine. Good.

 

[jjson775]

Not really, both are equally easy because the energy conservation solution morphs into the momentum conservation solution. One doesn't need the angles of the final momenta relative to the direction of the initial momentum, the Pythagorean theorem takes care of that. Of course, both approaches require finding the final speeds from the trajectories.
Energy conservation $$\frac{P_0^2}{2m}=\frac{P_1^2}{2m}+\frac{P_2^2}{2m}$$ Momentum conservation $$\frac{P_0^2}{\cancel{2m}}=\frac{P_1^2}{\cancel{2m}}+\frac{P_2^2}{\cancel{2m}}.$$

That’s right. I did not consider this.

 

[Delta2]

So if I 've understood well, a consequence of the collision being elastic is that the velocities after the collision are perpendicular, right?

 

[Delta2]

The question I want desperately but being afraid to ask: How two point particles collide but not with a "head on " collision

 

[TSny]

The question I want desperately but being afraid to ask: How two point particles collide but not with a "head on " collision

Shhh ...You're not allowed to ask that

 

[TSny]

So if I 've understood well, a consequence of the collision being elastic is that the velocities after the collision are perpendicular, right?

Yes. It's a consequence of

1. elastic collision
2. one particle is at rest before collision
3. the two particles have same mass

@kuruman has essentially provided a proof. If $\vec P_0$ is the momentum of the moving particle before the collsion and $\vec P_1$ and $\vec P_2$ the momenta after the collision then conservation of energy can be written as $$\frac{P_0^2}{2m} = \frac{P_1^2}{2m}+ \frac{P_2^2}{2m}$$ or $$P_0^2 = P_1^2 + P_2^2 \;\;\;\;\; [\rm {eq \; 1}]$$

"Squaring" the conservation of momentum equation gives $$(\vec P_0)^2 = \left( \vec P_1 + \vec P_2 \right)^2$$ or $$P_0^2 = P_1^2 + P_2^2 + 2 \vec P_1 \cdot \vec P_2 \;\;\;\;\; [\rm {eq \; 2}]$$

Comparing eq. 1 and eq. 2 shows that $\vec P_1 \cdot \vec P_2 = 0$

Edit: This proof assumes Newtonian mechanics. For relativistic speeds, the final momenta will not generally be perpendicular. For the numbers in this problem, I find that the incoming electron actually has a speed of $.56c$. So, we're encroaching into the relativistic domain. If I didn't make any mistakes, the relativistic calculation gives an answer for the initial KE of 107 keV while the Newtonian result is 115 keV. So, they agree pretty well to 2 significant figures.

 

[PeroK]

The question I want desperately but being afraid to ask: How two point particles collide but not with a "head on " collision

It's the influence of the electric fields that causes the deflection and the angle of deflection is determined by the impact parameter. A classical example is Rutherford scattering:

http://hyperphysics.phy-astr.gsu.edu/hbase/rutsca.html

And, in fact, a spaceship that slingshots round a planet is also an elastic collision.