Calculating Energy Change in Compton Scattering

[ajhunte]

Can Someone look over this and tell me if the work is correct.

Homework Statement

An object comes in with known velocity (v) and known mass (m) moving directly along the x-axis. It strikes another object of the same mass (m) which is stationary. Find the change in energy of the incoming object based on the angle at which it scattered ($$\theta$$) and the velocity of the target after the collision (u). The collision only occurs in 2 dimensions. (Note: The angle at which the target object is moving can be removed algebraically.)

$$E_{i}$$=Energy of Incoming Object before collision
$$E_{f}$$=Energy of Incoming Object after collision
$$E_{2}$$=Energy of Target Object after collision

$$p_{i}$$= Momentum of incoming object before collision.
$$p_{f}$$=Momentum of Incoming object after collision.
$$p_{2}$$=Momentum of Target object after collion.

$$\phi$$= Arbitrary Angle of Target object scattering (should not matter based on the note.

Homework Equations

$$E=1/2*m*v^{2}$$
p=m*v
$$p^{2}/(2*m)=E$$

The Attempt at a Solution

Energy Balance:
$$E_{i}=E_{f}+E_{2}$$

X-Momentum Balance:
$$p_{i}=p_{f}*Cos(\theta) +p_{2}*Cos(\phi)$$

Y-Momentum Balance: (This should be a zero momentum system in y-direction)
$$p_{f}*Sin(\theta)=p_{2}*Sin(\phi)$$

Squaring only the Momentum Equations and adding them together.

Y-Balance:
$$p_{f}^{2}*Sin^{2}(\theta)=p_{2}^{2}*Sin^{2}(\phi)$$

X-Balance: (after getting $$\phi$$ isolated on one side then squaring)
$$p_{2}^{2}*Cos^{2}(\phi)=p_{i}^{2} - p_{i}*p_{f}*Cos(\theta)+p_{f}^{2}*Cos^{2}(\theta)$$

Adding the two momentum Equations and using $$Sin^{2}+Cos^{2}=1$$
$$p_{2}^{2}=p_{i}^{2}-p_{i}*p_{f}*Cos(\theta)+p_{f}^{2}$$

Relating back to energy, using the relationship defined in section 2.
Since all masses are the same I divide the newly found momentum equation by 2m in order to get to energy:
$$E_{2}=E_{i}+E_{f}-\sqrt{E_{i}}*\sqrt{E_{f}}*Cos(\theta)$$

(Note: $$\sqrt{2m}*\sqrt{2m}=2m$$ and $$p/\sqrt{2m}=\sqrt{E}$$ )

Combining with the original Energy Balance Equation to Eliminate $$E_{2}$$. This involves subtracting the equation I just solved for and the original equation.
$$0=2*E_{f}-\sqrt{E_{i}}*\sqrt{E_{f}}*Cos(\theta)$$

Applying the Quadratic Equation
$$\sqrt{E_{f}}=\sqrt{E_{i}}*Cos(\theta) +\- \frac{\sqrt{\sqrt{E_{i}}^{2}*Cos^{2}(\theta)-0}}{4}$$


Minus Sign Answer Leads to 0, so nontrivial answer is:
$$E_{f}=\frac{E_{i}*Cos^{2}(\theta)}{4}$$

Is this the correct solution, or is there a step that I made a mistake or false assumption? It seems to me that this is wrong, because even a grazing trajectory decreases the initial energy by 3/4.

 

[Mindscrape]

This is essentially compton scattering. Why do you have to angles? Also, are you assuming it is inelastic, or were you told that it is inelastic? I would probably assume an elastic collision.

 

[ajhunte]

Your right, this is basically compton scattering without the relativisitic effects. That is why the first 4 steps are the same as the Compton Scattering derivation. The difference is that the since this does not use relativistic effects:
$$E=\frac{p^{2}}{2m}$$

For Compton Scattering, the conversion between energy and momentum for Photons and relativistic electron is different:

For Relativistic Electrons: $$pc=\sqrt{E_{0}^{2}-m_{0}^{2}c^{4}}$$

For Photons: $$\frac{E}{c}$$

This is where the difference is. That is why I asked the question, because this answer does not seem intuitive that if I have a grazing trajectory (i.e. $$\theta$$ goes to 0) I still lose 3/4 of the energy.

 

[alphysicist]

Hi ajhunte,

Can Someone look over this and tell me if the work is correct.

Homework Statement

An object comes in with known velocity (v) and known mass (m) moving directly along the x-axis. It strikes another object of the same mass (m) which is stationary. Find the change in energy of the incoming object based on the angle at which it scattered ($$\theta$$) and the velocity of the target after the collision (u). The collision only occurs in 2 dimensions. (Note: The angle at which the target object is moving can be removed algebraically.)

$$E_{i}$$=Energy of Incoming Object before collision
$$E_{f}$$=Energy of Incoming Object after collision
$$E_{2}$$=Energy of Target Object after collision

$$p_{i}$$= Momentum of incoming object before collision.
$$p_{f}$$=Momentum of Incoming object after collision.
$$p_{2}$$=Momentum of Target object after collion.

$$\phi$$= Arbitrary Angle of Target object scattering (should not matter based on the note.

Homework Equations

$$E=1/2*m*v^{2}$$
p=m*v
$$p^{2}/(2*m)=E$$

The Attempt at a Solution

Energy Balance:
$$E_{i}=E_{f}+E_{2}$$

X-Momentum Balance:
$$p_{i}=p_{f}*Cos(\theta) +p_{2}*Cos(\phi)$$

Y-Momentum Balance: (This should be a zero momentum system in y-direction)
$$p_{f}*Sin(\theta)=p_{2}*Sin(\phi)$$

Squaring only the Momentum Equations and adding them together.

Y-Balance:
$$p_{f}^{2}*Sin^{2}(\theta)=p_{2}^{2}*Sin^{2}(\phi)$$

X-Balance: (after getting $$\phi$$ isolated on one side then squaring)
$$p_{2}^{2}*Cos^{2}(\phi)=p_{i}^{2} - p_{i}*p_{f}*Cos(\theta)+p_{f}^{2}*Cos^{2}(\theta)$$

You are missing a factor of two here. The factor of two will keep the four from appearing in the denominator of the final answer.

 

[ajhunte]

Thank You, That is exactly what I was missing.