Inelastic Scattering - Calculate Energy Change

In summary, the conversation is discussing how to find the change in energy of an incoming object after colliding with a stationary object of the same mass at a certain angle. The conversation involves using equations for energy and momentum balance, and eventually leads to the solution that the energy of the incoming object after the collision is equal to half of its initial energy multiplied by the cosine of the angle at which it was scattered. However, there is a potential mistake in assuming that both objects deflect at the same angle, as objects always deflect at right angles to each other.
  • #1
ajhunte
12
0
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 ([tex]\theta[/tex]) 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.)

[tex]E_{i}[/tex]=Energy of Incoming Object before collision
[tex]E_{f}[/tex]=Energy of Incoming Object after collision
[tex]E_{2}[/tex]=Energy of Target Object after collision

[tex]p_{i}[/tex]= Momentum of incoming object before collision.
[tex]p_{f}[/tex]=Momentum of Incoming object after collision.
[tex]p_{2}[/tex]=Momentum of Target object after collion.

[tex]\phi[/tex]= Arbitrary Angle of Target object scattering (should not matter based on the note.


Homework Equations


[tex]E=1/2*m*v^{2}[/tex]
p=m*v
[tex]p^{2}/(2*m)=E[/tex]

The Attempt at a Solution



Energy Balance:
[tex]E_{i}=E_{f}+E_{2}[/tex]

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

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

Squaring only the Momentum Equations and adding them together.

Y-Balance:
[tex]p_{f}^{2}*Sin^{2}(\theta)=p_{2}^{2}*Sin^{2}(\phi)[/tex]

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

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

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:
[tex]E_{2}=E_{i}+E_{f}-\sqrt{E_{i}}*\sqrt{E_{f}}*Cos(\theta)[/tex]

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

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

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


Minus Sign Answer Leads to 0, so nontrivial answer is:
[tex]E_{f}=\frac{E_{i}*Cos(\theta)}{2}[/tex]

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 1/2.
 
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  • #2
I think your error is in assuming they both deflect at the same angle.

Objects always deflect at right angles to each other. That's how you remove the resultant theta of the object that is originally stationary.

https://www.physicsforums.com/latex_images/17/1773673-11.png [Broken] Right there, look at the p2 term.
 
Last edited by a moderator:
  • #3


Your solution looks correct to me. I don't see any mistakes or false assumptions. However, I would recommend double checking your calculations and equations just to be sure. Also, make sure to include units in your final answer. Good job!
 

What is inelastic scattering?

Inelastic scattering is a process in which a particle or photon collides with another particle and loses some of its energy. This results in a change in the wavelength or frequency of the scattered particle.

How is energy change calculated in inelastic scattering?

The energy change in inelastic scattering can be calculated using the formula: ΔE = Ei - Ef, where ΔE is the change in energy, Ei is the initial energy of the particle, and Ef is the final energy of the particle.

What factors affect the energy change in inelastic scattering?

The energy change in inelastic scattering is affected by the energy and mass of the particles involved, as well as the angle and speed of the collision. It can also be influenced by external factors such as temperature and pressure.

How is inelastic scattering different from elastic scattering?

Inelastic scattering results in a loss of energy, while elastic scattering does not. Additionally, inelastic scattering involves a change in the wavelength or frequency of the scattered particle, while elastic scattering does not.

What are some real-life applications of inelastic scattering?

Inelastic scattering is used in various fields such as nuclear physics, materials science, and medical imaging. It is used to study the structure of materials, identify unknown substances, and develop new technologies such as positron emission tomography (PET) scans.

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