Can Someone look over this and tell me if the work is correct.(adsbygoogle = window.adsbygoogle || []).push({});

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

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.

2. Relevant equations

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

p=m*v

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

3. 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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# Homework Help: Inelastic SCATTERING

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