Inelastic Scattering - Calculate Energy Change

AI Thread Summary
The discussion focuses on calculating the energy change during an inelastic scattering collision between two objects of equal mass, where one is initially stationary. The user presents equations for energy and momentum conservation, attempting to derive the final energy of the incoming object after the collision based on the scattering angle. Concerns are raised about the assumption that both objects deflect at the same angle, with suggestions that they actually deflect at right angles to each other. The user is seeking validation of their calculations and whether any assumptions made may lead to incorrect conclusions about energy loss.
ajhunte
Messages
12
Reaction score
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 (\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)<br />

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 [\b]
\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(\theta)}{2}

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.
 
Physics news on Phys.org
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 Right there, look at the p2 term.
 
Last edited by a moderator:
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top