Elastic Glancing Collision: Final Speeds and Scattering Angle

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SUMMARY

The discussion focuses on an elastic glancing collision involving two masses, m and 3m, moving towards each other with the same initial speed v. After the collision, mass m moves downwards at a right angle to its initial direction, while the final speed of both masses and the scattering angle of mass 3m are to be determined. The conservation of momentum and kinetic energy equations are applied, leading to the conclusion that the final speed of mass m is v and the velocity of mass 3m in the x-direction after the collision is (2/3)v. The challenge arises from the need to define new variables for the speeds post-collision due to the use of 'v' for initial speed.

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Homework Statement


Two masses (m and 3m) are moving towards each other along the x-axis with the same initial speed (v). Mass m is traveling to the left while mass 3m is traveling to the right. They undergo elastic glancing collision such that mass m is moving downwards after the collision at right angles from its initial direction. a) Find the final speed of the two masses and b) The angle at which the 3m mass is scattered.

Homework Equations


Ei=Ef
E=(0.5)mv^2
pi=pf
p=mv

The Attempt at a Solution


Pi=pf (x-direction)
3mv-mv = 3mv
2mv = 3mv
v=3mv/2m
v=mv
0=m

pi = pf (y-direction)
0=-mv+3m(1\3)v
mv=mv
 
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Pi=pf (x-direction)
3mv-mv = (3m)*Vx where Vx is the velocity of 3m in the x direction after collision
2mv = 3mv*Vx
Vx = (2/3)*v

pi = pf (y-direction)
0=-mv+3m(1\3)v
mv=mv
Oh, I made that mistake, too! But you can't use "v" for the downward speed of mass m after the collision as that letter is used for something else. Use another letter, perhaps "u". And Vy for the upward speed of 3m. Then you see that the equation can't be solved since you have two unknowns, u and Vy.

You can get another equation from the fact that the collision is elastic. That should be delightfully complicated.
 

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