What Speed and Angle Does m2 Leave After Collision?

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The discussion centers on a physics problem involving a collision between two masses, m1 and m2, where m1 strikes m2, which is initially at rest. The key focus is on determining the speed and angle at which m2 departs after the collision, given that half of the initial kinetic energy is lost. Participants are analyzing the conservation of momentum and kinetic energy equations to solve for the unknowns, specifically the angle θ and speed u2 of mass m2. There is some confusion regarding the distinction between kinetic energy and momentum in the calculations. Clarification on the correct equations and approach is sought to resolve the problem effectively.
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Homework Statement



A mass "m1" moves along the x-axis with velocity of magnitude v0 on a frictionless table. It strikes another mass m2 which is initially at rest. The mass m1 goes off along the y-axis. If half the original kinetic energy is lost in the collision, with what speed and at what angle does m2 leave the point of collision?

Homework Equations



1/2KE(initial) = KE(final)
(1/2)Pix=Pfx
(1/2)Piy=Pfy

The Attempt at a Solution



Pix=vom1
Pfx=u2m2cosθ
Piy=0
Pfy=u1m1 - u2m2sinθ

I think these are the initial and final components of momentum, but I don't know how to find θ and u2 in terms of known variables.
 
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hi texan14! :smile:
texan14 said:
1/2KE(initial) = KE(final)
(1/2)Pix=Pfx
(1/2)Piy=Pfy
…
Pix=vom1
Pfx=u2m2cosθ
Piy=0
Pfy=u1m1 - u2m2sinθ

you seem to be confusing KE and momentum :confused:

what is your KE equation?​
 

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