Glancing Collisions In 2D, help simplifying

AI Thread Summary
The discussion centers on a physics problem involving a glancing collision in 2D, where a mass m1 collides with a stationary mass 2m. The goal is to calculate the velocities V1 and V2 after the collision, given that both masses move off at 45 degrees. Participants clarify the application of momentum conservation equations in both x and y directions, correcting initial attempts at these equations. The correct equations involve the total initial momentum equating to the total final momentum for both axes. One participant successfully simplified the problem and achieved a high score on their assignment.
Visionary_
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Homework Statement


We are only given variables in this problem, that is:
m1 travels at V, collides with 2m which is at rest
results in m going off at 45° at V1
2m also goes off at 45° down at V2
Calculate V1 and V2 in terms of V


Homework Equations


Ʃρx-initial=Ʃρx-final
Ʃρy-initial=Ʃρy-final
Use Linear combinations; sin45= √2/2=cos45

The Attempt at a Solution


x direction:
mv1=mv1√2/2+2mv√2/2
y direction
0=√2/2(mv1+2mv)
 
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Visionary_ said:

Homework Statement


We are only given variables in this problem, that is:
m1 travels at V, collides with 2m which is at rest
results in m going off at 45° at V1
2m also goes off at 45° down at V2
Calculate V1 and V2 in terms of V


Homework Equations


Ʃρx-initial=Ʃρx-final
Ʃρy-initial=Ʃρy-final
Use Linear combinations; sin45= √2/2=cos45

The Attempt at a Solution


x direction:
mv1=mv1√2/2+2mv√2/2
y direction
0=√2/2(mv1+2mv)

The equations you have written out are not quite right.

You have:

x direction:
mv1=mv1√2/2+2mv√2/2
y direction
0=√2/2(mv1+2mv)

but were perhaps trying for

x direction:
m1V=m1V1√2/2+m2V2√2/2
y direction
0=√2/2(m1V1+m2V2)

Though perhaps you really did mean masses m and 2m ?
 
yes the second mass was 2m as in twice the mass,
i simplified the problem further and got a 90% on the paper so its fine :)
 
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