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Elastic collisions question

An atom of mass m moving in the x direction with speed v collides elastically with an atom of mass 3m at rest. After the collision the first atom moves in the y direction. Find the direction of motion of the second atom and speed of both atoms (in terms of v) after the collision?
Ans: mass m moves at v/sqrt(2), mass 3m moves at v/sqrt(6) in the direction theta=-35.3 degrees.

Momentum before = momentum after

MV1i + MV2i = MV1f + MV2f

x-direction MV1i = MV1fx + MV2fx

y-direction 0 = MV1fy + MV2fy

this translates into

mv = mV1fx + 3mV2fx ....................(i)

0 = mV1fy + 3mV2fy ....................(ii)

mv^2 = m(0 + V1fy^2) + 3m(V2fx^2 + V2fy^2) ........(iii)

where :
M, V1i, V2i, V1f, V2f, V1fx, V2fx, V1fy, V2fy , etc are variables

m,v are values

x = x-component , y = y-component, f = final , i = initial

The question is are these equations right, as I cannot get the above answers
Thanks and regards.
 
Last edited:

Doc Al

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John O' Meara said:
this translates into

mv = mV1fx + 3mV2fx ....................(i)

0 = mV1fy + 3mV2fy ....................(ii)

mv^2 = m(0 + V1fy^2) + 3m(V2fx^2 + V2fy^2) ........(iii)
OK, but realize that V1fx = 0, V1fy = v1f. That will simplify things, giving you three equations and three unknowns.
 

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