Finding Final Velocity of Mass 1 After Collision

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SUMMARY

The discussion centers on calculating the final velocity of mass 1 (2.77 kg) after a collision with mass 2 (1 kg), which initially moves at rest and subsequently moves with a velocity of (2i - 3j). The conservation of linear momentum equations are applied, specifically M1V1ix + M2V2ix = M1V1fx + M2V2fx and M1V1iy + M2V2iy = M1V1fy + M2V2fy. The participant attempted to solve for the final velocity components of mass 1 but expressed confusion regarding the need for kinetic energy equations in elastic collisions. It was clarified that while momentum is conserved, the loss of kinetic energy does not imply the objects stick together.

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Nicki
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Homework Statement


Mass 1 (2.77 kg) moves with an initial velocity (1i - 1j) and mass 2 (1 kg) starts at rest. After the collision, M2 moves with a velocity of (2i - 3j). What is the final velocity of m1?

Homework Equations


Since they don't stick together, KE and linear momentum are conserved.
M1V1ix + M2V2ix = M1V1fx + M2V2fx

M1V1iy + M2V2iy = M1V1fy + M2V2fy

1/2 M1V1i^2 + 1/2M2V2i^2 = 1/2 M1V1f^2 + 1/2M2V2f^2

V1f = (sqrt)V1xf^2 + V1yf^2

The Attempt at a Solution


I don't really understand these problems, and we had a substitute professor into teach this, so here goes nothing, literally.
2.77 (1) + 0 = 2.77 Vx + 1(2) Vx = 0.278 m/s ?
2.77(-1) + 0 = 2.77 Vy + 1(-3) Vy = 0.0.83 m/s ?

This doesn't seem right? Shouldn't I need to use th equations for KE for an elastic collision?
 
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Nicki said:
Since they don't stick together, KE and linear momentum are conserved.
No. Momentum is conserved, but loss of some KE does not necessarily mean the objects coalesce. That is only the extreme case, in which the KE lost is maximised.
 

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