Collision of Two Particles: Conservation of Momentum and Final Velocities

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Homework Statement

A particle of mass 5.0kg travels initially with a velocity of 8.0mˆı and then interacts with a particle of mass 6.0kg which was initially at rest. After the interaction the 5.0kg mass travels at a speed of 4.0m/s along a direction which makes an angle of 53◦ with the x-axis.

What is the change in momentum of the 5kg mass? -28i -16j
What is the speed of the 6kg mass after the interaction?

Homework Equations

m1v1=(m1v2)f+m2v2

The Attempt at a Solution

So I am thinking the momentum must be conserved. If the initial velocity was 8m/s in the i direction, the final velocity must be as well. The final velocity of the 5kg mass in component form is 4cos(307)=2.4i and 4sin(307)=-3.2j. This means the final velocity of the 6kg mass is 5.6i and 3.2j, giving 6.5 as the final velocity.

Apparently the answer is 5.4, can you show me what i should be doing instead. Thanks :)

 

[Qwertywerty]

So I am thinking the momentum must be conserved. If the initial velocity was 8m/s in the i direction, the final velocity must be as well.

These two statements, don't mean the same thing. How did you conclude that the final velocity must equal the initial?

 

[heartshapedbox]

Ok i see what i did wrong, lol, long day.
Total initial momentum: 40i
5kg mass final momentum: 12i, -16j
Due to conservation of momentum: 6kg final momentum: 28i, 16j
sqrt((28/6)^2+(-16/6)^2)
final speed: 5.4m/s