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## Homework Statement

A particle of mass 3.2kg is moving due west with a velocity of 6m/s. Another particle of mass 1.6kg is moving due north with a velocity of 5.0m/s. The two particles are interacting. After 2s the first particle is moving in the direction N 30° E with a velocity of 3m/s. Find the magnitude and direction of the second particle.

## Homework Equations

Δp

_{1}= -Δp

_{2}

## The Attempt at a Solution

Okay, so I have all the information I need except that I need to decompose the direction of v

_{1}' (i.e. the velocity of the first particle after the interaction) into coordinates along the x and y axes.

Here is my data:

v

_{1}= -6i

v

_{1}' = 3(cos(60)i+sin(60)j) = 1.5i +2.6j

v

_{2}= 5j

v

_{2}' = ?

I write m

_{1}(v

_{1}' - v

_{1}) = -m

_{2}(v

_{2}' - v

_{2}) (i.e. Δp

_{1}= -Δp

_{2})

Solving for v

_{2}' I get:

(m

_{1}(v

_{1}- v

_{1}')/m

_{2}) + v

_{2}= v

_{2}'

I put in the relevant data and get:

-15i -0.2j = v

_{2}'

But apparently this isn't correct. I keep checking my equation against the provided answer but I can't see where my mistake is even though I know my answer is wrong. I think I made a mistake when I decomposed the vector but I don't see how.

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