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.
Δp1 = -Δp2
The Attempt at a Solution
Okay, so I have all the information I need except that I need to decompose the direction of v1' (i.e. the velocity of the first particle after the interaction) into coordinates along the x and y axes.
Here is my data:
v1 = -6i
v1' = 3(cos(60)i+sin(60)j) = 1.5i +2.6j
v2 = 5j
v2' = ?
I write m1(v1' - v1) = -m2(v2' - v2) (i.e. Δp1 = -Δp2)
Solving for v2' I get:
(m1(v1 - v1')/m2) + v2 = v2'
I put in the relevant data and get:
-15i -0.2j = v2'
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.