physics question. please help!

Question

An unstable nucleus of mass 1.7 ✕ 10⁻²⁶ kg, initially at rest at the origin of a coordinate system, disintegrates into three particles. One particle, having a mass of m₁ = 1.8 ✕ 10⁻²⁷ kg, moves in the positive y-direction with speed v₁ = 5.0 ✕ 10⁶ m/s. Another particle, of mass m₂ = 7.0 ✕ 10⁻²⁷ kg, moves in the positive x-direction with speed v₂ = 3.6 ✕ 10⁶ m/s. Find the magnitude and direction of the velocity of the third particle. (Assume that the +x-axis is to the right and the +y-axis is up along the page.) 

a) magnitude: ____ m/s

b) direction: ____  ° counterclockwise from the +x-axis

Answer 1

Use conservation of momentum in x and y directions.

In positive x direction initial momentum = 0.

Mass of 3rd particle = 1.7 x 10^-26 - 1.8 x 10^ -27 - 7 x 10^-27 = 8.2 x 10 ^ -27

So if vx is speed of 3rd particle in positive x direction, (7 x 10 ^ -27) (3.6 x 10^6) + (8.2 x 10^-27) vx = 0

giving vx = -3 x 10^6 m/s (so 3 x 10^6 m/s in the negative x direction)

In positive y direction initial momentum is 0

So if vy is speed of 3rd particle in positive y direction

1,8 x 10^-27 x 5 x 10^6 + 8.2 x 10^-27 x vy = 0 giving vy = - 1 x 10^6 m/s, so 1 x 10^6 in negative y direction

So magnitude of velocity = √ (vx^2 + vy^2) [ you can find that ]

and if the angle is 180 + x° counterclockwise from x axis then tan x = (1 x 10^6 / 3 x 10^6) so you can find x