Ask Your Teacher 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 = 3.4 ✕ 10−27 kg, moves in the positive y-direction with speed v1 = 5.8 ✕ 106 m/s. Another particle, of mass m2 = 8.0 ✕ 10−27 kg, moves in the positive x-direction with speed v2 = 3.8 ✕ 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.)

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

Ithink the answer is fusion