In: Physics
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 = 1.0 ✕ 10−27 kg, moves in the positive y-direction with speed v1 = 5.4 ✕ 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.)
Initially the system is at rest, the initial momentum of the system is zero.
The first particle move in the positive y-direction, the momentum of the first particle is
The second particle move in the positive x-direction, the momentum of the second particle is
The mass of the third particle is
where M is the mass of the unstable nucleus.
Let v3 be the velocity of the third particle, its momentum is
Total final momentum of the system is
Using conservation of momentum
We get
Substituting , , , , we get
The magnitude of the Velocity is
Since both x and y-component of the velocity vector are negative, the velocity points in the third quadrant. The angle that the velocity makes with the positive x-axis is
Since the vector points in the third quadrant