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 = 1.0 ✕ 10−27 kg, moves in the positive y-direction with speed v1 = 5.0 ✕ 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.)  

Answer 1

Mass of third particle = mass of nucleus - ( mass of two particles)

Mass of third particle = 1.7e-26 - ( 1e-27 + 8e-27)

Mass of third particle = 8e-27 Kg

so,

we know

m1v1y + m2v2y + m3v3y = 0

v3y = - 1e-27 * 5e6 - 8e-27 * 0 / 8e-27

v3y = - 6.25e5 m/s

and

similarly

v3x = - 8e-27 * 3.8e6 / 8e-27

v3x = - 3.8e6 m/s

so,

magnitude

v3 = sqrt ( v3x² + v3y²)

v3 = 3.85e6 m/s

and

direction = 9.34 degree ( below negative x - axis)

or

direction = 189.34 degree ( counter clockwise from + x -axis)