Question

An unstable nucleus with a mass of 16.3 × 10−27 kg initially at rest disintegrates into three particles. One of the particles, of mass 4.9 × 10−27 kg, moves along the positive yaxis with a speed of 4.5 × 106 m/s. Another particle, of mass 8.7 × 10−27 kg, moves along the positive x-axis with a speed of 3.4 × 106 m/s.

a) Find the speed of the third particle. Answer in units of m/s.

b) At what angle does the third particle move?

Answer 1

Part A.

Since there was no external force applied, So total momentum before and after disintegration will remain conserved, So

Pi = Pf

Pi = 0, since initially unstable nucleus was at rest, So

Pf = 0

m1*v1 + m2*v2 + m3*v3 = 0

m1 = mass of 1st particle = 4.9*10^-27 kg

m2 = mass of 2nd particle = 8.7*10^-27 kg

m3 = mass of 3rd particle = M - (m1 + m2) = 16.3*10^-27 - (4.9*10^-27 + 8.7*10^-27) = 2.7*10^-27 kg

v1 = 4.5*10^6 m/s in +y direction = (4.5*10^6 j) m/s

v2 = 3.4*10^6 m/s in +x direction = (3.4*10^6 i) m/s

v3 = Velocity of 3rd particle = ?

So, Using given values:

4.9*10^-27*4.5*10^6 j + 8.7*10^-27*3.4*10^6 i + 2.7*10^-27*v3 = 0

v3 = -(4.9*10^-27*4.5*10^6 j + 8.7*10^-27*3.4*10^6 i)/(2.7*10^-27)

v3 = -8.7*10^-27*3.4*10^6/(2.7*10^-27) i - 4.9*10^-27*4.5*10^6/(2.7*10^-27) j

v3 = (-10.9556*10^6 i - 8.1667*10^6 j) m/s

Speed of 3rd particle will be:

|v3| = 10^6*sqrt ((-10.9556)^2 + (-8.1667)^2)

|v3| = 13.66*10^6 m/s

Part B.

Since Vx < 0 and Vy < 0 for 3rd particle, So particle is moving in 3rd quadrant,

Direction = arctan (Vy/Vx) = arctan (8.1667/10.9556)

Direction = 36.7 deg below -ve x-axis = 216.7 deg Counterclockwise from +ve x-axis

Let me know if you've any query.