Question

An unstable particle with a mass equal to 3.34 ✕ 10⁻²⁷ kg is initially at rest. The particle decays into two fragments that fly off with velocities of 0.979c and −0.851c, respectively. Find the masses of the fragments. (Hint: Conserve both mass–energy and momentum.)

m_(0.979c) =

m_(-0.851c) =

Answer 1

let mass of first fragment is m1 and second fragment mass is m2.

Then m1 + m2 = 3.34 x 10-27 kg

Using law of conservation of momentum ,

Initial momentum is zero as particle is at rest. Then final momentum of the system must be zero.

Final momentum = m1 x (0.979c) + m2 x (−0.851c) = 0

m1 = 0.87 m2 (nearly)

since m1 + m2 = 3.34 x 10-27 kg then

1.87 m2 = 3.34 x 10-27 kg

m2 = 1.786 x 10-27 kg

m1 = 1.554 x 10-27 kg

m(0.979c) = 1.55 x 10-27 kg

m(-0.851c) = 1.79 x 10-27 kg