An unstable nucleus of mass 17.0 Times 10^-20 kg initially at rest disintegrates

Question

An unstable nucleus of mass 17.0 Times 10^-20 kg initially at rest disintegrates into three particles moving in the x-y plane. One of the particles moves in the y-direction with a velocity of 5.00 Times 10^6 m/s (m_1 = 5.00 times 10^-20 kg), m_2 = 8.40 times 10^-20 kg is moving in the x-direction at 4.00 times 10^6 m/s. Find the velocity of the third particle. 1.16 Times 10^7 m/s; theta_3 = 36.7 degree Q3 How much energy was released when the atom fissured into three particles. 3.73 Times 10^-6 J

Explanation / Answer

Use conservation of momentum.
The nucleus is at rest, so the initial momentum is zero.
The final (total) momentum must also be zero.

First find the the mass of the third particle.
m = (17×10^-20 kg) - (5×10^-20 kg) - (8.4×10^-20 kg)
m = 3.6×10^-20 kg

The first particle moves along the y axis, so it doesn't have any x momentum.
The second particle moves along the x axis, so it doesn't have any y momentum.

The x component of the velocity of the third particle will be:
(8.4×10^-20 kg)×(4×10^6 m/s) + (3.6×10^-20 kg)×Vx = 0
Vx = -9.33×10^6 m/s

The y component of the velocity of the third particle will be:
(5×10^-20 kg)×(5×10^6 m/s) + (3.6×10^-20 kg)×Vy = 0
Vy = -6.944×10^6 m/s

The speed of the third particle is:
V = ( (-9.33×10^6 m/s)² + (-6.944×10^6 m/s)² )
V = 1.16 * 10^7 m/s < - - - - - - - - - - - - - - - - - - - - - - - - - - - - - answer a

The direction of the third particle is:
= arctan( (-6.944×10^6 m/s) / (-9.33×10^6 m/s) )
= 36.66°

b. Energy released =

[0.5 * 5*10^-20*(5*10^6)^2] + [0.5 * 8.4*10^-20*(4*10^6)^2] + [0.5 *3.6*10^-20*(1.16*10^7)^2]

= 3.73 * 10^-6 J

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