An unstable atomic nucleus of mass 1.58×10−26 kg initially at rest disintegrates into three particles. one of the particles, of mass 5.04×10−27 kg, moves in the y-direction with a speed of 6.00×106 m/s. another particle, of mass 8.50×10−27 kg, moves in the x-direction with a speed of 4.00×106 m/s. assume mass is conserved in this process. a. find the x-component of the velocity of the third particle. b. find the y-component of the velocity of the third particle. c. find the total kinetic energy increase in the process.

Explanation:

We can take conservation of linear momentum to find the velocities:

The initial momentum of the nucleus will be:

as is at rest.

After the decay, the first particle has a momentum

the second one has a momentum

By conservation of linear momentum we have:

for the third particle, we know that mass is conserved:

The velocity will be:

The kinetic energy is given by

And, as the initial kinetic energy is zero, this must be the increase in energy.

(a) - 4.9 x 10^6 m/s

(b) - 4.38 x 10^6 m/s

(c) 17.36 x 10^-14 J

Explanation:

M = 1.58 x 10^-26 kg

U = 0

m1 = 5.04 x 10^-27 kg

v1 = 6 x 10^6 m/s along Y axis

m2 = 8.5 x 10^-27 kg

v2 = 4 x 10^6 m/s along X axis

Let m3 be the mass of third particle which is moving with velocity v having X axis component is vx and y axis component is vy.

So, m3 = M - m1 - m2 = 1.58 x 10^-26 - 5.04 x 10^-27 - 8.5 x 10^-27

m3 = 6.9 x 10^-27 kg

(a) Use the conservation of momentum along X axis

M x U = m1 x 0 + m2 x v2 + m3 x vx

0 = 0 + 8.5 x 10^-27 x 4 x 10^6 + 6.9 x 10^-27 x vx

vx = - 4.9 x 10^6 m/s

(b) Use the conservation of momentum along Y axis

M x U = m1 x v1 + m2 x 0 + m3 x vy

0 = 5.04 x 10^-27 x 6 x 10^6 + 0 + 6.9 x 10^-27 x vy

vy = - 4.38 x 10^6 m/s

(c) The resultant velocity of third mass,

The formula for the kinetic energy is

Total kinetic energy =

K = 17.36 x 10^-14 J

answer: the answer is a

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