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A stationary Uranium nucleus decays, ejecting an alpha particle of mass roughly one-sixtieth of the Uranium...

A stationary Uranium nucleus decays, ejecting an alpha particle of mass roughly one-sixtieth of the Uranium nucleus' mass. The alpha particle travels horizontally to the right at a speed of 8.0×106 m/s. Which of the following is the best description of the motion of the remaining roughly 59/60 fraction of the Uranium nucleus (which is now a Thorium nucleus) after the decay?

Horizontally to the right with a speed of more than 8.0×106 m/s
Horizontally to the left with a speed of more than 8.0×106 m/s
Horizontally to the right with a speed equal to 8.0×106 m/s
Horizontally to the left with a speed equal to 8.0×106 m/s
Horizontally to the right with a speed of less than 8.0×106 m/s
Horizontally to the left with a speed of less than 8.0×106 m/s

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Homework Answers

Answer #1

Let M be the mass of the Uranium which is at rest initially and let M/60 is the mass of alpha particle which travels towards right with 8 x 106 m/s. As alpha particle is ejected from the uranium nucleus then the remaining mass is 59M/60.

Applying conservation of momentum,

M x 0 = M/60 x 8 x 106 + 59M/60 V, where V is the velocity of thorium nucleus.

V = - 8/59 x 106 = - 0.1356 x 106 m/s

The speed of thorium is less than 8 x 106 m/s and the negative sign indicates towards left.

Last option

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