How fast was the parent?

Classical Mechanics Level 2

Two particles of masses $m$ and $4m$ have velocities of $4v$ and $v,$ respectively. They are moving at an angle of $30^\circ$ with respect to each other.

If the particles have come from a decay of a single parent particle, and both mass and momentum are conserved, how fast was the parent particle moving?

Note: For the purposes of this problem, ignore energy conservation requirements and relativistic effects.

\frac{4}{5}v\sin 30^\circ

\frac{8}{5}v\cos 15^\circ

\frac{8}{5}v\sin 30^\circ

4v\cos 30^\circ

1 solution

Rohit Gupta
Jan 31, 2017

The decay of nuclei takes place due to the internal forces and hence the momentum of the nuclei and its fragments remain conserved during the decay.

The momentum of the particle of mass $m$ is $\vec p_m = 4mv (\cos 15 ^\circ \hat i + \sin 15 ^\circ \hat j).$ The momentum of the particle of mass $4m$ is $\vec p_{4m} = 4mv ( \cos 15 ^\circ \hat i - \sin 15 ^\circ \hat j ).$ The momentum of the parent nuclei is $\vec p_{\text{parent}} = 5mu \hat i .$

Applying Conservation of Linear Momentum , $\begin{aligned} \vec p_{\text{parent}} &= \vec p_m + \vec p_{4m} \\ 5mu \hat i &= 4mv ( \cos 15 ^\circ \hat i + \sin 15 ^\circ \hat j) + 4mv ( \cos 15 ^\circ \hat i - \sin 15 ^\circ \hat j ) \\ 5mu \hat i &= 8mv \cos 15 ^\circ \hat i \\ u &= \boxed{\frac{8}{5} v \cos 15 ^\circ}. \\ \end{aligned}$

It was convenient that the momentum for each of the post-decay particles was the same value of $4mv$ , ensuring that both their paths make the same (absolute) angle with the horizontal. If their momentums had been different then we would have had to done some additional trig to find the distinct angles each makes with the horizontal.

Brian Charlesworth - 4 years, 4 months ago

Yes, agreed the data given saved some calculations.

Rohit Gupta - 4 years, 4 months ago

How fast was the parent?

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