There is a bundle of identical nuclei moving with velocity v. The nuclei in the beam spontaneously divide into pairs of identical fragments. The velocity of the fragments moving in the direction of the beam is 3v. Find the velocity of the fragments moving in the direction perpendicular to the beam.

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    <h3> Statement </h3>
    <p>
        $1.4.15^*.$ There is a bundle of identical nuclei moving with velocity $v$. The nuclei in the beam spontaneously divide into pairs of identical fragments. The velocity of the fragments moving in the direction of the beam is $3v$. Find the velocity of the fragments moving in the direction perpendicular to the beam.
    </p>

    <h3>Solution</h3>
    <p>
        <p>Let's move to the beam center of mass frame.</p>

In this frame of reference, the relative velocity is related to the velocity in the NFR by the relation

$$\vec{v'} = \vec{v} + \vec{v}_{rel}$$

Where $\vec{v'}$ and $\vec{v}$ are the velocity in the inertial reference frame and the velocity of the reference frame of the system, respectively.

By the condition, when $\vec{v}_{rel}$ and $\vec{v}$ are co-directed $v' = 3 v$

From where

$$v_{rel} = 2v$$

Next, let's consider the fragments that flew with the speed of $\vec{v}_\perp$)

Going back to inertial reference frame, we get that

$$\vec{v}_\perp = \vec{v}_{rel}-\vec{v}$$

By the Pythagorean theorem

$$\fbox{${v}_\perp = 3{v}$}$$

    <h4>Answer</h4>
    <p>
        $$\sin\alpha = u/v$$
    </p>



    <p style="text-align: right; font-style: italic; font-size: 14;">   
      Aliaksandr Melnichenka<br>
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