$1.4.14.$ A nucleus traveling at speed $v$ splits into two identical fragments. Determine the maximum possible angle $\alpha$ between the velocities of one of the fragments and the vector $v$, if the fragments have a velocity $u < v$ during the decay of a resting nucleus.
The velocity and one of the accelerations of the nucleus are
$$ \vec{u} = \vec{v}_{0} + \vec{v} $$
where $ \vec{v}_{0} $ is its velocity before the decay of the nucleus, $ \vec{v} $ is the velocity of the fragment relative to the resting nucleus (that is, in the coordinate system in which the nucleus was at rest before the decay).
Let's arrange the vector $ \vec{v}_{0} $ as shown in the figure. The end of the vector can lie on a circle of radius $| \vec{v} |$.
It is clear from the figure that the vector and forms the maximum angle $ \alpha $ with the vector $ \vec{v}_{0} $ if the angle between the vectors $ \vec{u} $ and $ \vec{v} $ is $ \pi / 2 $. In this case
$$\fbox{$ \alpha = \arcsin \frac{v}{v_0} $}$$
$$\sin \alpha = u/v$$