Solution to Question 21
Let the mass of the shell be $m$ and its initial velocity in the laboratory frame be $\vec{u}$. The shell explodes into two identical fragments, each of mass $\frac{m}{2}$.
Consider the explosion in the Center of Mass (CM) frame. In this frame, the total momentum is zero. Therefore, after the explosion, the two fragments must move in opposite directions with equal speeds. Let this speed be $v’$.
The initial kinetic energy in the CM frame (before explosion) is zero. The increase in kinetic energy $\Delta W$ is entirely converted into the kinetic energy of the fragments in the CM frame.
$$ \Delta W = \text{KE}_{\text{final, CM}} $$ $$ \Delta W = \frac{1}{2} \left(\frac{m}{2}\right) v’^2 + \frac{1}{2} \left(\frac{m}{2}\right) v’^2 $$ $$ \Delta W = \frac{m}{2} v’^2 $$
From this, we can express the relative velocity magnitude $v’$ as: $$ v’^2 = \frac{2\Delta W}{m} \quad \dots(1) $$
The velocity of a fragment in the laboratory frame ($\vec{v}_{\text{lab}}$) is the vector sum of the velocity of the center of mass ($\vec{u}$) and the velocity of the fragment relative to the CM ($\vec{v}’$).
For the two fragments: $$ \vec{v}_1 = \vec{u} + \vec{v}’ $$ $$ \vec{v}_2 = \vec{u} – \vec{v}’ $$
We wish to find the maximum angle $\theta$ between the velocity vectors $\vec{v}_1$ and $\vec{v}_2$. Geometrically, the vectors $\vec{v}_1$ and $\vec{v}_2$ are formed by adding $\vec{v}’$ and $-\vec{v}’$ to $\vec{u}$.
The angle between $\vec{v}_1$ and $\vec{v}_2$ is maximized when the relative velocity vector $\vec{v}’$ is perpendicular to the drift velocity $\vec{u}$. If $\vec{v}’$ is not perpendicular, the “spread” of the resulting vectors is reduced in one direction, diminishing the included angle.
Fig 1: Vector addition when relative velocity is perpendicular to CM velocity.
From the diagram, the two velocity triangles are right-angled. Let the angle between $\vec{v}_1$ and $\vec{u}$ be $\alpha$. By symmetry, the total angle $\theta = 2\alpha$.
In the upper right-angled triangle: $$ \tan \alpha = \tan\left(\frac{\theta}{2}\right) = \frac{v’}{u} $$
We need to find $\cos \theta$. Using the trigonometric identity $\cos \theta = \frac{1 – \tan^2(\theta/2)}{1 + \tan^2(\theta/2)}$:
$$ \cos \theta = \frac{1 – \left(\frac{v’}{u}\right)^2}{1 + \left(\frac{v’}{u}\right)^2} $$ $$ \cos \theta = \frac{u^2 – v’^2}{u^2 + v’^2} $$
Substitute the value of $v’^2$ from Eq (1): $v’^2 = \frac{2\Delta W}{m}$.
$$ \cos \theta = \frac{u^2 – \frac{2\Delta W}{m}}{u^2 + \frac{2\Delta W}{m}} $$
Multiply numerator and denominator by $m$: $$ \cos \theta = \frac{mu^2 – 2\Delta W}{mu^2 + 2\Delta W} $$
Thus, the maximum angle is: $$ \theta = \cos^{-1}\left( \frac{mu^2 – 2\Delta W}{mu^2 + 2\Delta W} \right) $$