Solution to Question 33
The shell moves with velocity $\vec{u}$ towards the plane. Upon explosion, fragments are emitted with speed $v$ in all directions relative to the center of mass (CM). The velocity of any fragment in the laboratory frame is: $$ \vec{V}_{lab} = \vec{u} + \vec{v}_{rel} $$
If $v < u$, the fragments are confined within a cone. The maximum angle $\theta_{max}$ that a fragment's path makes with the central axis (normal to the plane) is determined by the condition where the relative velocity vector is tangent to the circle of possible velocities.
Figure 3: Left: Velocity vector geometry showing the maximum deviation angle $\theta$. Right: The cone of fragments intersecting the plane.
From the velocity vector triangle (Figure 3, Left), the maximum semi-vertical angle $\theta$ of the cone is given by:
$$ \sin \theta = \frac{v}{u} $$
The fragments travel in straight lines from the point of explosion (distance $l$ from the plane). The area on the plane is a circle of radius $r$:
$$ r = l \tan \theta $$
Using the trigonometric identity $\tan \theta = \frac{\sin \theta}{\sqrt{1 – \sin^2 \theta}}$:
$$ \tan \theta = \frac{v/u}{\sqrt{1 – (v/u)^2}} = \frac{v/u}{\sqrt{\frac{u^2 – v^2}{u^2}}} = \frac{v}{\sqrt{u^2 – v^2}} $$
Thus, the radius of the circle is:
$$ r = \frac{vl}{\sqrt{u^2 – v^2}} $$
If $v < u$, the area is a circle of radius $r = \frac{vl}{\sqrt{u^2 - v^2}}$ (Area $= \pi r^2$).
If $v \ge u$, the fragments are not confined to a forward cone and will eventually cover the entire plane.