1. Center of Mass Frame Analysis

Consider the shell of mass $m$ moving with velocity $u$. The explosion separates the shell into two identical fragments ($m/2$ each). The energy $\Delta W$ is converted entirely into the Kinetic Energy of the fragments in the Center of Mass (CM) frame.

Let $v_0$ be the speed of each fragment in the CM frame.

$$KE_{CM} = \frac{1}{2} \left(\frac{m}{2}\right) v_0^2 + \frac{1}{2} \left(\frac{m}{2}\right) v_0^2 = \frac{m}{2} v_0^2$$ $$\frac{m}{2} v_0^2 = \Delta W \Rightarrow v_0 = \sqrt{\frac{2 \Delta W}{m}}$$

2. Lab Frame Transformation

In the lab frame, the velocity of the fragments is the vector sum of the CM velocity ($\vec{u}$) and the relative velocity ($\vec{v}_0$).

$$\vec{v}_1 = \vec{u} + \vec{v}_0$$ $$\vec{v}_2 = \vec{u} – \vec{v}_0$$

We want to find the maximum angle $\theta$ between vectors $\vec{v}_1$ and $\vec{v}_2$. This occurs when the explosion direction is perpendicular to the initial velocity $\vec{u}$.

3. Calculation of Angle

From the geometry (where $\vec{v}_0 \perp \vec{u}$):

$$\tan(\theta/2) = \frac{v_0}{u}$$

Using the double angle identity for cosine:

$$\cos \theta = \frac{1 – \tan^2(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{1 – (v_0/u)^2}{1 + (v_0/u)^2} = \frac{u^2 – v_0^2}{u^2 + v_0^2}$$

Substitute $v_0^2 = \frac{2 \Delta W}{m}$:

$$\cos \theta = \frac{u^2 – \frac{2 \Delta W}{m}}{u^2 + \frac{2 \Delta W}{m}} = \frac{m u^2 – 2 \Delta W}{m u^2 + 2 \Delta W}$$