1. Momentum Conservation

The shell explodes at the highest point where its initial momentum is zero. $$\vec{P}_{initial} = 0$$ Sum of fragment momenta: $$2m \vec{v}_A + 3m \vec{v}_B + 4m \vec{v}_C = 0$$

Assumption: we assume the speed of each fragment is the same ($v$). $$|\vec{v}_A| = |\vec{v}_B| = |\vec{v}_C| = v$$

2. Calculating Angles and Relative Speeds

Let the velocity vectors be $\mathbf{v}_A, \mathbf{v}_B, \mathbf{v}_C$. The momentum equation is:

$$2\mathbf{v}_A + 3\mathbf{v}_B + 4\mathbf{v}_C = 0$$

Finding Angle between A and B ($\theta_{AB}$)

Isolate $\mathbf{v}_C$: $$2\mathbf{v}_A + 3\mathbf{v}_B = -4\mathbf{v}_C$$ Squaring both sides (dot product): $$|2\mathbf{v}_A + 3\mathbf{v}_B|^2 = |-4\mathbf{v}_C|^2$$ $$4v^2 + 9v^2 + 12 v^2 \cos\theta_{AB} = 16v^2$$ $$13 + 12 \cos\theta_{AB} = 16 \Rightarrow \cos\theta_{AB} = \frac{3}{12} = \frac{1}{4}$$ Relative velocity squared ($v_{AB}^2$): $$|\mathbf{v}_A – \mathbf{v}_B|^2 = v^2 + v^2 – 2v^2 \cos\theta_{AB} = 2v^2 – 2v^2(1/4) = 1.5 v^2$$

Finding Angle between A and C ($\theta_{AC}$)

Isolate $\mathbf{v}_B$: $$2\mathbf{v}_A + 4\mathbf{v}_C = -3\mathbf{v}_B$$ Squaring both sides: $$4v^2 + 16v^2 + 16 v^2 \cos\theta_{AC} = 9v^2$$ $$20 + 16 \cos\theta_{AC} = 9 \Rightarrow \cos\theta_{AC} = -\frac{11}{16}$$ Relative velocity squared ($v_{AC}^2$): $$|\mathbf{v}_A – \mathbf{v}_C|^2 = v^2 + v^2 – 2v^2 \cos\theta_{AC} = 2v^2 – 2v^2(-11/16)$$ $$|\mathbf{v}_A – \mathbf{v}_C|^2 = 2v^2 + \frac{11}{8}v^2 = \frac{16+11}{8}v^2 = \frac{27}{8}v^2$$

3. Distance Ratio

The distance between fragments after time $t$ is $l = v_{rel} t$. $$l_{AB} = \sqrt{1.5} \, v \, t$$ $$l_{AC} = \sqrt{\frac{27}{8}} \, v \, t$$ Taking the ratio $\frac{l_{AC}}{l_{AB}}$: $$ \text{Ratio}^2 = \frac{27/8}{1.5} = \frac{3.375}{1.5} = 2.25$$ $$\text{Ratio} = \sqrt{2.25} = 1.5$$ $$l_{AC} = 1.5 l_{AB}$$