1. Determine Explosion Time and Location

Shell velocity $\vec{u} = 2.5 \hat{i}$ m/s. It passes origin at $t=0$. Explosion occurs at time $t_e$. Location of explosion $\vec{R}_e = 2.5 t_e \hat{i}$.

Fragment A is at $\vec{r}_A = (85, 80, 64)$ at $t=10$. Velocity of A is $\vec{v}_A = 10\hat{i} + 10\hat{j} + 8\hat{k}$.
We can backtrack from A’s position at $t=10$ to find the explosion point.

$$\vec{r}_A = \vec{R}_e + \vec{v}_A (10 – t_e)$$ $$85 \hat{i} + 80 \hat{j} + 64 \hat{k} = (2.5 t_e \hat{i}) + (10\hat{i} + 10\hat{j} + 8\hat{k})(10 – t_e)$$

Equating x-components:

$$85 = 2.5 t_e + 10(10 – t_e)$$ $$85 = 2.5 t_e + 100 – 10 t_e$$ $$7.5 t_e = 15 \Rightarrow t_e = 2 \text{ s}$$

Location of explosion:

$$\vec{R}_e = 2.5(2) \hat{i} = 5 \hat{i} \text{ m}$$

2. Determine Velocity of B

Fragment B is at $\vec{r}_B = (-35, 0, -64)$ at $t=10$. It started at $\vec{R}_e = (5, 0, 0)$ at $t=2$. Time elapsed $\Delta t = 10 – 2 = 8$ s.

$$\vec{v}_B = \frac{\vec{r}_B – \vec{R}_e}{\Delta t} = \frac{(-35 – 5)\hat{i} + (0-0)\hat{j} + (-64-0)\hat{k}}{8}$$ $$\vec{v}_B = \frac{-40\hat{i} – 64\hat{k}}{8} = -5\hat{i} – 8\hat{k} \text{ m/s}$$

3. Conservation of Momentum

Total momentum before explosion ($P_i$) equals total momentum after ($P_f$). Initial mass $M=4.0$ kg. $\vec{u} = 2.5 \hat{i}$.

$$\vec{P}_i = 4(2.5 \hat{i}) = 10 \hat{i}$$ $$\vec{P}_f = m_A \vec{v}_A + m_B \vec{v}_B + m_C \vec{v}_C$$ Using $m_A=2, m_B=1, m_C=1$: $$10 \hat{i} = 2(10\hat{i} + 10\hat{j} + 8\hat{k}) + 1(-5\hat{i} – 8\hat{k}) + 1(\vec{v}_C)$$ $$10 \hat{i} = (20\hat{i} + 20\hat{j} + 16\hat{k}) + (-5\hat{i} – 8\hat{k}) + \vec{v}_C$$ $$10 \hat{i} = (15\hat{i} + 20\hat{j} + 8\hat{k}) + \vec{v}_C$$ $$\vec{v}_C = (10 – 15)\hat{i} – 20\hat{j} – 8\hat{k}$$ $$\vec{v}_C = -5\hat{i} – 20\hat{j} – 8\hat{k}$$