1. Analyzing the Motion of Fragment A to Find Explosion Time

Let the explosion occur at time $t_0$. The shell travels along the x-axis with velocity $\vec{v}_{shell} = 2.5\hat{i} \, \mathrm{m/s}$. Therefore, the position of the explosion is: $$ \vec{r}_{exp} = (2.5 t_0)\hat{i} $$

After the explosion ($t > t_0$), fragment A moves with constant velocity $\vec{v}_A = 10\hat{i} + 10\hat{j} + 8\hat{k}$. Its position at $t = 10 \, \mathrm{s}$ is given by: $$ \vec{r}_A(10) = \vec{r}_{exp} + \vec{v}_A (10 – t_0) $$

Substituting the given coordinates $(85, 80, 64)$:

$$ 85\hat{i} + 80\hat{j} + 64\hat{k} = 2.5 t_0 \hat{i} + (10\hat{i} + 10\hat{j} + 8\hat{k})(10 – t_0) $$

Comparing the y-components:

$$ 80 = 10(10 – t_0) \implies 8 = 10 – t_0 \implies t_0 = 2 \, \mathrm{s} $$

The explosion occurred at $t=2\,\mathrm{s}$. The position of the explosion is: $$ \vec{r}_{exp} = 2.5(2)\hat{i} = 5\hat{i} \, \mathrm{m} $$

2. Finding the Velocity of Fragment B

Fragment B is observed at $\vec{r}_B = (-35, 0, -64)$ at $t=10\,\mathrm{s}$. We use the same kinematic equation: $$ \vec{r}_B(10) = \vec{r}_{exp} + \vec{v}_B (10 – t_0) $$ $$ -35\hat{i} – 64\hat{k} = 5\hat{i} + \vec{v}_B (10 – 2) $$ $$ -40\hat{i} – 64\hat{k} = 8\vec{v}_B $$ $$ \vec{v}_B = -5\hat{i} – 8\hat{k} \, \mathrm{m/s} $$

3. Conservation of Momentum

The momentum of the system is conserved just before and after the explosion. $$ \vec{P}_{initial} = \vec{P}_{final} $$ $$ M_{shell} \vec{v}_{shell} = m_A \vec{v}_A + m_B \vec{v}_B + m_C \vec{v}_C $$

Given: $M_{shell} = 4\,\mathrm{kg}$, $m_A = 2\,\mathrm{kg}$, $m_B = 1\,\mathrm{kg}$, $m_C = 1\,\mathrm{kg}$.

$$ 4(2.5\hat{i}) = 2(10\hat{i} + 10\hat{j} + 8\hat{k}) + 1(-5\hat{i} – 8\hat{k}) + 1(\vec{v}_C) $$ $$ 10\hat{i} = (20\hat{i} + 20\hat{j} + 16\hat{k}) + (-5\hat{i} – 8\hat{k}) + \vec{v}_C $$ $$ 10\hat{i} = 15\hat{i} + 20\hat{j} + 8\hat{k} + \vec{v}_C $$

Solving for $\vec{v}_C$:

$$ \vec{v}_C = (10 – 15)\hat{i} – 20\hat{j} – 8\hat{k} $$