© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: State the Conservation of Linear Momentum
When a rock explodes into multiple pieces, the total momentum of all parts must remain equal to the initial momentum of the system (which is zero if the rock was initially at rest). Mathematically:
$ \overrightarrow{p}_1 + \overrightarrow{p}_2 + \overrightarrow{p}_3 = 0 $
Step 2: Express the Known Momenta
Let the three parts have masses $m_1 = 1 \, \text{kg}$, $m_2 = 2 \, \text{kg}$, and $m_3$ (unknown). Their velocities are $v_1 = 12 \, \text{m/s}$ (along the x-axis), $v_2 = 8 \, \text{m/s}$ (along the y-axis), and $v_3 = 4 \, \text{m/s}$ (direction to be determined by momentum balancing).
Thus:
$ \overrightarrow{p}_1 = m_1 \, \overrightarrow{v}_1 = (1 \, \text{kg})(12 \, \text{m/s}) \hat{i} = 12 \hat{i} \, \text{kg·m/s} $
$ \overrightarrow{p}_2 = m_2 \, \overrightarrow{v}_2 = (2 \, \text{kg})(8 \, \text{m/s}) \hat{j} = 16 \hat{j} \, \text{kg·m/s} $
Step 3: Find the Momentum of the Third Piece
According to the conservation of momentum:
$ \overrightarrow{p}_3 = - \bigl(\overrightarrow{p}_1 + \overrightarrow{p}_2 \bigr). $
Substituting the values:
$ \overrightarrow{p}_1 + \overrightarrow{p}_2 = 12 \hat{i} + 16 \hat{j} \, \text{kg·m/s}. $
Therefore,
$ \overrightarrow{p}_3 = - \bigl(12 \hat{i} + 16 \hat{j}\bigr) \, \text{kg·m/s} = -12 \hat{i} - 16 \hat{j} \, \text{kg·m/s}. $
Step 4: Calculate the Magnitude of the Third Piece’s Momentum
The magnitude of $ \overrightarrow{p}_3 $ is:
$ p_3 = \sqrt{(-12)^2 + (-16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \,\text{kg·m/s}. $
Step 5: Determine the Mass of the Third Piece
We know the speed of the third piece is $ v_3 = 4 \,\text{m/s} $. By definition,
$ p_3 = m_3 \, v_3 \implies m_3 = \frac{p_3}{v_3} = \frac{20 \,\text{kg·m/s}}{4 \,\text{m/s}} = 5 \,\text{kg}. $
Final Answer
The mass of the third part is 5 kg.
