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Step-by-Step Solution
Step 1: Identify the Principle Involved
The problem is based on the law of conservation of momentum, which states that the total momentum of a system remains constant if no external force acts on it. Here, initially, the rock is at rest, so total initial momentum is zero. After the explosion, the vector sum of the momenta of all three pieces must also be zero.
Step 2: Express the Known Data
• Mass of first piece, $m_1 = 1\,\text{kg}$, velocity $v_1 = 12\,\text{m s}^{-1}$.
• Mass of second piece, $m_2 = 2\,\text{kg}$, velocity $v_2 = 8\,\text{m s}^{-1}$.
• Mass of third piece, $m_3 = ?$ (unknown), velocity $v_3 = 4\,\text{m s}^{-1}$.
• Directions: the first two fragments move at right angles (90°) to each other.
Step 3: Calculate the Resultant Momentum of the First Two Pieces
Since the first and second pieces move at right angles, their momenta form the legs of a right-angled triangle in the momentum space.
Momentum of first piece $p_1 = m_1 \times v_1 = 1 \times 12 = 12\,\text{kg m s}^{-1}$.
Momentum of second piece $p_2 = m_2 \times v_2 = 2 \times 8 = 16\,\text{kg m s}^{-1}$.
Resultant momentum of these two pieces
$$
p_{12} = \sqrt{p_1^2 + p_2^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\,\text{kg m s}^{-1}.
$$
Step 4: Find the Momentum of the Third Piece
Because total initial momentum was zero, the vector sum of all three momenta must be zero. Thus, the third piece must have a momentum equal in magnitude and opposite in direction to the resultant momentum of the first two pieces:
$$
p_3 = 20\,\text{kg m s}^{-1}.
$$
Step 5: Calculate the Mass of the Third Piece
The third piece's velocity is given as $v_3 = 4\,\text{m s}^{-1}$, and its momentum is $p_3 = 20\,\text{kg m s}^{-1}$. Hence,
$$
m_3 \times v_3 = p_3 \quad \Rightarrow \quad m_3 = \frac{p_3}{v_3} = \frac{20}{4} = 5\,\text{kg}.
$$
Step 6: Final Answer
The mass of the third piece is 5 kg.
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