An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each

Question

An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass $1 	ext{ kg}$ moves with a speed of $12 	ext{ m s}^{-1}$ and the second part of mass $2 	ext{ kg}$ moves with $8 	ext{ m s}^{-1}$ speed. If the third part flies off with $4 	ext{ m s}^{-1}$ speed, then its mass is:

Accepted Answer

1. **Conservation of Momentum:** Since the explosion is caused by internal forces, the external force on the system is zero. Thus, the total linear momentum is conserved. The rock is initially at rest ($P_{initial} = 0$).
   $$ \vec{P}_{final} = \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \implies \vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) $$
   (Refer to NCERT Class 11, Section 5.7 regarding Conservation of Momentum ).
2. **Calculate Momenta of First Two Parts:**
   - Part 1: $p_1 = m_1 v_1 = 1 	imes 12 = 12 	ext{ kg m s}^{-1}$.
   - Part 2: $p_2 = m_2 v_2 = 2 	imes 8 = 16 	ext{ kg m s}^{-1}$.
3. **Resultant Momentum ($p_{12}$):** Since the first two parts move at right angles ($90^\circ$), the magnitude of their resultant momentum is:
   $$ p_{12} = \sqrt{p_1^2 + p_2^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 	ext{ kg m s}^{-1} $$
4. **Determine Mass of Third Part:** The momentum of the third part must balance this resultant ($p_3 = p_{12} = 20 	ext{ kg m s}^{-1}$). Given $v_3 = 4 	ext{ m s}^{-1}$:
   $$ m_3 = \frac{p_3}{v_3} = \frac{20}{4} = 5 	ext{ kg} $$

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