An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, $1\

Question

An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, $1	ext{ kg}$ first part moving with a velocity of $12	ext{ m s}^{-1}$ and $2	ext{ kg}$ second part moving with a velocity of $8	ext{ m s}^{-1}$. If the third part flies off with a velocity of $4	ext{ m s}^{-1}$, its mass would be:

Accepted Answer

1. **Identify the Principle:** In an explosion, internal forces act, but if no external forces are involved, the total linear momentum of the system is conserved [Class 11 Physics, Ch 6, Sec 6.4]. Since the rock was initially at rest, the total initial momentum is zero ($P_{initial} = 0$).
2. **Apply Conservation of Momentum:** The vector sum of the final momenta of the three parts must be zero.
   $$ \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \implies \vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) $$
   This means the magnitude of the momentum of the third part ($p_3$) must equal the magnitude of the resultant momentum of the first two parts.
3. **Calculate Momenta:**
   - Part 1: $p_1 = m_1 v_1 = 1	ext{ kg} 	imes 12	ext{ m/s} = 12	ext{ kg m/s}$.
   - Part 2: $p_2 = m_2 v_2 = 2	ext{ kg} 	imes 8	ext{ m/s} = 16	ext{ kg m/s}$.
4. **Calculate Resultant:** Since $\vec{p}_1$ and $\vec{p}_2$ are at right angles:
   $$ |\vec{p}_1 + \vec{p}_2| = \sqrt{p_1^2 + p_2^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20	ext{ kg m/s} $$
   Therefore, $p_3 = 20	ext{ kg m/s}$.
5. **Find Mass:**
   $$ p_3 = m_3 v_3 $$
   $$ 20 = m_3 	imes 4 $$
   $$ m_3 = \frac{20}{4} = 5	ext{ kg} $$

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