Solved: PHYSICS Question for NEET

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

5 kg

7 kg

17 kg

3 kg

Step-by-Step Solution

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 ( $P_{initial} = 0$ ).
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.
Calculate Momenta:

Part 1: $p_1 = m_1 v_1 = 1\text{ kg} \times 12\text{ m/s} = 12\text{ kg m/s}$ .
Part 2: $p_2 = m_2 v_2 = 2\text{ kg} \times 8\text{ m/s} = 16\text{ kg m/s}$ .

Calculate Resultant: Since $\vec{p}_1$ and $\vec{p}_2$ are at right angles: $|\vec{p}_1 + \vec{p}_2| = \sqrt{p_1^2 + p_2^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\text{ kg m/s}$ Therefore, $p_3 = 20\text{ kg m/s}$ .
Find Mass: $p_3 = m_3 v_3$ $20 = m_3 \times 4$ $m_3 = \frac{20}{4} = 5\text{ kg}$

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