In projectile motion two particles of masses m₁ and m₂ have velocities V₁, and V₂ respectively at time t = 0. Their velocities become V₁' and V₂' at time 2 t while still moving in air. The value of [(m₁V₁' + m₂V₂') - (m₁V₁ + m₂V₂)] is ( g = acceleration due to gravity) (a) zero (b) 1/2(m₁ + m₂)gt (c) (m₁ + m₂)gt (d) 2(m₁ + m₂)gt

Answer: (d). The external force on the pair is (m₁+m₂)g downward, so = (m₁+m₂)g Δ t. The time interval is from 0 to 2t, so Δ t = 2t. || = 2(m₁+m₂)gt (directed downward). So the answer is (d) 2(m₁+m₂)gt.

In projectile motion two particles of masses m₁ and m₂ have… | MHT CET 2024 (3 May, Shift 2)

Q4MHT CET 2024 · 3 MAY · SHIFT 2Medium

Motion of Centre of Mass

In projectile motion two particles of masses $m_1$ and $m_2$ have velocities $\vec{V}_1$ , and $\vec{V}_2$ respectively at time $t = 0$ . Their velocities become $\overrightarrow{V_1'}$ and $\overrightarrow{V_2'}$ at time 2 t while still moving in air. The value of $\left[\left(m_1\overrightarrow{V_1'} + m_2\overrightarrow{V_2'}\right) - \left(m_1\vec{V}_1 + m_2\vec{V}_2\right)\right]$ is ( $g =$ acceleration due to gravity)

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Answer: (d)

The external force on the pair is $(m_1+m_2)g$ downward, so $\Delta\vec{p} = (m_1+m_2)\vec{g}\,\Delta t$ .
The time interval is from $0$ to $2t$ , so $\Delta t = 2t$ .
$|\Delta\vec{p}| = 2(m_1+m_2)gt$ (directed downward).
So the answer is (d) $2(m_1+m_2)gt$ .

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