Q109 / 150 · MHT CET 2024 (03 May Shift 2) · Physics

In projectile motion two particles of masses $\mathrm{m}_1$ and $m_2$ have velocities $\vec{V}_1$, and $\vec{V}_2$ respectively at time $\mathrm{t}=0$. Their velocities become $\overrightarrow{\mathrm{V}_1{ }^{\prime}}$ and $\overrightarrow{\mathrm{V}_2{ }^{\prime}}$ at time 2 t while still moving in air. The value of $\left[\left(m_1 \overrightarrow{V_1{ }^{\prime}}+m_2 \overrightarrow{V_2{ }^{\prime}}\right)-\left(m_1 \overrightarrow{V_1}+m_2 \overrightarrow{V_2}\right)\right]$ is ( $\mathrm{g}=$ acceleration due to gravity)

  • A
    zero
  • B
    $\frac{1}{2}\left(\mathrm{~m}_1+\mathrm{m}_2\right) \mathrm{gt}$
  • C
    $\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{gt}$
  • Correct answer
    D
    $2\left(m_1+m_2\right) g t$

Solution

$\begin{aligned} & \mathrm{F}_{\mathrm{ext}}=\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{g} \\ & \frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}=\frac{\left[\left(\mathrm{m}_1 \overrightarrow{\mathrm{~V}_1^{\prime}}+\mathrm{m}_2 \overrightarrow{\mathrm{~V}_2^{\prime}}\right)-\left(\mathrm{m}_1 \overrightarrow{\mathrm{~V}_1}+\mathrm{m}_2 \overrightarrow{\mathrm{~V}_2}\right)\right]}{2 \mathrm{t}-0}\end{aligned}$ $\because \quad \mathrm{F}_{\mathrm{ext}}=\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}$ $\therefore \quad\left(m_1+m_2\right) g=\frac{\left[\left(m_1 \overrightarrow{\mathrm{~V}_1^{\prime}}+\mathrm{m}_2 \overrightarrow{\mathrm{~V}_2^{\prime}}\right)-\left(\mathrm{m}_1 \overrightarrow{\mathrm{v}_1}+\mathrm{m}_2 \overrightarrow{\mathrm{~V}_2}\right)\right]}{2 \mathrm{t}}$ $\therefore \quad\left[\left(\mathrm{m}_1 \overrightarrow{\mathrm{~V}_1^{\prime}}+\mathrm{m}_2 \overrightarrow{\mathrm{~V}_2^{\prime}}\right)-\left(\mathrm{m}_1 \overrightarrow{\mathrm{~V}_1}+\mathrm{m}_2 \overrightarrow{\mathrm{~V}_2}\right)\right]$ $=2\left(m_1+m_2\right) g t$

Hence, the correct answer is option: 'D'.

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