Question 323: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

Considering $\mathrm{m}_1$ and $\mathrm{m}_2$ together as system. The only external force acting on the system is
$$\begin{aligned}
& \mathrm{F}_{\mathrm{ext}}=\left(\mathrm{m}_1+\mathrm{m}_2ight) \mathrm{g} \
& \frac{\Delta P}{\Delta t}=\frac{\left[\left(m_1 \overline{V_1}+m_2 \overline{V_2^{\prime}}ight)-\left(m_1 \overline{V_1}+m_2 \overline{V_2}ight)ight]}{2 \mathrm{t}-0} \
& \because \quad \mathrm{~F}_{\mathrm{ext}}=\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}} \
& 	herefore \quad\left(\mathrm{~m}_1+\mathrm{m}_2ight) \mathrm{g}=\frac{\left[\left(\mathrm{m}_1 \overline{\mathrm{~V}_1^{\prime}}+\mathrm{m}_2 \overline{\mathrm{~V}_2^{\prime}}ight)-\left(\mathrm{m}_1 \overrightarrow{\mathrm{~V}}_1+\mathrm{m}_2 \overline{\mathrm{~V}_2}ight)ight]}{2 \mathrm{t}} \
& 	herefore \quad\left[\left(\mathrm{~m}_1 \overline{\mathrm{~V}_1^{\prime}}+\mathrm{m}_2 \overline{\mathrm{~V}_2^{\prime}}ight)-\left(\mathrm{m}_1 \overline{\mathrm{~V}}_1+\mathrm{m}_2 \overline{\mathrm{~V}_2}ight)ight]=2\left(\mathrm{~m}_1+\mathrm{m}_2ight) \mathrm{gt}
\end{aligned}$$

Answer

zero

Answer

$\frac{1}{2}\left(\mathrm{~m}_1+\mathrm{m}_2\right) \mathrm{gt}$

Answer

$\left(m_1+m_2\right) g t$

Answer

$2\left(m_1+m_2\right) g t$

Centre of Mass and Rotational Motion

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