ExamBro
ExamBro
MHT CET · Physics · Center of Mass Momentum and Collision

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)

  1. A zero
  2. B \(\frac{1}{2}\left(\mathrm{~m}_1+\mathrm{m}_2\right) \mathrm{gt}\)
  3. C \(\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{gt}\)
  4. D \(2\left(m_1+m_2\right) g t\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

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