Question

In a system of particles $8kg$ mass is subjected to a force of $16N$ along $+vex-$axis and another 8 kg mass is subjected to a force of $8N$ along $+vey-$axis. The magnitude of the acceleration of center of mass and the angle made by it with $x-$axis are given, respectively, by :

• A. $\frac{\sqrt{5}}{2}m{s}^{-2},\theta ={45}^0$
• B. $3\sqrt{5}m{s}^{-2},\theta =ta{n}^{-1}\left(\frac{2}{3}\right)$
• C. $\frac{\sqrt{5}}{2}m{s}^{-2},\theta =ta{n}^{-1}\left(\frac{1}{2}\right)$
• D. $1m{s}^{-2},\theta =ta{n}^{-1}\sqrt{3}$

Answer 1

The correct option is C $\frac{\sqrt{5}}{2}m{s}^{-2},\theta =ta{n}^{-1}\left(\frac{1}{2}\right)$

Acceleration center of mass, $\overrightarrow{a_{CM}}=\frac{m_1\overrightarrow{a_1}+m_2\overrightarrow{a_2}}{m_1+m_2}$

here, $m_1\overrightarrow{a_1}=\overrightarrow{F_1}=16\hat{i}$ and

$m_2\overrightarrow{a_1}=\overrightarrow{F_2}=8\hat{j}$

$\overrightarrow{a_{CM}}=\frac{16\hat{i}+8\hat{j}}{8+8}=\hat{i}+\frac{1}{2}\hat{j}........\left(1\right)$

$\therefore |\overrightarrow{a_{CM}}|=\sqrt{1^2+(1/2{)}^2}=\frac{\sqrt{5}}{2}m/s^2$

If $\overrightarrow{a_{CM}}$ will make angle $\theta$ with $x-$ axis, then

$\overrightarrow{a_{CM}}=|\overrightarrow{a_{CM}}|(\cos \theta \hat{i}+\sin \theta \hat{j})...........\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right),|\overrightarrow{a_{CM}}|\cos \theta =1$ and $|\overrightarrow{a_{CM}}|\sin \theta =1/2$

$\therefore \tan \theta =1/2\Rightarrow \theta ={\tan }^{-1}\left(1/2\right)$