91Ó°ÊÓ

Given that a system consists of two particles.

At t=0 one particle say 1 is at origin; the other, say 2 has a mass of 0.50 kg , is on the y-axis at y =6.0 m .

At t =0 the center of mass of the system is on the y-axis at y =2.4 m . The velocity of the center of the mass is given by $(0.75m/s^3)t^2\stackrel{Áåœ}{i}$.

Center of mass, y =2.4 m

Velocity of center of mass,$v=(0.75m/s^3)t^2\stackrel{Áåœ}{i}$

$m_2=0.5kg$

$$\begin{gathered} y_1=0m \\ {y}_2=6m \end{gathered}$$

Center of mass is

${\mathbf{x}}_{\mathbf{i}}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{3}}{\mathbf{x}}_{\mathbf{3}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{3}}}$

Where role="math" localid="1668077549585" ${\mathbf{m}}_{\mathbf{i}\mathbf{\text{'}}\mathbf{s}}$ are masses and ${\mathit{x}}_{\mathbf{i}}\mathbf{\text{'}}$are positions.

Now center of mass is $y=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$

So

$$\begin{gathered} 2.4\mathrm{m}=\frac{m_1\left(0\right)+\left(0.5\mathrm{kg}\right)\left(6\mathrm{m}\right)}{m_1+m_2} \\ {m}_1+m_2=\frac{3.0\mathrm{kg}â‹\dots \mathrm{m}}{2.4\mathrm{m}} \\ {m}_1+m_2=1.25\mathrm{kg} \end{gathered}$$

Also,

$$\begin{gathered} m_1=1.25kg-0.5kg \\ {m}_1=0.75kg \end{gathered}$$

Hence, total mass of the system is 1.25 kg.

Acceleration of center of mass is$a=\frac{dv}{dt}$

So

$$\begin{gathered} a=\frac{d}{dt}\left(0.75m/s^3\right)t^2\stackrel{Áåœ}{i} \\ =2\left(0.75m/s^3\right)t\stackrel{Áåœ}{i} \\ =\left(1.5m/s^3\right)t\stackrel{Áåœ}{i} \end{gathered}$$

Hence, acceleration of center of mass at any time is$(1.5m/s^3)t\stackrel{Áåœ}{i}$

Net external force is$F=(m_1+m_2)a$

So

$$\begin{gathered} F=\left(1.25kg\right)\left(1.5tm/s^3\right) \\ =1.875t \end{gathered}$$

At t=3 s,

$$\begin{gathered} F=\left(1.875kgm/s^3\right)\left(3s\right) \\ =5.6N \end{gathered}$$

Hence, net external force is (5.6 N) .