Question

At time $t=0$, force ${\overrightarrow{F}}_1=(-4.00\hat{i}+5.00\hat{j})N$ acts on an initially stationary particle of mass $2.00\times {10}^{-3}kg$ and force ${\overrightarrow{F}}_2=(2.00\hat{i}-4.00\hat{j})N$ acts on an initially stationary particle of mass $4\times {10}^{-3}kg$, then at $t=0$ to $t=2ms$

• A. The centre of mass of two particles system displace by $0.745mm$
• B. The centre of mass of two particle system will displace in $ta{n}^{-1}\left(\frac{-1}{2}\right)$ measured counterclock wise from x-axis.
• C. The kinetic energy of centre of mass at $t=2.00ms$ is $1.67\times {10}^{-3}J$
• D. None of above

Answer 1

Answer: (A), (B), (C)

The kinetic energy of centre of mass at $t=2.00ms$ is $1.67\times {10}^{-3}J$

Let ${\overrightarrow{F}}_1$ be the force acting on $m_1,$ and ${\overrightarrow{F}}_2$ the force acting on $m_2$. According to Newton's second law, their displacements are
${\overrightarrow{d}}_1=\frac{1}{2}{\overrightarrow{a}}_1{t}^2=\frac{1}{2}\left(\frac{{\overrightarrow{F}}_1}{m_1}\right)t^2,{\overrightarrow{d}}_2=\frac{1}{2}{\overrightarrow{a}}_2{t}^2=\frac{1}{2}\left(\frac{{\overrightarrow{F}}_2}{m_2}\right)t^2$
The corresponding displacement of the centre of mass is
${\overrightarrow{d}}_{cm}=\frac{m_1{\overrightarrow{d}}_1+m_2{\overrightarrow{d}}_2}{m_1+m_2}=\frac{1}{2}\frac{m_1}{m_1+m_2}\left(\frac{{\overrightarrow{F}}_1}{m_1}\right)t^2+\frac{1}{2}\frac{m_2}{m_1+m_2}\left(\frac{{\overrightarrow{F}}_2}{m_2}\right)t^2=\frac{1}{2}\left(\frac{{\overrightarrow{F}}_1+{\overrightarrow{F}}_2}{m_1+m_2}\right)t^2$

(a) The two masses are $m_1=2.00\times {10}^{-3}$ kg and $m_2=4.00\times {10}^{-3}$ kg. With the forces given by ${\overrightarrow{F}}_1=(-4.00N)\hat{i}+\left(5.00N\right)\hat{j}$ and ${\overrightarrow{F}}_2=\left(2.00N\right)\hat{i}-\left(4.00N\right)\hat{j}$, and $t=2.00\times {10}^{-3}$ s, we obtain
${\overrightarrow{d}}_{cm}=\left(\frac{{\overrightarrow{F}}_1+{\overrightarrow{F}}_2}{m_1+m_2}\right)t^2/2=\frac{1}{2}\frac{(-4.00N+2.00N)\hat{i}+(5.00N-4.00N)\hat{j}}{2.00\times {10}^{-3}kg+4.00\times {10}^{-3}kg}(2.00\times {10}^{-3}s{)}^2=(-6.67\times {10}^{-4}m)\hat{i}+(3.33\times {10}^{-4}m)\hat{j}$
The magnitude of ${\overrightarrow{d}}_{cm}$ is
$d_{cm}=\sqrt{(-6.67\times {10}^{-4}m{)}^2+(3.33\times {10}^{-4}m{)}^2}=7.45\times {10}^{-4}m$

or $0.745mm$.
(b) The angle of ${\overrightarrow{d}}_{cm}$ is given by
$\theta =ta{n}^{-1}$$\left(\frac{3.33\times {10}^{-4}m}{-6.67\times {10}^{-4}m}\right)=ta{n}^{-1}(-\frac{1}{2})$
measured counterclock wise from x-axis.
(c) The velocities of the two masses are
${\overrightarrow{v}}_1={\overrightarrow{a}}_{1}t=\frac{{\overrightarrow{F}}_{1}t}{m_1},{\overrightarrow{v}}_2={\overrightarrow{a}}_{2}t=\frac{{\overrightarrow{F}}_{2}t}{m_2}$
and the velocity of center of mass is

${\overrightarrow{v}}_{cm}=\frac{m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2}{m_1+m_2}=\frac{m_1}{m_1+m_2}\left(\frac{{\overrightarrow{F}}_{1}t}{m_1}\right)+\frac{m_2}{m_1+m_2}\left(\frac{{\overrightarrow{F}}_{2}t}{m_2}\right)=\left(\frac{{\overrightarrow{F}}_1+{\overrightarrow{F}}_2}{m_1+m_2}\right)t$
The corresponding kinetic energy of the center of mass is
$K_{cm}=\frac{1}{2}(m_1+m_2)v_{cm}^2=\frac{1}{2}\frac{|{\overrightarrow{F}}_1+{\overrightarrow{F}}_2{|}^2}{m_1+m_2}{t}^2$
With $|{\overrightarrow{F}}_1+{\overrightarrow{F}}_2|=|(-2.00N)\hat{i}+\left(1.00N\right)\hat{j}|=\sqrt{5}N$, we get
$K_{cm}=\frac{1}{2}\frac{{\overrightarrow{F}}_1+{\overrightarrow{F}}_2{|}^2}{m_1+m_2}{t}^2=\frac{1}{2}\frac{(\sqrt{5}N{)}^2}{2.00\times {10}^{-3}kg+4.00\times {10}^{-3}kg}(2.00\times {10}^{-3}s{)}^2=1.67\times {10}^{-3}J$