At time t-0- force F1-4-00-5-00- N acts on an initially stationary particle of mass 2-00- 10-3 kg and force F2-2-00-4-00- N acts on an initially stationary particle of mass 4- 10-3 kg- then at t-0 to t-2 ms

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Standard XII

Physics

Inertial vs Non Inertial Frames

At time t=0, ...

Question

At time $t = 0$ , force ${\to F}_{1} = (- 4.00^i + 5.00^j) N$ acts on an initially stationary particle of mass $2.00 \times 10^{- 3} k g$ and force ${\to F}_{2} = (2.00^i - 4.00^j) N$ acts on an initially stationary particle of mass $4 \times 10^{- 3} k g$ , then at $t = 0$ to $t = 2 m s$

The centre of mass of two particles system displace by

0.745 m m

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The centre of mass of two particle system will displace in

t a n^{- 1} (\frac{- 1}{2})

measured counterclock wise from x-axis.

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The kinetic energy of centre of mass at

t = 2.00 m s

1.67 \times 10^{- 3} J

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None of above

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Solution

The correct option is C The kinetic energy of centre of mass at $t = 2.00 m s$ is $1.67 \times 10^{- 3} J$
Let ${\to F}_{1}$ be the force acting on $m_{1},$ and ${\to F}_{2}$ the force acting on $m_{2}$ . According to Newton's second law, their displacements are
${\to d}_{1} = \frac{1}{2} {\to a}_{1} t^{2} = \frac{1}{2} (\frac{{\to F}_{1}}{m_{1}}) t^{2}, {\to d}_{2} = \frac{1}{2} {\to a}_{2} t^{2} = \frac{1}{2} (\frac{{\to F}_{2}}{m_{2}}) t^{2}$
The corresponding displacement of the centre of mass is
${\to d}_{c m} = \frac{m_{1} {\to d}_{1} + m_{2} {\to d}_{2}}{m_{1} + m_{2}} = \frac{1}{2} \frac{m_{1}}{m_{1} + m_{2}} (\frac{{\to F}_{1}}{m_{1}}) t^{2} + \frac{1}{2} \frac{m_{2}}{m_{1} + m_{2}} (\frac{{\to F}_{2}}{m_{2}}) t^{2} = \frac{1}{2} (\frac{{\to F}_{1} + {\to F}_{2}}{m_{1} + m_{2}}) t^{2}$

(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 ${\to F}_{1} = (- 4.00 N)^i + (5.00 N)^j$ and ${\to F}_{2} = (2.00 N)^i - (4.00 N)^j$ , and $t = 2.00 \times 10^{- 3}$ s, we obtain
${\to d}_{c m} = (\frac{{\to F}_{1} + {\to F}_{2}}{m_{1} + m_{2}}) t^{2} / 2 = \frac{1}{2} \frac{(- 4.00 N + 2.00 N)^i + (5.00 N - 4.00 N)^j}{2.00 \times 10^{- 3} k g + 4.00 \times 10^{- 3} k g} (2.00 \times 10^{- 3} s)^{2} = (- 6.67 \times 10^{- 4} m)^i + (3.33 \times 10^{- 4} m)^j$
The magnitude of ${\to d}_{c m}$ is
$d_{c m} = \sqrt{(- 6.67 \times 10^{- 4} m)^{2} + (3.33 \times 10^{- 4} m)^{2}} = 7.45 \times 10^{- 4} m$

or $0.745 m m$ .
(b) The angle of ${\to d}_{c m}$ is given by
$θ = t a n^{- 1}$ $(\frac{3.33 \times 10^{- 4} m}{- 6.67 \times 10^{- 4} m}) = t a n^{- 1} (- \frac{1}{2})$
measured counterclock wise from x-axis.
(c) The velocities of the two masses are
${\to v}_{1} = {\to a}_{1} t = \frac{{\to F}_{1} t}{m_{1}}, {\to v}_{2} = {\to a}_{2} t = \frac{{\to F}_{2} t}{m_{2}}$
and the velocity of center of mass is

${\to v}_{c m} = \frac{m_{1} {\to v}_{1} + m_{2} {\to v}_{2}}{m_{1} + m_{2}} = \frac{m_{1}}{m_{1} + m_{2}} (\frac{{\to F}_{1} t}{m_{1}}) + \frac{m_{2}}{m_{1} + m_{2}} (\frac{{\to F}_{2} t}{m_{2}}) = (\frac{{\to F}_{1} + {\to F}_{2}}{m_{1} + m_{2}}) t$
The corresponding kinetic energy of the center of mass is
$K_{c m} = \frac{1}{2} (m_{1} + m_{2}) v_{c m}^{2} = \frac{1}{2} \frac{| {\to F}_{1} + {\to F}_{2} |^{2}}{m_{1} + m_{2}} t^{2}$
With $| {\to F}_{1} + {\to F}_{2} | = | (- 2.00 N)^i + (1.00 N)^j | = \sqrt{5} N$ , we get
$K_{c m} = \frac{1}{2} \frac{{\to F}_{1} + {\to F}_{2} |^{2}}{m_{1} + m_{2}} t^{2} = \frac{1}{2} \frac{(\sqrt{5} N)^{2}}{2.00 \times 10^{- 3} k g + 4.00 \times 10^{- 3} k g} (2.00 \times 10^{- 3} s)^{2} = 1.67 \times 10^{- 3} J$

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Inertial vs Non Inertial Frames

Standard XII Physics

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At time t=0, force →F1=(−4.00^i+5.00^j) N acts on an initially stationary particle of mass 2.00×10−3 kg and force →F2=(2.00^i−4.00^j) N acts on an initially stationary particle of mass 4×10−3 kg, then at t=0 to t=2 ms

At time $t = 0$ , force ${\to F}_{1} = (- 4.00^i + 5.00^j) N$ acts on an initially stationary particle of mass $2.00 \times 10^{- 3} k g$ and force ${\to F}_{2} = (2.00^i - 4.00^j) N$ acts on an initially stationary particle of mass $4 \times 10^{- 3} k g$ , then at $t = 0$ to $t = 2 m s$