First slide

Motion of centre of mass

Question

Acceleration of two identical particles moving in a straight line are as shown in figure. The corresponding a-t graph of their centre of mass will be:

Easy

A
B
C
D

Solution

$\large {{\text{a}}_{\text{1}}} = \frac{{10}}{{10}}{\text{t}} = {\text{t, }}{{\text{a}}_{\text{2}}} = 10\,{\text{m/}}{{\text{s}}^{\text{2}}}$

$\large {{\text{a}}_{{\text{CM}}}} = \frac{{{{\text{m}}_{\text{1}}}{{\text{a}}_{\text{1}}} + {{\text{m}}_{\text{2}}}{{\text{a}}_{\text{2}}}}}{{{{\text{m}}_{\text{1}}} + {{\text{m}}_{\text{2}}}}}({{\text{m}}_{\text{1}}} = {{\text{m}}_{\text{2}}})$

$\large = \frac{{({{\text{a}}_{\text{1}}} + {{\text{a}}_{\text{2}}})}}{2} = \left( {5 + \frac{{\text{t}}}{{\text{2}}}} \right)$
At t = 0, $a_{C M} = 5 m s^{- 2}$ and t = 10 s, $a_{C M} = 10 m s^{- 2}$
So, the graph is a straight line with intercept $5 m s^{- 2}$ and slope $\frac{1}{2} m s^{- 2}$ .

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