# Two particles of equal mass have velocities $\overrightarrow{V_1}=4 \hat {i}$ and $V_2= 4 \hat {j} m/s$. First particle has an accleration $\overrightarrow{a_1}=(5 \hat i+5 \hat{j})m/s^2$ While the acceleration of the other particle is zero. The center of mass of the two particles moves in a path of

a) Ellipse

b) Parabola

c) Circle

d) Straight line

## 1 Answer

$X_{cm}= \large\frac{m_1x_1+m_2x_2}{m_1+m_2}$
$X_1=(4 t+\large\frac{1}{2}$$5t^2) X_2=0 X_{cm}=\large\frac{4t+\large\frac{1}{2} 5t^2}{2} Y_{cm}=\large\frac{m_1y_1+m_2y_2}{m_1+m_2} y_1= \large\frac{1}{2}$$5t^2$
$y_2= 4t$
$Y_{cm}=\large\frac{m\bigg(\Large\frac{5}{2} \normalsize+2\bigg)+m(4t)}{m+m}$
$Y_{cm}=\large\frac{4t+\Large\frac{1}{2} \normalsize 5t^2}{2}$
$X_{cm}=Y_{cm}$
Hence centre of mass moves in a straight line

answered Nov 21, 2013 by
edited Nov 21, 2013 by meena.p

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