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Two particles of equal mass are projected from the ground with speed $v_1$ and $v_2$ at angles $\theta_1$ and $\theta_2$ as shown. The center of mass of the two particles

a) will move in parabolic path for any value of $v_1,v_2, \theta_1$ and $\theta_2$

b) can move in a vertical line.

c) can move in a horizontal line.

d) will move in a straight line for any values of $v_1,v_2, \theta_1$ and $ \theta_2$

1 Answer

$V_{x\;COM}=\large\frac{m_1v_{1 x} +m_2v_{2x}}{(m_1+m_2)}$
$\qquad= \large\frac{m_1V_1 \cos \theta_1-m_2 v_2 \cos \theta_2}{m_1+m_2}$
$\qquad= \large\frac{V_1 \cos \theta_1- v_2 \cos \theta_2}{m}$$\qquad[m_1=m_2]$
$V_{y\;COM}= \large\frac{m_1[u_1 \sin \theta_1 - gt]+m_2 [u_2 \sin \theta_2-gt]}{m_1+m_2}$
$\qquad= \large\frac{u_1 \sin \theta_1+u_2 \sin \theta_2- 2 gt}{2}$
If $v_1 \cos \theta_1=v_2 \cos \theta_2$
$V_x COM=0.$
$V_y COM \neq 0.$
Hence b is the correct one.
answered Nov 22, 2013 by meena.p
edited Jun 17, 2014 by lmohan717

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