Question

Two particles of equal mass have velocities $\vec{\upsilon_{1}}=4\hat{i}\,m\,s^{-1}$ and $\vec{\upsilon_{2}}=4\hat{j}\,m\,s^{-1}$. First particles has an acceleration $\vec{a}_{1}=\left(5\hat{i}+5\hat{j}\right)m\,s^{-2}$ while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of

• A. straight line
• B. parabola
• C. circle
• D. ellipse

Answer 1

Answer: (A)

$
\begin{array}{l}
X _{ cm }=\frac{ m _{1} x _{1}+ m _{2} x _{2}}{ m _{1}+ m _{2}} \\
X _{1}=\left(4 t +\frac{1}{2} 5 t ^{2}\right) \\
X _{2}=0 \\
X _{ cm }=\frac{4 t +\frac{1}{2} 5 t ^{2}}{2} \\
Y _{ cm }=\frac{ m _{1} y _{1}+ m _{2} y _{2}}{ m _{1}+ m _{2}} \\
y _{1}=\frac{1}{2} 5 t ^{2} \\
y _{2}=4 t \\
Y _{ cm }=\frac{ m \left(\frac{5}{2}+2\right)+ m (4 t )}{ m + m } \\
Y _{ cm }=\frac{4 t +\frac{1}{2} 5 t ^{2}}{2} \\
X _{ cm }= Y _{ cm }
\end{array}
$
Hence centre of mass moves in a straight line