MHT CET · Physics · Center of Mass Momentum and Collision
A system consists of three particles each of mass ' \(\mathrm{m}_1\) ' placed at the corners of an equilateral triangle of side ' \(\frac{\mathrm{L}}{3}\), A particle of mass ' \(\mathrm{m}_2\) ' is placed at the mid point of any one side of the triangle. Due to the system of particles, the force acting on \(\mathrm{m}_2\) is
- A \(\frac{3 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2}\)
- B \(\frac{6 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2}\)
- C \(\frac{9 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2}\)
- D \(\frac{12 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2}\)
Answer & Solution
Correct Answer
(D) \(\frac{12 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2}\)
Step-by-step Solution
Detailed explanation
From the fig., we can see that the forces due to masses at Q and R cancel each other as they are equal and opposite. The force at \(\mathrm{P}\) is only due to \(\mathrm{m}_1\).
In \(\triangle P Q S\),
\(\mathrm{h}=\frac{\mathrm{L}}{3} \cos 30^{\circ}=\frac{\mathrm{L} \sqrt{3}}{6}\)
\(\therefore \quad\) Force on \(\mathrm{m}_2\) due to \(\mathrm{m}_1\) at \(\mathrm{P}\) is
\(\begin{aligned} F & =\frac{G m_1 m_2}{\left(\frac{L \sqrt{3}}{6}\right)^2} \\ & =\frac{G m_1 m_2 \cdot 12}{L^2} \\ & =\frac{12 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2}\end{aligned}\)
In \(\triangle P Q S\),
\(\mathrm{h}=\frac{\mathrm{L}}{3} \cos 30^{\circ}=\frac{\mathrm{L} \sqrt{3}}{6}\)
\(\therefore \quad\) Force on \(\mathrm{m}_2\) due to \(\mathrm{m}_1\) at \(\mathrm{P}\) is
\(\begin{aligned} F & =\frac{G m_1 m_2}{\left(\frac{L \sqrt{3}}{6}\right)^2} \\ & =\frac{G m_1 m_2 \cdot 12}{L^2} \\ & =\frac{12 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~L}^2}\end{aligned}\)
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