AP EAMCET · PHYSICS · Center of Mass Momentum and Collision
Two particles of masses in the ratio \(1: 2\) are placed along a vertical line. The lighter particle is raised through a height of \(9 \mathrm{~cm}\). To raise the centre of mass of the system by 2 \(\mathrm{cm}\), the heavier particle should be
- A moved 1.5 cm downward
- B moved 2 cm upward
- C moved 1.5 cm upward
- D moved 2 cm downward
Answer & Solution
Correct Answer
(A) moved 1.5 cm downward
Step-by-step Solution
Detailed explanation
Let initially particles are at origin, their masses are \(m\) and \(2 m\). \( \begin{aligned} 2 & =\frac{m \times 9+2 m \times y_2}{3 m} \\ 2 & =\frac{9+2 y_2}{3} \\ y_2 & =-1.5 \mathrm{~cm} \end{aligned} \) So, second particle must be moved \(1.5 \mathrm{~cm}\) downward.
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