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JEE Main 201912 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Evening Shift
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Main 2019 · 12 April · Evening Shift · Physics · Center of Mass, Momentum and Collision

Three particles of masses 50g , 100g and 150g are placed at the...

Three particles of masses 50g50g, 100gand150g100g and 150g are placed at the vertices of an equilateral triangle of side 1m1m (as shown in the figure). The (x,y)\left(x, y\right) coordinates of the centre of mass will be:

Three particles of masses 50g , 100g and 150g are placed at the vertices of an equilateral triangle of side 1m (as

Options

  1. A(712m,34m)\left(\dfrac{7}{12}m, \dfrac{\sqrt{3}}{4}m\right)
  2. B(712m,38m)\left(\dfrac{7}{12}m, \dfrac{\sqrt{3}}{8}m\right)
  3. C(34m,512m)\left(\dfrac{\sqrt{3}}{4}m, \dfrac{5}{12}m\right)
  4. D(38m,712m)\left(\dfrac{\sqrt{3}}{8}m, \dfrac{7}{12}m\right)

Correct answer

A. (712m,34m)\left(\dfrac{7}{12}m, \dfrac{\sqrt{3}}{4}m\right)

Step-by-step solution

Coordinates of A:(0,0)A: \left(0, 0\right)
B:(12m,32m)B: \left(\dfrac{1}{2}m, \dfrac{\sqrt{3}}{2}m\right)
C:(1m,0)C: ( 1m, 0 )
Coordinates of center of mass
\left({x}_{cm}, {y}_{cm}\right) = (50×0+150×12+100×150+150+100m,50×0+150×32+100×050+150+100m)\left(\dfrac{50 \times 0 + 150 \times \dfrac{1}{2} + 100 \times 1}{50 + 150 + 100}m, \dfrac{50 \times 0 + 150 \times \dfrac{\sqrt{3}}{2} + 100 \times 0}{50 + 150 + 100}m\right)
=(712m,34m)= \left(\dfrac{7}{12}m, \dfrac{\sqrt{3}}{4}m\right)
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