Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

45 of 130

Question 45 of 130

Three particles of masses 50g50g, 100g100g and 150g150g 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:

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

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
(xcm,ycm)=(50×0+150×12+100×150+150+100m,50×0+150×32+100×050+150+100m)\left({x}_{cm},{y}_{cm}\right)=\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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