JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Conservation of Momentum

Three particles of masses

50 g

100 g

and

150

g are placed at the vertices of an equilateral triangle of side 1 m (as shown in the figure). The

(x, y)

coordinates of the centre of mass will be :

(A) $(\frac{3}{7} m, \frac{7}{12} m)$
(B) $(\frac{7}{12} m, \frac{3}{8} m)$
(C) $(\frac{3}{4} m, \frac{5}{12} m)$
(D) $(\frac{7}{12} m, \frac{3}{4} m)$ ✓ Correct

Answer: (D)

Solution

Place the

50

g at the origin, the

100

g at

(1, 0)

and the

150

g at the apex

(\frac{1}{2}, \frac{3}{2})

. With a total mass of

300

x_{c m} = \frac{50 ( 0 ) + 100 ( 1 ) + 150 ( \frac{1}{2} )}{300} = \frac{175}{300} = \frac{7}{12}

m and

y_{c m} = \frac{150 ( \frac{3}{2} )}{300} = \frac{3}{4}

m, so the centre of mass sits at

(\frac{7}{12}, \frac{3}{4})

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option

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