A two particle system with masses $2 kg$ and $3 kg$ are present at (15,0) and (0, 20) in a $x, y$ plane. Then its centre of mass lies on lines

y = 2 x

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\frac{x}{12} + \frac{y}{24} = 1

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3 x - y = 6

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y = x

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Solution

The correct options are
A $y = 2 x$
B $\frac{x}{12} + \frac{y}{24} = 1$
C $3 x - y = 6$
$X_{C M} = \frac{m_{1} \times x_{1} + m_{2} \times x_{2}}{m_{1} + m_{2}}$
$X_{C M} = \frac{2 \times 15}{5} = 6$
$Y_{C M} = \frac{m_{1} \times y_{1} + m_{2} \times y_{2}}{m_{1} + m_{2}}$
$Y_{C M} = \frac{3 \times 20}{5} = 12$
As $(6, 12)$ passes through $\frac{x}{12} + \frac{y}{24} = 1$ , $3 x - y = 6$ , $y = 2 x$ , centre of mass lies on these lines.

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A two particle system with masses 2 kg and 3 kg are present at (15,0) and (0, 20) in a x, y plane. Then its centre of mass lies on lines

The correct options are A y=2x B x12+y24=1 C 3x−y=6XCM=m1×x1+m2×x2m1+m2 XCM=2×155=6 YCM=m1×y1+m2×y2m1+m2 YCM=3×205=12 As (6,12) passes through x12+y24=1, 3x−y=6, y=2x, centre of mass lies on these lines.

A two particle system with masses $2 kg$ and $3 kg$ are present at (15,0) and (0, 20) in a $x, y$ plane. Then its centre of mass lies on lines