JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Conservation of Momentum

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The identical spheres each of mass 2 M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is

\frac{4 2}{x}

, where the value of

x

is ______.

Answer: 3

Solution

Place the right-angle vertex at the origin, with the other two spheres at

(4, 0)

and

(0, 4)

(metres), each of mass

2 M

(total mass

6 M

r_{C O M} = \frac{2 M ( 0 , 0 ) + 2 M ( 4 , 0 ) + 2 M ( 0 , 4 )}{6 M} = \frac{( 8 M , 8 M )}{6 M} = (\frac{4}{3}, \frac{4}{3})

∣ r_{C O M} ∣ = (\frac{4}{3})^{2} + (\frac{4}{3})^{2} = \frac{4}{3} 2 = \frac{4 2}{3}

Comparing with

\frac{4 2}{x}

x = 3

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Integer answer

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