Three identical spheres of mass m , are placed at the vertices of an equilateral triangle of length a . When released, they interact only through gravitational force and collide after a time T=4 seconds. If the sides of the triangle are increased to length 2 a and also the masses of the spheres are made 2 m , then they will collide after \underline\hspace2cm seconds.

Gravitation PYQ — JEE Main 2025 · 3 Apr · Shift 1 · Q74 with Solution

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Correct answer: 8

Use symmetry of motion

Three identical masses placed at the vertices of an equilateral triangle will always remain at the vertices of a shrinking equilateral triangle because of symmetry.

So we only need to study the motion of one sphere toward the center.

Force on one sphere when side length is $r$

At any instant, let the side of the triangle be $r$ .

Each of the other two spheres exerts gravitational force

F_0 = \frac{Gm^2}{r^2}

on the chosen sphere.

The angle between these two forces is $60^\circ$ , so the resultant is

F = \sqrt{F_0^2 + F_0^2 + 2F_0^2\cos 60^\circ} = \sqrt{3}\,F_0 = \sqrt{3}\frac{Gm^2}{r^2}.

Hence acceleration of each sphere is

A = \frac{F}{m} = \sqrt{3}\frac{Gm}{r^2}.

Thus the acceleration has the form

A \propto \frac{m}{r^2}.

How collision time scales

Suppose the size scale of the system is $r_0$ initially. Let collision time be $T$ .

From dimensional/scaling analysis for motion under acceleration

\ddot r \sim \frac{Gm}{r^2},

we get

\frac{r_0}{T^2} \sim \frac{Gm}{r_0^2}.

Therefore,

T^2 \sim \frac{r_0^3}{Gm} \quad \Rightarrow \quad T \propto \sqrt{\frac{r_0^3}{m}}.

Since the initial side length plays the role of initial size,

T \propto \sqrt{\frac{a^3}{m}}.

Apply to the changed system

Initially,

T_1 = 4 \text{ s}, \quad a_1 = a, \quad m_1 = m.

In the new case,

a_2 = 2a, \quad m_2 = 2m.

\frac{T_2}{T_1} = \sqrt{\frac{a_2^3/m_2}{a_1^3/m_1}} = \sqrt{\frac{(2a)^3/(2m)}{a^3/m}} = \sqrt{\frac{8a^3}{2m}\cdot \frac{m}{a^3}} = \sqrt{4} = 2.

Hence,

T_2 = 2T_1 = 2 \times 4 = 8 \text{ s}.

Final answer

The new collision time is

\boxed{8}

seconds.

More from Gravitation

View written solutionFree

Correct answer: 8

Use symmetry of motion

Three identical masses placed at the vertices of an equilateral triangle will always remain at the vertices of a shrinking equilateral triangle because of symmetry.

So we only need to study the motion of one sphere toward the center.

Force on one sphere when side length is $r$

At any instant, let the side of the triangle be $r$ .

Each of the other two spheres exerts gravitational force

F_0 = \frac{Gm^2}{r^2}

on the chosen sphere.

The angle between these two forces is $60^\circ$ , so the resultant is

F = \sqrt{F_0^2 + F_0^2 + 2F_0^2\cos 60^\circ} = \sqrt{3}\,F_0 = \sqrt{3}\frac{Gm^2}{r^2}.

Hence acceleration of each sphere is

A = \frac{F}{m} = \sqrt{3}\frac{Gm}{r^2}.

Thus the acceleration has the form

A \propto \frac{m}{r^2}.

How collision time scales

Suppose the size scale of the system is $r_0$ initially. Let collision time be $T$ .

From dimensional/scaling analysis for motion under acceleration

\ddot r \sim \frac{Gm}{r^2},

we get

\frac{r_0}{T^2} \sim \frac{Gm}{r_0^2}.

Therefore,

T^2 \sim \frac{r_0^3}{Gm} \quad \Rightarrow \quad T \propto \sqrt{\frac{r_0^3}{m}}.

Since the initial side length plays the role of initial size,

T \propto \sqrt{\frac{a^3}{m}}.

Apply to the changed system

Initially,

T_1 = 4 \text{ s}, \quad a_1 = a, \quad m_1 = m.

In the new case,

a_2 = 2a, \quad m_2 = 2m.

\frac{T_2}{T_1} = \sqrt{\frac{a_2^3/m_2}{a_1^3/m_1}} = \sqrt{\frac{(2a)^3/(2m)}{a^3/m}} = \sqrt{\frac{8a^3}{2m}\cdot \frac{m}{a^3}} = \sqrt{4} = 2.

Hence,

T_2 = 2T_1 = 2 \times 4 = 8 \text{ s}.

Final answer

The new collision time is

\boxed{8}

seconds.

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