Q4. Multiple Correct Two objects of mass m and 4 m are at rest at an infinite separation. They move towards each other under mutual attraction. If G is the universal gravitational constant. Then at separation r . (1) The sum of energy of the two objects is negative. (2) Their relative velocity of approach is in magnitude (3) The sum of kinetic energy of the objects is . (4) The sum of angular momenta of both the objects is zero about any point.

Concepts you need

1. Gravitational Potential Energy (GPE) between two masses $m_1$ and $m_2$ separated by distance $r$ is $U = -\frac{G m_1 m_2}{r}$ Here the zero level of potential energy is taken at $r=\infty$.

2. Conservation of Mechanical Energy $\text{Initial Energy} = \text{Final Kinetic Energy} + \text{Final Potential Energy}$

3. Conservation of Linear Momentum (because no external force acts on the whole two-mass system). If they start from rest, the centre of mass (CM) stays at rest, hence $m v = 4m V \qquad \Rightarrow \qquad v = 4V$ where $v$ is the speed of the lighter mass and $V$ is the speed of the heavier one.

4. Relative Speed of approach is simply the sum of the individual speeds when they head straight toward each other: $v_{\text{rel}} = v + V$

5. Angular Momentum about any point is zero because

   • their motion is strictly along the line joining them, and
   • the net external torque is zero.

Logical chain to solve

1. Write momentum relation $v = 4V$ from the CM being at rest.
2. Write energy conservation using the above velocity relation; solve for $V$ (or $v$).
3. Compute
   • relative speed $v_{\text{rel}}$,
   • total kinetic energy $K$,
   • check sign of total mechanical energy $E = K + U$,
   • reason about angular momentum.
4. Compare each computed/ deduced quantity with the statements (1)–(4).

What’s happening here?

Imagine two ice-skaters standing very far apart on a super-slippery lake. One skater is light (mass $m$) and the other is heavier (mass $4m$). At first they are both standing still. Suddenly they feel a magical rope pulling them toward each other (this rope is like gravity).

Because the rope only pulls straight along the line connecting them, they will simply glide toward each other – no turns, no curves.

Key ideas:

1. Energy can change form but the total stays the same. At the start, there is no kinetic energy (nobody is moving) and the gravitational energy is zero because they are infinitely far apart. As they move closer, gravitational energy becomes negative, and the lost energy turns into their motion energy (kinetic).
2. Momentum of the pair stays zero. They begin with zero total momentum, so the light skater must move faster than the heavy one to keep the momenta equal and opposite.
3. Because they glide along one straight line, they never spin about any point – so the system has no angular momentum.

These simple facts let us write easy equations to find their speeds, kinetic energy, and check which of the given statements are right.

Step-by-step calculation

Let

• Masses: $m_1 = m$, $m_2 = 4m$
• Speeds when separated by distance $r$: $v$ (for $m$), $V$ (for $4m$)
• Initial separation: $\infty$ (both at rest)

1. Conservation of Linear Momentum

Total initial momentum is zero, so at any time m v = 4m V \;\; \Longrightarrow \;\; v = 4V \tag{1}

2. Conservation of Mechanical Energy

Initial mechanical energy $E_i = 0$ (no kinetic, zero potential at $\infty$).

Final energy $K = \frac12 m v^2 + \frac12 (4m) V^2$ $U = -\frac{G m (4m)}{r} = -\frac{4 G m^2}{r}$

Set $E_i = E_f$: $0 = K + U$ Substitute $v = 4V$ into $K$: $K = \frac12 m (4V)^2 + 2 m V^2 = 8 m V^2 + 2 m V^2 = 10 m V^2$ So 10 m V^2 - \frac{4 G m^2}{r} = 0 \;\; \Longrightarrow \;\; V^2 = \frac{2 G m}{5 r} \tag{2} From (1): $v^2 = 16 V^2 = 16\left(\frac{2 G m}{5 r}\right) = \frac{32 G m}{5 r}$

3. Relative Speed of Approach

$v_{\text{rel}} = v + V = 4V + V = 5V$ Square it: $v_{\text{rel}}^2 = 25 V^2 = 25\left(\frac{2 G m}{5 r}\right) = \frac{10 G m}{r}$ Hence $v_{\text{rel}} = \sqrt{\frac{10 G m}{r}}$

4. Total Kinetic Energy

Using $V^2$ from (2): $K = 10 m V^2 = 10 m \left(\frac{2 G m}{5 r}\right) = \frac{4 G m^2}{r}$

5. Total Mechanical Energy

$E = K + U = \frac{4 G m^2}{r} - \frac{4 G m^2}{r} = 0$ So the sum of energies is zero, not negative.

6. Angular Momentum

Since both masses move directly along the line joining them, the position and momentum vectors are collinear, giving zero angular momentum about any reference point.

Checking Statements

1. "Sum of energy is negative" ⟹ False (it is zero).
2. "Relative velocity magnitude ..." ⟹ True; it equals $\sqrt{\frac{10 G m}{r}}$.
3. "Sum of kinetic energy is …" ⟹ True; it is $\frac{4 G m^2}{r}$.
4. "Sum of angular momenta is zero about any point" ⟹ True.

Hence, statements (2), (3), (4) are correct.