TWO SMALL BODIES OF MASSES 10KG AND 20 KG ARE KEPT AT A DISTANCE 1.0 m APART ANS RELEASED . ASSUMING THAT ONLY MUTUAL GRAVITATIONAL FORCES ARE ACTING , FIND THE SPEEDS OF THE PARTICLES WHEN THE SEPRATION DECREASES TO .5 m .

To solve the problem of two small bodies with masses of 10 kg and 20 kg that are initially 1.0 m apart and then released, we need to analyze the gravitational forces acting between them and how these forces affect their speeds as they move closer together. We'll apply the principles of gravitational attraction and conservation of energy to find the speeds of the particles when the separation decreases to 0.5 m.

Understanding Gravitational Forces

The gravitational force between two masses can be calculated using Newton's law of universal gravitation, which states:

F = G * (m1 * m2) / r²

Where:

• F is the gravitational force between the masses.
• G is the gravitational constant, approximately 6.674 × 10⁻¹¹ N(m/kg)².
• m1 and m2 are the masses (10 kg and 20 kg).
• r is the distance between the centers of the two masses.

Calculating Initial Gravitational Force

Initially, when the distance is 1.0 m, we can calculate the gravitational force:

F_initial = G * (10 kg * 20 kg) / (1.0 m)²

Substituting the values:

F_initial = (6.674 × 10⁻¹¹) * (200) / 1 = 1.3348 × 10⁻⁸ N

Energy Considerations

As the two masses move closer together, the gravitational potential energy is converted into kinetic energy. The potential energy (U) at a distance r is given by:

U = -G * (m1 * m2) / r

Initially, at 1.0 m:

U_initial = -G * (10 kg * 20 kg) / 1.0 m = -1.3348 × 10⁻⁸ J

When the distance decreases to 0.5 m:

U_final = -G * (10 kg * 20 kg) / 0.5 m = -2.6696 × 10⁻⁸ J

Change in Potential Energy

The change in potential energy (ΔU) as the masses move from 1.0 m to 0.5 m is:

ΔU = U_final - U_initial = -2.6696 × 10⁻⁸ J - (-1.3348 × 10⁻⁸ J) = -1.3348 × 10⁻⁸ J

Applying Conservation of Energy

The total mechanical energy is conserved, meaning the loss in potential energy equals the gain in kinetic energy. The kinetic energy (K) of each mass can be expressed as:

K = 0.5 * m * v²

Let v1 be the speed of the 10 kg mass and v2 be the speed of the 20 kg mass. The total kinetic energy when they are at 0.5 m apart is:

K_total = 0.5 * 10 * v1² + 0.5 * 20 * v2²

Setting the change in potential energy equal to the total kinetic energy gives us:

1.3348 × 10⁻⁸ J = 0.5 * 10 * v1² + 0.5 * 20 * v2²

Relating the Speeds

Since the two masses are moving towards each other, we can relate their speeds using the conservation of momentum:

m1 * v1 = m2 * v2

Substituting the masses:

10 * v1 = 20 * v2

This simplifies to:

v2 = 0.5 * v1

Substituting Back to Find Speeds

Now we can substitute v2 back into the kinetic energy equation:

1.3348 × 10⁻⁸ J = 0.5 * 10 * v1² + 0.5 * 20 * (0.5 * v1)²

Expanding this gives:

1.3348 × 10⁻⁸ J = 5 * v1² + 0.5 * 20 * 0.25 * v1²

1.3348 × 10⁻⁸ J = 5 * v1² + 2.5 * v1²

1.3348 × 10⁻⁸ J = 7.5 * v1²

Solving for v1²:

v1² = 1.3348 × 10⁻⁸ J / 7.5

v1² = 1.7797 × 10⁻⁹

v1 = √(1.7797 × 10⁻⁹) ≈ 4.22 × 10⁻⁵ m/s

Finding v2

Now, substituting back to find v2:

v2 = 0.5 * v1 ≈ 0.5 * 4.22 × 10⁻⁵ ≈ 2.11 × 10⁻⁵ m/s

Final Speeds

Thus, the speeds of the particles when the separation decreases to 0.5 m are approximately:

• Speed of the 10 kg mass (v1): 4.22 × 10⁻⁵ m/s
• Speed of the