Two particles of masses 20 kg and 10 kg are initially at a distance of 1.0m. Find the speeds of the particles when the separation between them decreases to 0.5 m, if only gravitational forces are acting.

To solve the problem of two particles with masses 20 kg and 10 kg moving towards each other under the influence of gravitational forces, we can use the principles of conservation of energy and Newton's law of universal gravitation. Let's break it down step by step.

Understanding the Forces at Play

According to Newton's law of universal gravitation, the gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:

F = G \frac{m_1 m_2}{r^2}

where \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).

Initial and Final Conditions

Initially, the two particles are 1.0 m apart. We want to find their speeds when they are 0.5 m apart. The gravitational potential energy (U) at a distance \( r \) is given by:

U = -G \frac{m_1 m_2}{r}

Calculating Initial Potential Energy

At the initial separation of 1.0 m:

• Mass \( m_1 = 20 \, \text{kg} \)
• Mass \( m_2 = 10 \, \text{kg} \)
• Distance \( r = 1.0 \, \text{m} \)

The initial potential energy \( U_i \) is:

U_i = -G \frac{20 \times 10}{1} = -200G

Calculating Final Potential Energy

At the final separation of 0.5 m:

The final potential energy \( U_f \) is:

U_f = -G \frac{20 \times 10}{0.5} = -400G

Energy Conservation Principle

The total mechanical energy in the system is conserved. Thus, the change in potential energy will equal the change in kinetic energy:

U_i - U_f = K_f - K_i

Initially, both particles are at rest, so \( K_i = 0 \). Therefore:

-200G + 400G = K_f

This simplifies to:

K_f = 200G

Finding Kinetic Energy

The kinetic energy \( K \) of a particle is given by:

K = \frac{1}{2}mv^2

Let \( v_1 \) be the speed of the 20 kg mass and \( v_2 \) be the speed of the 10 kg mass. The total kinetic energy is:

K_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2

Substituting the values:

200G = \frac{1}{2} (20)v_1^2 + \frac{1}{2} (10)v_2^2

This simplifies to:

200G = 10v_1^2 + 5v_2^2

Using Conservation of Momentum

Since no external forces are acting on the system, the momentum is conserved. The initial momentum is zero, so:

m_1 v_1 + m_2 v_2 = 0

Substituting the masses gives:

20v_1 + 10v_2 = 0

From this, we can express \( v_2 \) in terms of \( v_1 \):

v_2 = -2v_1

Substituting Back into Kinetic Energy Equation

Now, substitute \( v_2 \) into the kinetic energy equation:

200G = 10v_1^2 + 5(-2v_1)^2

This expands to:

200G = 10v_1^2 + 20v_1^2

Combining terms gives:

200G = 30v_1^2

Solving for \( v_1^2 \) yields:

v_1^2 = \frac{200G}{30} = \frac{20G}{3}

Thus, the speed of the 20 kg mass is:

v_1 = \sqrt{\frac{20G}{3}}

Finding the Speed of the Second Particle

Now, substituting back to find \( v_2 \):

v_2 = -2\sqrt{\frac{20G}{3}} = -\frac{4\sqrt{20G}}{\sqrt{3}}

Final Speeds

In summary, the speeds of the particles when the separation decreases to 0.5 m are:

• Speed of the 20 kg particle: v_1 = \sqrt{\frac{20G}{3}} \, \text{m/s}
• Speed of the 10 kg particle: v_2 = -\frac{4\sqrt{20G}}{\sqrt{3}} \, \text{m/s}

These equations give you the speeds of both particles as they move towards each other due to