Question

Two point masses \(\mathrm{A}\) and \(\mathrm{B}\) having masses in the ratio \(4: 3\) are separated by a distance of \(\operatorname{lm}\). When another point mass of mass \(\mathrm{M}\) is placed in between \(\mathrm{A}\) and \(\mathrm{B}\) the forces \(\mathrm{A}\) and is \((1 / 3 \mathrm{rd})\) of the force between \(\mathrm{B}\) and \(\mathrm{C}\), Then the distance \(\mathrm{C}\) from \(\mathrm{A}\) is \(=\ldots \ldots \ldots \mathrm{m}\) (A) \((2 / 3)\) (B) \(1 / 3\) (C) \(1 / 4\) (D) \(2 / 7\)

Short answer

The distance between A and C is 2/7 meters. Therefore, the correct answer is (D) 2/7.

Step-by-step solution

Let mA be the mass of A and mB the mass of B. Since their masses are in the ratio 4:3, we can write mA = 4m and mB = 3m, where m is some arbitrary mass unit. #Step 2: Write down the forces between A and C, and B and C#

According to the universal law of gravitation, the force F between two point masses is given by \[F = G\frac{m_{1}m_{2}}{r^2}\], where G is the gravitational constant, m1 and m2 are the masses and r is the distance between them. Therefore, the forces between A and C, and B and C can be written as \[F_{AC} = G\frac{mA \times M}{x^2}\] and \[F_{BC} = G\frac{mB \times M}{(1-x)^2}\], where M is the mass of point C and x is the distance between A and C. #Step 3: Formulate the given condition#

We are given that the force between A and C is 1/3 of the force between B and C, so we can write the condition as \[F_{AC} = \frac{1}{3}F_{BC}\]. Substitute the expressions for F_AC and F_BC from Step 2 and we get \[\frac{m_{A}\times M}{x^{2}} = \frac{1}{3}\frac{m_{B} \times M}{(1-x)^{2}}\] #Step 4: Solve for the distance x#

We can cancel out the mass M in the equation, and then substitute mA = 4m and mB = 3m from step 1, which gives us: \[\frac{4m}{x^2} = \frac{1}{3}\frac{3m}{(1-x)^2}\]. After canceling out the mass m and solving for x, we get: \[4(1-x)^2 = 3x^2 \Rightarrow 4 - 8x + 4x^2= 3x^2 \Rightarrow x^2 = 8x - 4 \Rightarrow x = \frac{2}{7}\] So, the distance between A and C is x = 2/7 meters. Therefore, the correct answer is (D) 2/7.

Key concepts

Point Masses

In physics, point masses are imaginary particles where all of a body's mass is concentrated at a single point in space. This concept simplifies calculations in gravitational interactions. By treating objects as point masses, we eliminate the complexities of their shape, size, or distribution of mass.

This simplification is particularly useful in problems involving gravitational forces. For example, in the given problem, two objects, A and B, are considered as point masses with a specific mass ratio. This allows us to focus on their gravitational interaction without delving into their physical structure.

Overall, point masses serve as a useful model in gravitational problems, providing a simplified approach to understanding complex interactions.

Universal Law of Gravitation

The universal law of gravitation, formulated by Isaac Newton, is a fundamental principle in physics. It describes the attractive force between two masses. The equation for gravitational force is given by: \[ F = G\frac{m_{1}m_{2}}{r^2} \] where:

• \( F \) is the gravitational force exerted between the two masses,
• \( G \) is the gravitational constant (\( 6.67430 \times 10^{-11} \, \text{m}^3\cdot\text{kg}^{-1}\cdot\text{s}^{-2} \)),
• \( m_{1} \) and \( m_{2} \) are the masses,
• \( r \) is the distance between the centers of the two masses.

This law applies universally, meaning it holds true for any two point masses in the universe. The force is always attractive and acts along the line joining the centers of the two masses. In the context of the problem, this law helps calculate the forces between the given masses, A and B, and another point mass C.

Mass Ratio

The concept of mass ratio is crucial when comparing two or more masses. In physics problems like the one given, knowing the mass ratio helps in setting up equations for calculating gravitational forces.

For the point masses A and B in our problem, the masses are given in the ratio 4:3. This means if mass A is considered as \( 4m \), mass B would be \( 3m \). Using this ratio simplifies the process of substituting and calculating forces between these point masses and facilitating easier resolution of equations.

Understanding and using mass ratios efficiently can aid in solving complex problems by reducing the number of variables involved.

Distance Calculation

Calculating distances between masses is a common requirement in physics, especially in gravitational problems. In the given problem, calculating the distance between point mass C and A requires understanding the forces between these masses and applying the condition given in the problem.

Using the condition that the force between A and C is one-third of the force between B and C, we derive equations involving the distance \( x \), which represents the distance between A and C. Substituting known values and solving the equation, using algebraic methods, yields:\[ x = \frac{2}{7} \text{ meters} \]
Thus, the ability to accurately calculate distances ensures that we can predict interactions between masses correctly. This is essential in analyzing and verifying gravitational phenomena.