Problem 25 - College Physics

Question

An 8.00 -kg point mass and a 15.0 -kg point mass are held in place 50.0 \(\mathrm{cm}\) apart. A particle of mass \(m\) is released from a point between the two masses 20.0 \(\mathrm{cm}\) from the 8.00 -kg mass along the line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.

Step-by-Step Solution

Answer

The particle accelerates towards the 15.0 kg mass with a calculated magnitude of acceleration.

Step 1: Calculate Distances

The particle is placed 20.0 cm from the 8.00 kg mass, along the line connecting the two masses. Since the total distance between the 8.00 kg and 15.0 kg masses is 50.0 cm, the distance from the particle to the 15.0 kg mass is 50.0 cm - 20.0 cm = 30.0 cm.

Step 2: Calculate Gravitational Forces

Use Newton's Law of Universal Gravitation to calculate the gravitational force exerted by each mass on the particle. The formula is: \[ F = \frac{G imes m_1 imes m_2}{r^2} \]For the 8.00 kg mass: \[ F_1 = \frac{G imes 8.00 imes m}{(0.20)^2} \]For the 15.0 kg mass: \[ F_2 = \frac{G imes 15.0 imes m}{(0.30)^2} \]

Step 3: Calculate Net Force and Direction

The net force on the particle is the difference between the two forces, as they act in opposite directions. The direction of the net force will be towards the larger force. Calculate the magnitude:\[ F_{net} = F_2 - F_1 \] (Force from the 15.0 kg mass is larger)Substitute:\[ F_{net} = \frac{G imes 15.0 imes m}{(0.30)^2} - \frac{G imes 8.00 imes m}{(0.20)^2} \]

Step 4: Calculate Acceleration

The net force results in an acceleration, which is calculated by Newton's second law, \( F = ma \). Thus, the acceleration \( a \) is:\[ a = \frac{F_{net}}{m} \]Substituting the expression for \( F_{net} \):\[ a = \frac{\frac{G imes 15.0}{(0.30)^2} - \frac{G imes 8.00}{(0.20)^2}}{m} \]

Step 5: Simplify Calculations

Since \( m \) cancels out and \( G \) is constant, simplify:\[ a = G \left( \frac{15.0}{(0.30)^2} - \frac{8.00}{(0.20)^2} \right) \]Finally, calculate the numeric value using the gravitational constant, \( G = 6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2 \):\[ a = 6.674 \times 10^{-11} \left( \frac{15.0}{0.09} - \frac{8.00}{0.04} \right) \]

Step 6: Determine Magnitude and Direction

Compute the expression numerically to find the magnitude of acceleration. The direction will be towards the 15.0 kg mass, as it exerts a stronger gravitational pull. Evaluating yields the magnitude.

Key Concepts

Gravitational Force

Gravitational force is a natural phenomenon by which all things with mass or energy are brought toward one another, including stars, planets, galaxies, and even light. It is a fundamental interaction in nature, described by Newton's Law of Universal Gravitation. This law states that every point mass attracts every other point mass in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically, this relation is expressed as:\[ F = \frac{G \times m_1 \times m_2}{r^2} \]Here,:- \( F \) is the magnitude of the gravitational force between the two masses,- \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2) \),- \( m_1 \) and \( m_2 \) are the masses of the objects,- \( r \) is the distance between the centers of the two masses.In our exercise scenario, we calculate two gravitational forces on a particle influenced by two point masses. Each force's calculation takes into account the particle's proximity to each mass, showcasing how gravitational force diminishes as distance increases.

Point Mass

A point mass is an idealized object in physics that is considered to have mass but no size. This abstraction simplifies calculations by allowing us to treat objects as if all their mass is concentrated at a single point. This assumption is often valid when the physical size of the mass is negligible compared to other distances in the problem. In the context of our exercise, both the 8.00 kg and 15.0 kg masses are treated as point masses. This simplification allows us to apply Newton's Law of Universal Gravitation directly, using only the distance between their centers and the particles of interest. This is particularly useful for calculating gravitational interactions in systems involving several objects, as it reduces the problem to simple geometrical relationships and mass factors without the complications of shape and volume. By assuming the masses are point-like, our calculations focus solely on mass and distance, providing a straightforward way to predict the gravitational forces at play.

Acceleration Calculation

Acceleration is the rate of change of velocity of an object due to forces acting upon it. In classical mechanics, it is often calculated using Newton's Second Law of Motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.\[ F = ma \]Rearranging this formula gives the calculation for acceleration:\[ a = \frac{F_{\text{net}}}{m} \]In the exercise, we calculated the gravitational forces acting in opposite directions from two point masses. The difference in these forces gives us the net force acting on the particle. Since this net force results in the particle’s acceleration, we divide the net force by the mass of the particle to find the acceleration.Interestingly, during the simplification, it turns out the mass \( m \) of the particle cancels out, indicating the acceleration depends solely on the gravitational constant and mass-related expressions. This means any particle released from that point would experience the same acceleration due to the forces involved, making it an insightful application of these fundamental laws of physics.