Two metal balls of mass \(m_{1}=5.00 \mathrm{~g}\) (diameter = \(5.00 \mathrm{~mm}\) ) and \(m_{2}=8.00 \mathrm{~g}\) (diameter \(=8.00 \mathrm{~mm}\) ) have positive charges of \(q_{1}=5.00 \mathrm{nC}\) and \(q_{2}=8.00 \mathrm{nC},\) respectively. A force holds them in place so that their centers are separated by \(8.00 \mathrm{~mm}\). What will their velocities be after the force is removed and they are separated by a large distance?

Short Answer

Expert verified
Answer: The final velocities of the two metal balls are \(v_1 = 0.163\,\text{m/s}\) for ball 1 and \(v_2 = 0.102\,\text{m/s}\) for ball 2.

Step by step solution

01

Convert Units

Before working on any formulas, we should first convert the given quantities into SI-units. This step ensures the consistency of the units in the calculations. Masses: \(m_1 = 5.00\,\text{g} = 5.00 \times 10^{-3}\,\text{kg}\), \(m_2 = 8.00\,\text{g} = 8.00 \times 10^{-3}\,\text{kg}\) \\ Charges: \(q_1 = 5.00\,\text{nC} = 5.00 \times 10^{-9}\,\text{C}\), \(q_2 = 8.00\,\text{nC} = 8.00 \times 10^{-9}\,\text{C}\) \\ Distance: \(r = 8.00\,\text{mm} = 8.00 \times 10^{-3}\,\text{m}\)
02

Calculate Initial Potential Energy

Now, we calculate the initial potential energy (PE) of the system using the formula for the electric potential energy between two charged objects. \(\text{PE} = k\frac{q_1 \times q_2}{r}\) In this formula, \(k\) is the electrostatic constant \((k = 8.99 \times 10^{9}\,\text{N}\,\text{m}^2\,\text{C}^{-2})\). \(\text{PE} = 8.99 \times 10^{9}\,\text{N}\,\text{m}^2\,\text{C}^{-2} \times \frac{(5.00 \times 10^{-9}\,\text{C})(8.00 \times 10^{-9}\,\text{C})}{(8.00 \times 10^{-3}\,\text{m})} = 4.493 \times 10^{-5}\,\text{N}\,\text{m}\)
03

Conservation of Mechanical Energy

Since there are no external forces acting on the system, the total mechanical energy is conserved. The initial mechanical energy is only the potential energy as the initial velocities are both zero. The final mechanical energy will only be in the form of kinetic energy, as the balls move far apart. \(\text{Initial Mechanical Energy} = \text{Final Mechanical Energy}\) \\ \(\text{PE}_\text{initial} = \text{KE}_1 + \text{KE}_2\) \(4.493 \times 10^{-5}\,\text{N}\,\text{m} = \frac{1}{2} m_1v_1^2 + \frac{1}{2} m_2v_2^2\)
04

Conservation of Linear Momentum

Linear momentum is also conserved in the system, since no external forces are acting on the system. Initially, both objects have zero momentum, so the final momenta should also add up to zero: \(m_1v_1 = m_2v_2\) We can get expressions for one of the velocities in terms of the other one from the conserved momentum equation: \(v_1 = \frac{m_2}{m_1}v_2\)
05

Solve for Final Velocities

Now, we can substitute the expression for \(v_1\) in terms of \(v_2\) from Step 4 into the mechanical energy conservation equation in Step 3, and solve for \(v_2\): \(4.493 \times 10^{-5}\,\text{N}\,\text{m} = \frac{1}{2}(5.00 \times 10^{-3}\,\text{kg})\left(\frac{8.00 \times 10^{-3}\,\text{kg}}{5.00 \times 10^{-3}\,\text{kg}}v_2\right)^2 + \frac{1}{2} (8.00 \times 10^{-3}\,\text{kg})v_2^2\) Solving for \(v_2\) gives: \(v_2 = 0.102\,\text{m/s}\) Now we can use the previously found expression for \(v_1\) to find its value: \(v_1 = \frac{m_2}{m_1}v_2 = \frac{8.00 \times 10^{-3}\,\text{kg}}{5.00 \times 10^{-3}\,\text{kg}}(0.102\,\text{m/s})\) \(v_1 = 0.163\,\text{m/s}\)
06

Results

The final velocities of the two metal balls after they are separated by a large distance are: For ball 1: \(v_1 = 0.163\,\text{m/s}\) \\ For ball 2: \(v_2 = 0.102\,\text{m/s}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Mechanical Energy
When considering phenomena like two charged metal balls in the absence of external forces, the principle of conservation of mechanical energy becomes applicable. Mechanical energy can be categorized into two forms: kinetic energy (KE), which is energy due to motion, and potential energy (PE), the stored energy based on position or configuration.

The conservation of mechanical energy states that in a closed system with no external work being done, the total mechanical energy remains constant. This concept is evident in our problem, where we initially have only electric potential energy due to the electrostatic force between the charges. As the balls are released and move apart, we expect their velocities to increase. This increase in kinetic energy must come from a decrease in potential energy since there is no external force doing work on the system.

By setting the initial potential energy equal to the sum of kinetic energies when the balls are far apart, we see energy transform from one form to the other while the total remains unchanged. This concept helps us to solve for the final velocities of the balls after enough time has passed that they are considered to be infinitely far from one another.
Linear Momentum Conservation
Another fundamental principle at play in our textbook example is the conservation of linear momentum, which states that in a system isolated from external forces, the total momentum remains constant over time.

Initially, both metal balls are at rest, which means the system has a total momentum of zero. Since no external forces act on the system when the balls are released, their momenta in opposite directions must cancel out to maintain that initial momentum of zero. Essentially, this means that the product of mass and velocity for one ball will be equal in magnitude and opposite in direction to the product for the other ball.

Using the linear momentum conservation, along with the information about the balls' masses, we deduce a relationship between their velocities. This relationship propels us towards resolving the unknown velocities by connecting them mathematically, thus allowing us to apply one equation (energy conservation) to solve for one variable at a time.
Electrostatic Force
The interaction between two charged objects, like our metal balls, is governed by electrostatic force, a fundamental force arising from their electric charges. Coulomb's law quantitatively describes this force as directly proportional to the product of the charges and inversely proportional to the square of the distance between their centers.

In our scenario, the electric potential energy, which is effectively the stored energy of the system due to the electrostatic force, is given by the initial separation distance and the quantity of charge on each ball. When the electrostatic force is the only force doing work (as external forces are being held back), this potential energy will eventually convert completely into kinetic energy. Understanding the electrostatic force and how it affects potential energy is crucial for predicting the system's behavior once the force holding the balls together is removed - allowing the conservation of energy principle to take full effect and enabling us to calculate the final velocities.

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Most popular questions from this chapter

A \(2.50-\mathrm{mg}\) dust particle with a charge of \(1.00 \mu \mathrm{C}\) falls at a point \(x=2.00 \mathrm{~m}\) in a region where the electric potential varies according to \(V(x)=\left(2.00 \mathrm{~V} / \mathrm{m}^{2}\right) x^{2}-\left(3.00 \mathrm{~V} / \mathrm{m}^{3}\right) x^{3}\) With what acceleration will the particle start moving after it touches down?

A particle with a charge of \(+5.0 \mu C\) is released from rest at a point on the \(x\) -axis, where \(x=0.10 \mathrm{~m}\). It begins to move as a result of the presence of a \(+9.0-\mu C\) charge that remains fixed at the origin. What is the kinetic energy of the particle at the instant it passes the point \(x=0.20 \mathrm{~m} ?\)

A point charge \(Q\) is placed a distance \(R\) from the center of a conducting sphere of radius \(a,\) with \(R>a\) (the point charge is outside the sphere). The sphere is grounded, that is, connected to a distant, unlimited source and/or sink of charge at zero potential. (Neither the distant ground nor the connection directly affects the electric field in the vicinity of the charge and sphere.) As a result, the sphere acquires a charge opposite in sign to \(Q\), and the point charge experiences an attractive force toward the sphere. a) Remarkably, the electric field outside the sphere is the same as would be produced by the point charge \(Q\) plus an imaginary mirror-image point charge \(q\), with magnitude and location that make the set of points corresponding to the surface of the sphere an equipotential of potential zero. That is, the imaginary point charge produces the same field contribution outside the sphere as the actual surface charge on the sphere. Calculate the value and location of \(q\). (Hint: By symmetry, \(q\) must lie somewhere on the axis that passes through the center of the sphere and the location of \(Q .)\) b) Calculate the force exerted on point charge \(Q\) and directed toward the sphere, in terms of the original quantities \(Q, R,\) and \(a\) c) Determine the actual nonuniform surface charge distribution on the conducting sphere.

A proton is placed midway between points \(A\) and \(B\). The potential at point \(A\) is \(-20 \mathrm{~V}\), and the potential at point \(B\) \(+20 \mathrm{~V}\). The potential at the midpoint is \(0 \mathrm{~V}\). The proton will a) remain at rest. b) move toward point \(B\) with constant velocity. c) accelerate toward point \(A\). d) accelerate toward point \(B\). e) move toward point \(A\) with constant velocity.

A solid metal ball with a radius of \(3.00 \mathrm{~m}\) has a charge of \(4.00 \mathrm{mC}\). If the electric potential is zero far away from the ball, what is the electric potential at each of the following positions? a) at \(r=0 \mathrm{~m},\) the center of the ball b) at \(r=3.00 \mathrm{~m},\) on the surface of the ball c) at \(r=5.00 \mathrm{~m}\)

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