Question

Two identical \(20.0-\mathrm{kg}\) spheres of radius \(10.0 \mathrm{~cm}\) are \(30.0 \mathrm{~cm}\) apart (center-to-center distance). a) If they are released from rest and allowed to fall toward one another, what is their speed when they first make contact? b) If the spheres are initially at rest and just touching, how much energy is required to separate them to \(1.00 \mathrm{~m}\) apart? Assume that the only force acting on each mass is the gravitational force due to the other mass.

Short answer

Answer: The speed of the spheres when they first make contact is approximately 4.27 x 10^-4 m/s. The energy required to separate them to 1.00 m apart is approximately 1.07 x 10^-7 J.

Step-by-step solution

Identify the initial conditions and find the initial potential energy

The initial conditions are when the two spheres are \(30.0 \mathrm{~cm}\) apart (center-to-center distance) and are at rest. We can then find the initial potential energy using the gravitational force formula: \(U_i = -G \frac{m_1 m_2}{r_i}\) Given the provided values, we have: \(U_i = - G \frac{(20.0 \,\mathrm{kg})^2}{(0.30\, \mathrm{m})}\)

Calculate the initial potential energy

Using the gravitational constant \(G = 6.674 \times 10^{-11} \,\mathrm{Nm^2/kg^2}\), now calculate the initial potential energy: \(U_i = -6.674 \times 10^{-11} \,\mathrm{Nm^2/kg^2} \cdot \frac{(20.0 \,\mathrm{kg})^2}{0.30\,\mathrm{m}} = -8.93 \times 10^{-8} \,\mathrm{J}\)

Determine the final conditions and find the final potential energy

The final conditions are when the two spheres just make contact. At this point, the center-to-center distance between the spheres is twice the radius of each sphere (\(2 \cdot 0.1 \mathrm{m} = 0.2 \,\mathrm{m}\)). We can find the final potential energy using the gravitational force formula: \(U_f = -G \frac{m_1 m_2}{r_f}\) Given the center-to-center distance, we have: \(U_f = - G \frac{(20.0 \,\mathrm{kg})^2}{(0.20 \,\mathrm{m})}\)

Calculate the final potential energy

Calculate the final potential energy with the given values: \(U_f = -6.674 \times 10^{-11} \,\mathrm{Nm^2/kg^2} \cdot \frac{(20.0 \,\mathrm{kg})^2}{0.20 \,\mathrm{m}} = -1.34 \times 10^{-7} \,\mathrm{J}\)

Use conservation of mechanical energy to find the final kinetic energy

Since the initial kinetic energy \(K_i = 0\), the conservation of mechanical energy equation becomes: \(K_f = U_i - U_f\) Now calculate the final kinetic energy: \(K_f = -8.93 \times 10^{-8} \,\mathrm{J} - (-1.34 \times 10^{-7} \,\mathrm{J}) = 4.47 \times 10^{-8} \,\mathrm{J}\)

Find the speed of the spheres when they first make contact

The final kinetic energy is shared between the two spheres, so we can write: \(K_f = \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2\) Divide both sides by \(m_1\) (since \(m_1=m_2\)) and solve for \(v\): \(v = \sqrt{\frac{2K_f}{m_1}} = \sqrt{\frac{2(4.47 \times 10^{-8} \,\mathrm{J})}{20.0 \,\mathrm{kg}}} \approx 4.27 \times 10^{-4}\,\mathrm{m/s}\) Thus, the speed of the spheres when they first make contact is approximately \(4.27 \times 10^{-4} \,\mathrm{m/s}\).

Calculate the energy required to separate the spheres to \(1.00\, \mathrm{m}\) apart

We need to find the energy required to change their potential energy from the initial value when they are just touching (at \(0.20\,\mathrm{m}\)) to when they are \(1.00\,\mathrm{m}\) apart: \(W = \Delta U = U_{1.00\,\mathrm{m}} - U_{0.20\,\mathrm{m}}\) We can find the potential energy when they are \(1.00\,\mathrm{m}\) apart using the gravitational force formula: \(U_{1.00\,\mathrm{m}} = - G \frac{m_1 m_2}{r} = -6.674 \times 10^{-11} \,\mathrm{Nm^2/kg^2} \cdot \frac{(20.0 \,\mathrm{kg})^2}{1.00 \mathrm{~m}} = -2.67 \times 10^{-8} \,\mathrm{J}\) Now calculate the energy required to separate them: \(W = -2.67 \times 10^{-8} \,\mathrm{J} - (-1.34 \times 10^{-7} \,\mathrm{J}) = 1.07 \times 10^{-7} \,\mathrm{J}\) Therefore, the energy required to separate the spheres to \(1.00\, \mathrm{m}\) apart is approximately \(1.07 \times 10^{-7}\,\mathrm{J}\).