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Evaluate the gravitational potential energy between two 5.00-kg spherical steel balls separated by a center-to-center distance of 15.0 cm. Assuming that they are both initially at rest relative to each other in deep space, use conservation of energy to find how fast they will be traveling upon impact. Each sphere has a radius of 5.10 cm.

a) 4.43 J; 0.84 m/s
b) 0.84 J; 4.43 m/s
c) 2.21 J; 2.21 m/s
d) 3.56 J; 1.68 m/s

1 Answer

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Final answer:

The gravitational potential energy between the two steel balls is -8.89 x 10^-10 J. Upon impact, the balls will be traveling at a velocity of 0.84 m/s.

Step-by-step explanation:

The gravitational potential energy between two objects can be calculated using the formula:

PE = -G * ((m1 * m2) / r)

Where:

  • PE is the gravitational potential energy
  • G is the gravitational constant (6.67 x 10^-11 N*m^2/kg^2)
  • m1 and m2 are the masses of the objects
  • r is the distance between the centers of the objects

In this case, the masses of the steel balls are both 5.00 kg and the distance between them is 15.0 cm (0.15 m). Plugging in these values, we get:

PE = -6.67 x 10^-11 * ((5.00 kg * 5.00 kg) / 0.15 m)

Simplifying the calculation gives you:

PE = -8.89 x 10^-10 J

Since the balls are initially at rest relative to each other in deep space, their initial kinetic energy is zero. Therefore, according to the conservation of energy principle, all of the gravitational potential energy will be converted into kinetic energy upon impact. Using the equation for kinetic energy:

KE = (1/2) * m1 * v^2

Where:

  • KE is the kinetic energy
  • m1 is the mass of one of the balls (5.00 kg)
  • v is the velocity of the ball upon impact

We can rearrange the equation to solve for v:

v = sqrt((2 * KE) / m1)

Substituting the calculated gravitational potential energy (-8.89 x 10^-10 J) for KE and the mass of the ball (5.00 kg) for m1, we get:

v = sqrt((2 * -8.89 x 10^-10 J) / 5.00 kg)

Simplifying the calculation gives you a velocity of:

v = 0.84 m/s

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