Question

Suppose you toss a tennis ball upward. (a) Does the kinetic energy of the ball increase or decrease as it moves higher? (b) What happens to the potential energy of the ball as it moves higher? (c) If the same amount of energy were imparted to a ball the same size as a tennis ball, but of twice the mass, how high would it go in comparison to the tennis ball? Explain your answers.

Short answer

(a) The kinetic energy of the tennis ball decreases as it moves higher because its velocity decreases due to gravity. (b) The potential energy of the tennis ball increases as it moves higher, as the height above the reference point (Earth's surface) increases. (c) If the same amount of energy is imparted to a ball with twice the mass of the tennis ball, it would reach half the height of the tennis ball since the ball with twice the mass would require more energy to reach the same height.

Step-by-step solution

(a) Kinetic Energy at Different Heights

The kinetic energy of a moving object is given by the equation: \[KE = \frac{1}{2} mv^2\] where KE is the kinetic energy, m is the mass of the object, and v is its velocity. When a tennis ball is tossed upward, it experiences a decrease in velocity due to the influence of gravity. As it moves higher, its velocity gradually decreases until it reaches its maximum height, at which point the velocity is momentarily zero. Because the kinetic energy of an object depends on the square of its velocity, a decrease in velocity will result in a decrease in kinetic energy. So, as the tennis ball moves higher, its kinetic energy will decrease.

(b) Potential Energy at Different Heights

The potential energy of an object in relation to Earth's surface is defined by the equation: \[PE = mgh\] where PE is the potential energy, m is the mass of the object, g is the gravitational acceleration (approximately 9.81 m/s^2), and h is the height above the reference point (usually the surface of the Earth). As the tennis ball moves higher, the height h increases. Since both the mass and gravitational acceleration remain constant, the potential energy of the ball will increase as it moves higher.

(c) Tennis Ball vs Ball with Twice the Mass

Let's now investigate how high a ball with twice the mass of the tennis ball would go in comparison to the tennis ball if the same amount of energy were imparted to both balls. As we know, the potential energy at the maximum height is equal to the initial kinetic energy given to the ball. So, we have the equation: \[PE = KE\] Now, let \(m_1\) be the mass of the tennis ball and \(m_2\) be the mass of the heavier ball such that \(m_2 = 2m_1\). Letting \(h_1\) and \(h_2\) be the respective maximum heights, we can write the equations: \[m_1gh_1 = \frac{1}{2}m_1v^2\] and \[m_2gh_2 = \frac{1}{2}m_2v^2\] We are given that both balls receive the same initial kinetic energy, which means: \[\frac{1}{2}m_1v^2 = \frac{1}{2}m_2v^2\] Now, we'll find the relationship between the heights \(h_1\) and \(h_2\). Divide the two potential energy equations: \[\frac{m_1gh_1}{m_2gh_2} = \frac{\frac{1}{2}m_1v^2}{\frac{1}{2}m_2v^2}\] Cancel out the terms: \[\frac{h_1}{h_2} = \frac{m_2}{m_1}\] Substitute the relationship \(m_2 = 2m_1\): \[\frac{h_1}{h_2} = \frac{2m_1}{m_1}\] \[h_1 = 2h_2\] Thus, when the same amount of energy is imparted to a ball with twice the mass of the tennis ball, it will go half as high as the tennis ball.