Suppose a 0.17-kg billiard ball is rolling down a pool table with an initial speed of 4.5 m/s. As it travels, it loses some of its energy as heat. The ball slows down to 3.5 m/s and then collides straight-on with a second billiard ball of equal mass. The first billiard ball completely stops and the second one rolls away with a velocity of 3.5 m/s. Assume the first billiard ball is the system.
Calculate w for the process.
Calculate q for the process.
Calculate delta E for the process.
What is w? How can only one ball be "the system" ?
The collision is elastic since the balls have equal mass and exchange velocities. The only time kinetic energy is lost is when the originally moving ball slows down. The lost energy as the ball slows from 4.5 to 3.5 m/s is converted to heat.
w is work
The work is done against friction while the ball slows down. Consider it equal to the loss of kinetic energy
Transfer of energy from one ball to another during collision is an interesting process. It is done by transferring stress waves, but there must be some motion of the balls during contact to allow work to be done on the other ball.
To calculate the work (w) for the process, we need to determine the change in kinetic energy. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.
The initial kinetic energy of the first billiard ball can be calculated using the formula: KE = 0.5 * mass * velocity^2. Plugging in the values, we get:
KE_initial = 0.5 * 0.17 kg * (4.5 m/s)^2 = 0.5 * 0.17 kg * 20.25 m^2/s^2 = 1.71975 J (rounded to four decimal places)
The final kinetic energy of the first billiard ball is zero since it comes to a complete stop. Therefore, the change in kinetic energy (w) is:
w = KE_final - KE_initial = 0 - 1.71975 J = -1.71975 J
So, the work for the process is approximately -1.71975 J.
To calculate the heat transfer (q) for the process, we need to apply the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
Since the first billiard ball loses some of its kinetic energy as heat, the change in internal energy is equal to the negative of the work:
ΔE = -w
Therefore, the heat transfer (q) for the process is also -1.71975 J.
Finally, to calculate the change in energy (ΔE) for the process, we simply substitute the calculated values:
ΔE = q = -1.71975 J
So, the change in energy for the process is approximately -1.71975 J.