The kinetic energy of a rolling billiard ball is given by KE=1/2mv2. Suppose a 0.17-kg billiard ball is rolling down a pool table with an initial speed of 4.7m/s . As it travels, it loses some of its energy as heat. The ball slows down to 3.7m/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 3.7m/s . Assume the first billiard ball is the system

Calculate W, Calculate q, Calculate ΔE .

Typo:

W=∆KE=Kf-Ki= 1/2(0.17kg)(3.7m/s)-1/2(0.17)(4.7m/s)=1/2(0.17kg)[3.7m/s-4.7m/s]

Well, this problem seems like a ball...er, I mean, a breeze! Let's break it down.

First, we need to find the work done on the first billiard ball. Work (W) is defined as the change in kinetic energy, and it can be calculated using the equation:

W = ΔKE = KE_final - KE_initial

The initial kinetic energy (KE_initial) of the first billiard ball is given by KE = 1/2 * m * v^2, where m is the mass and v is the initial velocity. Plugging in the values, we get:

KE_initial = 1/2 * 0.17 kg * (4.7 m/s)^2

Now, the final kinetic energy (KE_final) of the first billiard ball is zero since it comes to a complete stop. Therefore:

W = ΔKE = 0 - KE_initial = -KE_initial

So the work done on the first billiard ball is negative and is equal to the initial kinetic energy of the ball.

Next, let's move on to finding the heat (q) transferred to the surroundings. From the conservation of energy, we know that the change in internal energy (ΔE) of the system should equal the sum of the work done on the system (W) and the heat gained or lost (q).

Since the first billiard ball is losing energy, the change in internal energy (ΔE) will be negative. We can calculate it using the equation:

ΔE = q + W

Rearranging the equation, we find:

q = ΔE - W

Since the first billiard ball is losing energy, the change in internal energy (ΔE) is equal to -KE_initial. Plugging in the values, we get:

q = -(-KE_initial) - KE_initial = -KE_initial + KE_initial = 0

Therefore, the heat transferred to the surroundings is zero.

Finally, let's calculate the change in energy (ΔE) of the first billiard ball. Since the initial kinetic energy (KE_initial) was already calculated and the final kinetic energy (KE_final) is zero, we can find the change in energy using the equation:

ΔE = KE_final - KE_initial = 0 - KE_initial = -KE_initial

So, the change in energy of the first billiard ball is negative and equal to the initial kinetic energy.

To summarize:
- The work done on the first billiard ball (W) is equal to the negative of the initial kinetic energy.
- The heat transferred to the surroundings (q) is zero.
- The change in energy of the first billiard ball (ΔE) is negative and equal to the initial kinetic energy.

Hope that helps, and remember, when it comes to billiards, make sure you don't put too much spin on your break shot, or you might end up pocketing the cue ball!

To calculate the work done (W), heat transfer (q), and change in energy (ΔE) in this scenario, we need to consider the conservation of energy. Here's how you can calculate each of these values:

1. Work done (W):
The work done is the transfer of energy due to an external force acting on an object. In this case, the first billiard ball is not subjected to any external forces, so no work is done on it. Therefore, W = 0.

2. Heat transfer (q):
Since the billiard ball loses energy as heat, we can calculate the amount of heat transferred using the equation q = ΔE - W. Since W is 0, q = ΔE.

3. Change in energy (ΔE):
To calculate the change in energy, we need to calculate the initial and final kinetic energies of the first billiard ball.

Initial kinetic energy (KEi):
KEi = 1/2 * m * v^2,
where m is the mass of the ball (0.17 kg) and v is the initial velocity (4.7 m/s).
KEi = 1/2 * 0.17 kg * (4.7 m/s)^2.

Final kinetic energy (KEf):
Since the billiard ball stops completely, its final kinetic energy is 0.

Change in energy (ΔE):
ΔE = KEf - KEi,
where ΔE represents the energy lost as heat.
ΔE = 0 - (1/2 * 0.17 kg * (4.7 m/s)^2).

Now, let's calculate each value:

1. W = 0 J (No work is done)
2. q = ΔE = -0.799 J (Energy lost as heat)
3. ΔE = -0.799 J (Change in energy)

Note: The negative sign indicates that the ball loses energy as heat.

Hopefully someone double checks what I am going to tell you, but here it goes:

Work=KE

W=∆KE=Kf-Ki= 1/2(0.17kg)(4.7m/s)-1/2(0.17)(3.7m/s)=1/2(0.17kg)[4.7m/s-3.7m/s]

heat=q=∆KE, which is lost due to the billboard rolling down the pool table. What is the difference between the KEintial and KEfinal?

∆E=q + W

I am not sure about what I have told you, but someone will check and hopefully give you the correct response if I am wrong.

Read the question wrong

Work=KE

W=∆KE=Kf-Ki= 1/2(0.17kg)(0m/s)-1/2(0.17)(4.7m/s)=1/2(0.17kg)[0 m/s-4.7m/s]

**Note: Work will be negative once you calculate it.

heat=q=∆KE, which is lost due to the billboard rolling down the pool table. What is the difference between the KEintial and KE once it slows down.

q=∆KE=Kf-Ki= 1/2(0.17kg)(3.7m/s)-1/2(0.17)(4.7m/s)=1/2(0.17kg)[3.7m/s-4.7m/s]

**Note: q will also be negative, KE is lost meaning heat is loss

∆E=q + W