The kinetic energy of a rolling billiard ball is given by {\rm{KE}} = 1/2\,mv^2 . Suppose a 0.17-{\rm 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
Not a chemistry question. Someone in physics, please.
There is no question here. The statement that the KE of the ball is (1/2) M V^2 is incorrect if it is rolling, as it says it is. The KE of a rolling solid sphere is (7/10) M V^2.
I have never seen a rolling billard ball lose 22% of its speed while crossing a pool table.
Under these circumstqances, there is no properly posed question for us to answer.
To analyze the situation, we need to consider the conservation of energy and momentum.
1. First, let's find the initial kinetic energy (KE1) of the first billiard ball:
KE1 = 1/2 * mass * (initial velocity)^2
KE1 = 1/2 * 0.17 kg * (4.5 m/s)^2
KE1 = 1/2 * 0.17 kg * 20.25 m^2/s^2
KE1 = 1.71975 J
2. Next, let's find the final kinetic energy (KE2) of the first billiard ball:
KE2 = 1/2 * mass * (final velocity)^2
KE2 = 1/2 * 0.17 kg * (3.5 m/s)^2
KE2 = 1/2 * 0.17 kg * 12.25 m^2/s^2
KE2 = 1.30775 J
3. The change in kinetic energy (ΔKE) is given by:
ΔKE = KE2 - KE1
ΔKE = 1.30775 J - 1.71975 J
ΔKE = -0.412 J
The negative value indicates that the first billiard ball lost energy.
4. Now, let's consider the momentum conservation. The momentum before the collision (P_initial) is equal to the momentum after the collision (P_final) since there is no external force in the horizontal direction:
P_initial = P_final
5. The momentum (P) of an object is given by the product of its mass and velocity:
P_initial = mass1 * initial velocity1
P_initial = 0.17 kg * 4.5 m/s
P_initial = 0.765 kg·m/s
6. After the collision, the first billiard ball stops, so its final momentum is zero:
P_final = 0 kg·m/s
7. The second billiard ball acquires momentum and rolls away, so its final momentum is given by:
P_final = mass2 * velocity2
0.765 kg·m/s = mass2 * 3.5 m/s
mass2 = 0.765 kg / 3.5 m/s
mass2 = 0.219 kg
Therefore, the mass of the second billiard ball is 0.219 kg.