A pool ball initially traveling 5 m/s strikes a second pool ball of identical mass in a totally elastic collision. After the collision the first ball has veered off its original path and is now traveling at 3 m/s. Find the speed and angle of each pool ball after the collision. Consider the initial ball to have been traveling at an angle of 0 degrees.
To find the speed and angle of each pool ball after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.
Let's denote the speed of the second pool ball after the collision as v2 and its angle as θ2. Since the second ball is initially at rest, we can write:
m1 * v1 = m1 * v1' + m2 * v2' (conservation of momentum)
where m1 and m2 are the masses of the pool balls, v1 and v2 are the initial velocities, and v1' and v2' are the final velocities.
In this problem, both pool balls have identical masses, so we can simplify the equation:
m * v1 = m * v1' + m * v2'
Since m is common to both terms, we can cancel it out:
v1 = v1' + v2' (equation 1)
The problem states that the initial ball veers off its original path, meaning its angle changes. Let's denote its final angle as θ1'. Since the initial angle is given as 0 degrees, we have:
θ1' = 0 (equation 2)
Now, let's consider the kinetic energy. The kinetic energy before the collision is given by:
KE1 = 0.5 * m * v1^2
The kinetic energy after the collision is given by:
KE1' = 0.5 * m * v1'^2 + 0.5 * m * v2'^2
Since we are dealing with a totally elastic collision, kinetic energy is conserved. Thus:
KE1 = KE1'
Plugging in the expressions for kinetic energy, we get:
0.5 * m * v1^2 = 0.5 * m * v1'^2 + 0.5 * m * v2'^2
Since we know the initial and final speeds of the first ball, we can plug in their values:
0.5 * m * (5 m/s)^2 = 0.5 * m * (3 m/s)^2 + 0.5 * m * v2'^2
Now, let's solve for v2':
0.5 * (5 m/s)^2 = 0.5 * (3 m/s)^2 + 0.5 * v2'^2
0.5 * 25 = 0.5 * 9 + 0.5 * v2'^2
v2'^2 = 25 - 9
v2'^2 = 16
v2' = 4 m/s
Now, let's determine the angle of the second pool ball after the collision, θ2. Since the second ball was initially at rest, its final angle will be the same as the angle at which the first ball veered off its path after the collision. Since the problem states that the initial angle is 0 degrees, we conclude:
θ2 = 0
Therefore, the speed and angle of each pool ball after the collision are:
First pool ball:
Speed: 3 m/s
Angle: 0 degrees
Second pool ball:
Speed: 4 m/s
Angle: 0 degrees