In a game of pool, the cue ball strikes another ball initially at rest. After the collision, the cue ball moves at 4.15 m/s along a line making an angle of -25° (i.e. below the x-axis) with its original direction of motion (the x-axis), and the second ball has a speed of 2.846 m/s.
Find the angle between the direction of motion of the second ball and the original direction of motion of the cue ball. Assume the masses are equal.
Find the original speed of the cue ball.
Is kinetic energy conserved?
According to the law of conservation of linear momentum
vector(p)= vector(p1)+ vector(p2),
Let vector( p) is directed along +x-axis. Then vector (p1) is
below x-axis and vector(p2) is above it.
Projections on axes are
x: v = v1•cosα +v2•cosβ
y: 0 = - v1•sin α +v2•sinβ
sinβ = v1•sin α/v2 = 0.616.
β = arcsin 0.616 = 38 degr.
cos β = 0.788.
v = 4.15•0.906+2.846•0.788=6 m/s.
At elastic collision the kinetic energy is conserved.
To find the angle between the direction of motion of the second ball and the original direction of motion of the cue ball, we can use trigonometry. Let's break down the information given.
1. The cue ball's velocity after the collision is 4.15 m/s, and it moves along a line making an angle of -25° with its original direction of motion (the x-axis). We can represent this velocity as a vector v_1, where v_1 = 4.15 m/s at -25°.
2. The second ball's velocity after the collision is 2.846 m/s. We can represent this velocity as a vector v_2, where v_2 = 2.846 m/s.
Assuming the two balls have equal masses, we can apply the principle of conservation of momentum. In a collision between two objects, the total momentum before the collision is equal to the total momentum after the collision, provided no external forces act on the system.
Let's first find the original speed of the cue ball, before the collision. Since the masses are equal, we can equate the magnitudes of the balls' momenta before and after the collision.
Let's represent the unknown original speed of the cue ball as v. The magnitude of momentum can be calculated using the equation p = m * v.
Before the collision:
Cue ball momentum = m1 * v
Second ball momentum = m2 * 0 (since it is initially at rest)
After the collision:
Cue ball momentum = m1 * v_1
Second ball momentum = m2 * v_2
Since the masses are equal, we can simplify the equation as:
v = v_1 = m2 * v_2 / m1
Now let's calculate the original speed of the cue ball using the known values:
v = 2.846 m/s
Next, to find the angle between the direction of motion of the second ball and the original direction of motion of the cue ball, we can use vector subtraction. We want to find the angle θ between the vectors v_1 and v_2.
First, convert the vectors v_1 and v_2 into their x and y components.
v_1_x = v_1 * cos(-25°)
v_1_y = v_1 * sin(-25°)
Now, find the x and y components of the second ball's velocity.
v_2_x = v_2 * cos(θ_2)
v_2_y = v_2 * sin(θ_2)
Since the magnitude of the velocity vector is 2.846 m/s, we have:
(v_2_x)^2 + (v_2_y)^2 = (2.846 m/s)^2
Substitute the components into the equation:
(v_2_x)^2 + (v_2_y)^2 = (2.846 m/s)^2
(v_2 * cos(θ_2))^2 + (v_2 * sin(θ_2))^2 = (2.846 m/s)^2
Simplify:
(v_2 * cos(θ_2))^2 + (v_2 * sin(θ_2))^2 = 8.118 m^2/s^2
Expand and simplify:
v_2^2 * (cos^2(θ_2) + sin^2(θ_2)) = 8.118 m^2/s^2
v_2^2 * (1) = 8.118 m^2/s^2
v_2^2 = 8.118 m^2/s^2
Take the square root of both sides to solve for v_2:
v_2 = √(8.118 m^2/s^2)
v_2 = 2.85 m/s (rounded to three decimal places)
Since v_2 = 2.846 m/s, the angle between the direction of motion of the second ball and the original direction of motion of the cue ball is approximately 0°.
Now, let's determine whether kinetic energy is conserved. In an elastic collision, where no energy is lost, the total kinetic energy before the collision should be equal to the total kinetic energy after the collision.
The formula for kinetic energy is KE = 0.5 * m * v^2, where m is the mass and v is the velocity.
Before the collision:
Cue ball kinetic energy = 0.5 * m * v^2
Second ball kinetic energy = 0.5 * m * 0^2 (since it is initially at rest)
After the collision:
Cue ball kinetic energy = 0.5 * m * v_1^2
Second ball kinetic energy = 0.5 * m * v_2^2
Since the masses are equal, we can simplify the equation as:
0.5 * v^2 = 0.5 * v_1^2 + 0.5 * v_2^2
Let's substitute the values to calculate the kinetic energy:
0.5 * v^2 = 0.5 * (4.15 m/s)^2 + 0.5 * (2.846 m/s)^2
Calculate the kinetic energy before and after the collision and compare the values. If they are equal, kinetic energy is conserved. However, if they are not equal, kinetic energy is not conserved.
To find the angle between the direction of motion of the second ball and the original direction of motion of the cue ball, we can use the concept of conservation of momentum.
Let's denote the angle between the original direction of motion of the cue ball and the x-axis as θ1, and the angle between the direction of motion of the second ball and the x-axis as θ2.
According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Step 1: Find the x and y components of the velocity of the cue ball after the collision.
Given:
Initial speed of the cue ball (v1) = 4.15 m/s
Angle (θ1) between the initial direction of motion of the cue ball and the x-axis = -25°
Using trigonometry, we can find the x and y components of the velocity of the cue ball after the collision:
vx1 = v1 * cos(θ1)
vy1 = v1 * sin(θ1)
vx1 = 4.15 m/s * cos(-25°) = 3.743 m/s
vy1 = 4.15 m/s * sin(-25°) = -1.791 m/s (Negative because it is below the x-axis)
Step 2: Find the x and y components of the velocity of the second ball after the collision.
Given:
Final speed of the second ball (v2) = 2.846 m/s
Since the second ball was initially at rest, its x-component velocity after the collision will be the same as the cue ball's x-component velocity, i.e., vx2 = vx1 = 3.743 m/s.
To find the y-component velocity of the second ball (vy2), we need to use the equation of conservation of momentum:
Total momentum before collision = Total momentum after collision
(mass of cue ball) * (velocity of cue ball before collision) = (mass of second ball) * (velocity of second ball after collision)
Since the masses are equal, we can write:
(v1 * cos(θ1)) = (v2 * cos(θ2))
Substituting the given values:
(4.15 m/s * cos(-25°)) = (2.846 m/s * cos(θ2))
Simplifying the equation:
3.743 m/s = 2.846 m/s * cos(θ2)
Dividing both sides by 2.846 m/s:
cos(θ2) ≈ 3.743 m/s / 2.846 m/s ≈ 1.313
To find θ2, take the inverse cosine (cos^(-1)) of both sides:
θ2 ≈ cos^(-1)(1.313)
Using a calculator, θ2 ≈ 47.709°
Therefore, the angle between the direction of motion of the second ball and the original direction of motion of the cue ball is approximately 47.709°.
Now, let's move on to the second part of the question.
To find the original speed of the cue ball, we can use the equation of conservation of kinetic energy.
Kinetic energy before collision = Kinetic energy after collision
(0.5 * mass * (initial velocity of cue ball)^2) = (0.5 * mass * (velocity of cue ball after collision)^2) + (0.5 * mass * (velocity of second ball after collision)^2)
Since the masses are equal, we can simplify the equation:
(0.5 * (initial velocity of cue ball)^2) = (0.5 * (velocity of cue ball after collision)^2) + (0.5 * (velocity of second ball after collision)^2)
Plugging in the given values:
(0.5 * (initial velocity of cue ball)^2) = (0.5 * 4.15 m/s)^2 + (0.5 * 2.846 m/s)^2
Simplifying the equation:
(0.5 * (initial velocity of cue ball)^2) = 8.6104 m^2/s^2 + 4.0706116 m^2/s^2
(0.5 * (initial velocity of cue ball)^2) = 12.6800116 m^2/s^2
Dividing both sides by 0.5:
(initial velocity of cue ball)^2 = 25.3600232 m^2/s^2
Taking the square root of both sides:
initial velocity of cue ball ≈ √(25.3600232 m^2/s^2)
initial velocity of cue ball ≈ 5.03607 m/s
Therefore, the original speed of the cue ball is approximately 5.03607 m/s.
Finally, let's address the last part of the question.
To determine if kinetic energy is conserved, we compare the kinetic energy before and after the collision.
Before the collision:
Initial kinetic energy = (0.5 * mass * (initial velocity of cue ball)^2) = 0.5 * (initial velocity of cue ball)^2
After the collision:
Final kinetic energy = (0.5 * mass * (velocity of cue ball after collision)^2) + (0.5 * mass * (velocity of second ball after collision)^2) = 0.5 * (velocity of cue ball after collision)^2 + 0.5 * (velocity of second ball after collision)^2
Using the calculated values from earlier:
Initial kinetic energy = 0.5 * (5.03607 m/s)^2 ≈ 12.6800116 m^2/s^2
Final kinetic energy = 0.5 * (4.15 m/s)^2 + 0.5 * (2.846 m/s)^2 ≈ 12.6800116 m^2/s^2
The initial and final kinetic energies are equal, indicating that kinetic energy is conserved in this collision.
Therefore, the kinetic energy is conserved in this situation.