Consider the line passing through the center of the cue ball as the x axis. At the point of collision with the solid ball on the table, a line passing through the centers of the cue ball and the solid ball form a 45° angle with the X axis. Assume no external forces acting on the system and both balls have equal masses. At what angle will the cue ball move after the collision?

it’s 315

To determine the angle at which the cue ball will move after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.

1. Conservation of momentum: The total momentum before the collision is equal to the total momentum after the collision.
- Before the collision: The cue ball is moving toward the solid ball, while the solid ball is stationary. Thus, the initial momentum of the system is only due to the cue ball.
- After the collision: Both balls will be moving. Let's represent the angle at which the cue ball moves after the collision as θ.

2. Conservation of kinetic energy: The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
- Before the collision: The cue ball has some initial velocity, while the solid ball is stationary, so only the cue ball has kinetic energy.
- After the collision: Both balls will have kinetic energy.

Now let's calculate the angles using the given information:

1. Conservation of momentum:
- Before the collision: Momentum of the system = mass of cue ball × initial velocity of cue ball = m × V (assuming mass = m)
- After the collision: Momentum of the system = (mass of cue ball × velocity of cue ball) + (mass of solid ball × velocity of solid ball)
= m × Vθ (since the solid ball is initially stationary, its velocity is 0)

Since the total momentum before and after the collision is equal, we can write:
m × V = m × Vθ

2. Conservation of kinetic energy:
- Before the collision: Kinetic energy of the system = (1/2) × mass of cue ball × (initial velocity of cue ball)^2 = (1/2) × m × V^2
- After the collision: Kinetic energy of the system = (1/2) × (mass of cue ball × (velocity of cue ball)^2) + (1/2) × (mass of solid ball × (velocity of solid ball)^2)
= (1/2) × m × V^2θ^2 (since the solid ball is initially stationary, its velocity is 0)

Since the total kinetic energy before and after the collision is equal, we can write:
(1/2) × m × V^2 = (1/2) × m × V^2θ^2

Now let's solve these equations to find the angle θ:

From the conservation of momentum equation:
V = Vθ

From the conservation of kinetic energy equation:
V^2 = V^2θ^2

Taking the square root of both sides:
V = Vθ

Since the velocities are equal, we can conclude that the angle at which the cue ball will move after the collision is 45°.

To determine the angle at which the cue ball moves after the collision, we need to understand the concept of conservation of momentum and conservation of kinetic energy in an elastic collision.

In an elastic collision between two objects, the total momentum and total kinetic energy of the system are conserved.

Let's denote the initial velocity of the cue ball as "v1" and the initial velocity of the solid ball as "v2". Since the balls have equal masses and no external forces are acting, we can assume that the magnitude of their initial velocities is the same.

After the collision, let's assume the final velocity of the cue ball is "v1f" and the final velocity of the solid ball is "v2f". We want to find the angle at which the cue ball moves after the collision, so let's call it "θ".

To solve for the angle θ, we need to break down the velocities into their x and y components.

Considering the line passing through the center of the cue ball as the x-axis, we can resolve the initial and final velocities of the cue ball and solid ball into their x-components and y-components using trigonometry:

v1x = v1 * cos(45°)
v1y = v1 * sin(45°)

v2x = v2 * cos(45°)
v2y = v2 * sin(45°)

v1fx = v1f * cos(θ)
v1fy = v1f * sin(θ)

v2fx = v2f * cos(45°) (assuming the solid ball moves in the x-axis direction after the collision)
v2fy = v2f * sin(45°) (assuming the solid ball moves in the y-axis direction after the collision)

Since momentum is conserved, we have:

m * v1x + m * v2x = m * v1fx + m * v2fx (in the x-axis direction)
m * v1y + m * v2y = m * v1fy + m * v2fy (in the y-axis direction)

Simplifying the equations by dividing both sides by m, we get:

v1x + v2x = v1fx + v2fx
v1y + v2y = v1fy + v2fy

Substituting the trigonometric expressions for x and y components, we have:

v1 * cos(45°) + v2 * cos(45°) = v1f * cos(θ) + v2f * cos(45°)
v1 * sin(45°) + v2 * sin(45°) = v1f * sin(θ) + v2f * sin(45°)

Now, since we assume no external forces act on the system, the total kinetic energy of the system is conserved:

(1/2) * m * v1^2 + (1/2) * m * v2^2 = (1/2) * m * v1f^2 + (1/2) * m * v2f^2

Simplifying, we get:

v1^2 + v2^2 = v1f^2 + v2f^2

Using the above equations, we can now solve for the angle θ at which the cue ball moves after the collision. By substituting the values of v1, v2, v1fx, v1fy, v2fx, and v2fy, we can solve the equations to find the angle θ.

Please provide specific values for v1 and v2 if you would like to proceed with the calculations to find the angle θ.

Momentum has to be conserved. So the cue ball has to go the direction such as to cancel the momentum to the 45 degree side of the first ball.

Most likely, the cue ball will go to the other side of the x axis at 45 degrees. That makes momentum conserved.