A 0.165kg cue ball with an inital speed of 2 m/s bounces off the rail in a game of pool. For x and y axses located the counce reverses the y component of the ball's velocity but does not alter the x component(the angle is 30 in the negative x axis and it bounces across the y axis into an unknown angle).

a) what is the theda ?
b) What is the change in the ball's linear momentum in unit vector notation?

what are the equations I would use? I don't understand what to do . I tries find theda by arctan(v_x/v_y) but I don't know what v_x and v_y are... Please help me.

To solve this problem, let's break it down into smaller steps:

Step 1: Understanding the scenario
We have a cue ball with an initial speed of 2 m/s. It bounces off a rail, reversing its y-component of velocity but leaving its x-component unchanged. The initial angle of the ball is 30 degrees in the negative x-axis direction.

Step 2: Determine the initial velocity components
The initial velocity of the cue ball can be represented in terms of its x and y components. Since the angle is 30 degrees in the negative x-axis direction, we can use trigonometry to find these components.

The initial velocity in the x-direction (v_x) can be calculated as:
v_x = |v| * cos(theta)

The initial velocity in the y-direction (v_y) can be calculated as:
v_y = |v| * sin(theta)

Given that the initial speed of the ball is 2 m/s, we substitute this value into the equations:
v_x = 2 * cos(30)
v_y = 2 * sin(30)

Calculating these values gives us:
v_x = 2 * 0.866 = 1.732 m/s
v_y = 2 * 0.5 = 1 m/s

Step 3: Determine the y-component of velocity after the bounce
According to the problem, the ball bounces off the rail, reversing its y-component of velocity. Therefore, the new y-component of velocity (-v_y') will be the negative of the initial y-component of velocity (-v_y).

v_y' = -v_y = -1 m/s

Step 4: Determine the angle after the bounce
To find the new angle (theta'), we can use the inverse tangent function (arctan). The tangent of an angle is the ratio of the opposite side (v_y') to the adjacent side (v_x), so:

theta' = arctan(v_y' / v_x)

Substituting the known values:
theta' = arctan(-1 / 1.732)

Using a calculator, we find that:
theta' ≈ -30.96 degrees

Therefore, the new angle after the bounce is approximately -30.96 degrees.

To calculate the Change in momentum, we can use the following equation:
Change in momentum = Final momentum - Initial momentum

Since the mass of the cue ball is given as 0.165 kg, we can focus on the velocity components.

Step 5: Calculate the change in linear momentum in unit vector notation
The initial momentum in the x-direction (p_x) is given by:
p_x_initial = m * v_x_initial

Substituting the values, we get:
p_x_initial = 0.165 kg * 1.732 m/s

Similarly, the initial momentum in the y-direction (p_y) is given by:
p_y_initial = m * v_y_initial

Substituting the values, we get:
p_y_initial = 0.165 kg * 1 m/s

The final momentum in the x-direction (p_x_final) remains the same as the initial momentum since the x-component of velocity is not altered.

Similarly, the final momentum in the y-direction can be calculated as:
p_y_final = m * v_y_final

Since the ball reverses its y-component of velocity, the final y-component of velocity (v_y_final) is -v_y'.

Substituting the values, we get:
p_y_final = 0.165 kg * (-1 m/s)

Calculating these values gives us:
p_x_initial ≈ 0.286 kg·m/s
p_y_initial ≈ 0.165 kg·m/s
p_x_final ≈ 0.286 kg·m/s
p_y_final ≈ -0.165 kg·m/s

Now, we can express the change in momentum in unit vector notation:

Change in momentum = (Δp_x , Δp_y)

The change in momentum in the x-direction (Δp_x) is given by:
Δp_x = p_x_final - p_x_initial

Substituting the known values, we get:
Δp_x ≈ 0.286 kg·m/s - 0.286 kg·m/s = 0 kg·m/s

Similarly, the change in momentum in the y-direction (Δp_y) is given by:
Δp_y = p_y_final - p_y_initial

Substituting the known values, we get:
Δp_y ≈ -0.165 kg·m/s - 0.165 kg·m/s = -0.33 kg·m/s

Therefore, the change in the ball's linear momentum in unit vector notation is approximately (0 kg·m/s, -0.33 kg·m/s).