A ping pong ball (m= 6g) bounces off a table. Before striking the table the ball has a velocity of 4m/s in a direction pheta below the x-axis. It leaves the table with the same speed moving in a direction pheta above the x axis. What impulse does the table exert on the tennis ball. If the ball is in contact with the table for .o1 seconds what average force does the table exert on the ball?

I know I=p1-p2, but i don't understand how to deal with finding the angles

The impulse is dealing with the changed momentum, it only changes in the vertical. So, figure the vertical components>

v=4sinTheta in each case. One velocity is downward, one is upward. So subtracting them changes the sign, so they add.

deltamomentum=2mvsinTheta
Impulse = deltamomentum

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

Step 1: Finding the change in momentum (impulse):
In order to find the impulse exerted by the table on the ping pong ball, we need to find the change in momentum. Recall that momentum (p) is given by the product of mass (m) and velocity (v): p = m * v.

Given that the mass of the ping pong ball (m) is 6g (which is equal to 0.006 kg) and the initial velocity (v1) is 4 m/s, we can calculate the initial momentum (p1) by multiplying the mass by the magnitude of velocity: p1 = m * v1.

Similarly, after bouncing off the table, the ping pong ball leaves with the same speed but in the opposite direction. So, the final momentum (p2) is given by: p2 = m * v2, where v2 is the final velocity.

Since the magnitude of the velocity remains the same, the change in momentum (impulse, I) is given by: I = p1 - p2 = m * v1 - m * v2 = m * (v1 - v2).

Step 2: Finding the angles:
To handle the angles, we'll use trigonometry. Let's assume the angle below the x-axis is θ1, and the angle above the x-axis is θ2. Since the magnitudes of the velocities are the same, the angles are mirrored with respect to the x-axis.

Using the trigonometric relationships, we can express the vertical and horizontal components of the velocities as follows:
v1x = v1 * cos(θ1)
v1y = v1 * sin(θ1)

v2x = v2 * cos(θ2)
v2y = v2 * sin(θ2)

Step 3: Finding the final velocity and solving for the impulse:
Since the initial and final vertical velocities are equal in magnitude but opposite in direction, we have: v1y = -v2y.

Furthermore, since the magnitude of the velocity remains the same, we have: v1x = v2x.

We can rewrite the equations for the vertical and horizontal components of the velocities as follows:
v1 * sin(θ1) = -v2 * sin(θ2)
v1 * cos(θ1) = v2 * cos(θ2)

Solving these two equations simultaneously, we can find v2 in terms of θ1 and θ2:
v2 = v1 * (sin(θ1) / sin(θ2))

Now, substituting the expression for v2 into the change in momentum equation (I = m * (v1 - v2)), we can solve for the impulse.

Step 4: Calculating average force:
The average force (F) exerted by the table on the ping pong ball can be calculated using the equation: F = I / t, where t is the duration of contact between the ball and the table.

Given that t = 0.01 seconds (or 0.01 s), we can calculate the average force by dividing the impulse (I) by the contact time (t).

Note: Remember to convert all mass and time units to SI units (kilograms and seconds) for consistent calculations.

I hope this explanation helps you solve the problem!