A 2.7 kg ball strikes a wall with a velocity of

7.6 m/s to the left. The ball bounces off with
a velocity of 5.2 m/s to the right.
If the ball is in contact with the wall for
0.23 s, what is the constant force exerted on
the ball by the wall?
Answer in units of N

change in momentum / time

= 2.7(7.6+5.2)/.23

To find the constant force exerted on the ball by the wall, we can use Newton's second law of motion, which states that the force (F) applied to an object is equal to its mass (m) multiplied by its acceleration (a). The formula is given as:

F = m * a

In this case, the ball is in contact with the wall for a specific duration of time (0.23 s) and undergoes a change in velocity. We can calculate the acceleration using the equation:

a = (Change in velocity) / (Time)

First, let's calculate the change in velocity. The initial velocity of the ball is 7.6 m/s to the left, and the final velocity is 5.2 m/s to the right. Since the direction has changed, we need to take the absolute value of both velocities before calculating the change.

Change in velocity = |final velocity - initial velocity|
= |5.2 m/s - (-7.6 m/s)|
= |12.8 m/s|

Therefore, the change in velocity is 12.8 m/s.

Next, we can calculate the acceleration:

a = (Change in velocity) / (Time)
= 12.8 m/s / 0.23 s
= 55.65 m/s^2

Now that we have the acceleration, we can calculate the force exerted on the ball using Newton's second law of motion:

F = m * a
= 2.7 kg * 55.65 m/s^2
= 150.4555 N

Therefore, the constant force exerted on the ball by the wall is approximately 150.46 N.