A 2.50-kg steel ball strikes a massive wall at 10.0 m/s at an angle of θ = 60.0° with the plane of the wall. It bounces off the wall with the same speed and angle (see the figure below). If the ball is in contact with the wall for 0.224 s, what is the average force exerted by the wall on the ball? (Take to the right as the +x-direction and up as the +y-direction.)

To find the average force exerted by the wall on the ball, we can use the impulse-momentum principle. According to this principle, the change in momentum of an object is equal to the impulse applied to it.

First, let's find the initial momentum of the ball before it strikes the wall. The momentum of an object is given by the product of its mass and velocity. So, the initial momentum of the ball will be:

P1 = m * v1

Where:
m = mass of the ball = 2.50 kg
v1 = initial velocity of the ball = 10.0 m/s

Substituting the values into the equation, we get:

P1 = 2.50 kg * 10.0 m/s
P1 = 25 kg·m/s

Now, let's find the final momentum of the ball after it bounces off the wall. Since the ball bounces off the wall with the same speed and angle, the magnitude of its velocity remains the same. However, the direction is opposite. So, the final velocity of the ball will be:

v2 = -10.0 m/s

The final momentum of the ball will be:

P2 = m * v2

Substituting the values into the equation, we get:

P2 = 2.50 kg * -10.0 m/s
P2 = -25 kg·m/s

Next, let's calculate the change in momentum, which is equal to the impulse applied to the ball:

ΔP = P2 - P1

Substituting the values into the equation, we get:

ΔP = (-25 kg·m/s) - (25 kg·m/s)
ΔP = -50 kg·m/s

Now, we know that impulse is equal to the average force multiplied by the time interval over which it acts:

ΔP = Favg * Δt

Where:
Favg = average force exerted by the wall on the ball (what we need to find)
Δt = time interval = 0.224 s

Rearranging the equation, we can solve for Favg:

Favg = ΔP / Δt

Substituting the values into the equation, we get:

Favg = (-50 kg·m/s) / (0.224 s)

Calculating this, we find:

Favg ≈ -223.2 N

Therefore, the average force exerted by the wall on the ball is approximately -223.2 N. The negative sign indicates that the force is in the opposite direction as the positive x-axis, which is to the left.