A 1 kg steel ball strikes a wall with a speed of 9.99 m/s at an angle of 55.1 ◦ with the normal to the wall. It bounces off with the same speed and angle.

If the ball is in contact with the wall for 0.292 s, what is the magnitude of the average force exerted on the ball by the wall? Answer in units of N.

Ok, what is the change in momentum. Since the momentum parallel to the wall is the same, it does not change. So the momentum perpendicular (along the normal) is changed, it actually reverses, so one knows the change there, twice the initial momentum.

Momentum normal=mvcos55.1
Double that, is the change in momentum. then the Impluse must equal that.

Force*time=change in momentum.

I can confirm that the answer above is correct.

The impulse equals that, and then you divide the impulse by change in time (.292 s) in order to obtain your answer, the average force exerted on the ball.

To find the magnitude of the average force exerted on the ball by the wall, we can use Newton's second law of motion, which states that the force exerted on an object is equal to its mass multiplied by its acceleration.

Given:
- Mass of the steel ball = 1 kg
- Initial speed of the ball = 9.99 m/s
- Angle of incidence with the normal to the wall = 55.1 degrees
- Time of contact with the wall = 0.292 s

First, let's calculate the change in velocity of the ball during the collision.

The initial velocity vector of the ball can be split into two components:
- The perpendicular component (V_perp): V_initial * sin(angle)
- The parallel component (V_parallel): V_initial * cos(angle)

Since the ball bounces off with the same speed and angle, the magnitude of the perpendicular component of the final velocity after the collision is the same as the magnitude of the initial perpendicular component. So, V_perp_final = V_perp_initial = V_initial * sin(angle).

The change in the magnitude of the parallel component of the velocity is equal to two times the magnitude of the initial perpendicular component (assuming the collision is elastic):
ΔV_parallel = 2 * V_perp_initial = 2 * V_initial * sin(angle).

Now, we can calculate the change in velocity:
ΔV = sqrt((V_perp_final)^2 + (ΔV_parallel)^2).

Using the given values, we have:
ΔV = sqrt((V_initial * sin(angle))^2 + (2 * V_initial * sin(angle))^2).

Next, to calculate the average acceleration during the collision, we use the following formula:
Average acceleration = ΔV / Δt, where Δt is the time of contact with the wall.

Using the given value of Δt = 0.292 s, we can calculate the average acceleration.

Finally, using Newton's second law, we can determine the magnitude of the average force exerted on the ball:
Force = mass * average acceleration.

Plugging in the values into the formula, we can find the magnitude of the average force exerted on the ball by the wall, in units of Newtons (N).