A 2.80 kg steel ball strikes a massive wall at 12.8 m/s at an angle of alpha = 50.9° with the plane of the wall. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.156 s, what is the magnitude of the average force exerted on the ball by the wall?
To find the magnitude of the average force exerted on the ball by the wall, 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 calculate 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 (p_initial) is:
p_initial = mass * velocity_initial
Where mass is 2.80 kg and velocity_initial is 12.8 m/s.
Next, we need to calculate the final momentum of the ball after it bounces off the wall. The final momentum (p_final) can be calculated using the same formula as before, but with the final velocity (velocity_final) instead of the initial velocity.
p_final = mass * velocity_final
Since the ball bounces off with the same speed and angle, the magnitude of the final velocity will be equal to the magnitude of the initial velocity. Therefore, the magnitude of the final velocity (velocity_final) is also 12.8 m/s.
Now, according to the impulse-momentum principle, the change in momentum (Δp) is equal to the impulse (J) applied to the ball. In equation form:
Δp = J
Since momentum is a vector quantity, we will consider the direction of the momentum change due to the angle of incidence (alpha). The impulse can be calculated using the formula:
J = Δp = p_final - p_initial
To calculate the impulse, we need to find the change in momentum by subtracting the initial momentum from the final momentum. However, we need to consider the direction of the momentum change due to the angle of incidence.
To account for the angle of incidence, we can use the trigonometric property that says the momentum change in the direction perpendicular to the wall is zero, and the momentum change in the direction parallel to the wall is equal to the total momentum change.
In this case, the momentum change in the direction perpendicular to the wall is zero, and the momentum change in the direction parallel to the wall is equal to the magnitude of the momentum change.
Therefore, the change in momentum (Δp) is equal to the magnitude of the momentum change:
Δp = |p_final - p_initial|
Finally, to find the magnitude of the average force exerted on the ball by the wall, we divide the impulse (J) by the contact time (Δt):
Average force = J / Δt
Given that the contact time (Δt) is 0.156 s, we can now calculate the magnitude of the average force.