A 3.56-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 0.180 s, what is the average force exerted by the wall on the ball?
Well, this situation sounds like quite the ball-game! Let's see if we can figure out the average force the wall exerts on the ball.
Given that the ball is in contact with the wall for 0.180 s, we can start by finding the change in momentum of the ball during that time. Momentum, as you may know, is the product of mass and velocity.
The mass of the ball is 3.56 kg, and it strikes the wall at a speed of 10.0 m/s. Since the velocity is at an angle θ = 60.0° with the plane of the wall, we can break it down into its x and y components.
The x-component of the velocity is v_x = v * cos(θ), and the y-component is v_y = v * sin(θ). Plugging in the values, we get v_x = 10.0 m/s * cos(60.0°) and v_y = 10.0 m/s * sin(60.0°).
Now, when the ball bounces off the wall, it retains the same speed and angle. So, the change in momentum is equal to twice the initial momentum.
The initial momentum in the x-direction is p_initial_x = mass * v_x, and the initial momentum in the y-direction is p_initial_y = mass * v_y.
Therefore, the change in momentum in the x-direction is Δp_x = 2 * p_initial_x, and the change in momentum in the y-direction is Δp_y = 2 * p_initial_y.
Now that we have the change in momentum in both directions, we can use the formula for average force, which is F_average = ∆p / ∆t.
In our case, ∆p = sqrt(Δp_x^2 + Δp_y^2), and ∆t = 0.180 s.
Plugging in the values, we can calculate the average force exerted by the wall on the ball. But hey, if you're looking for a simpler answer, let's just say it's like being slapped by a steel octopus with a sense of humor. Ouch!
To find the average force exerted by the wall on the ball, we will use the impulse-momentum principle. Here are the steps to solve the problem:
Step 1: Calculate the initial momentum of the ball before it strikes the wall.
The initial momentum, p_initial, is given by the formula:
p_initial = m * v_initial
where m is the mass of the ball and v_initial is its initial velocity.
Given:
m = 3.56 kg (mass of the ball)
v_initial = 10.0 m/s (initial velocity of the ball)
Substituting the values:
p_initial = 3.56 kg * 10.0 m/s
p_initial = 35.6 kg·m/s
Step 2: Calculate the final momentum of the ball after it bounces off the wall.
The final momentum, p_final, can be calculated using the formula:
p_final = m * v_final
where v_final is the final velocity of the ball after bouncing off the wall. Since the ball's speed and angle of projection remain the same, v_final will also be 10.0 m/s.
Substituting the values:
p_final = 3.56 kg * 10.0 m/s
p_final = 35.6 kg·m/s
Step 3: Calculate the change in momentum of the ball.
The change in momentum, Δp, is given by the formula:
Δp = p_final - p_initial
Substituting the values:
Δp = 35.6 kg·m/s - 35.6 kg·m/s
Δp = 0 kg·m/s
Step 4: Calculate the average force exerted by the wall on the ball.
The average force, F_avg, can be calculated using the formula:
F_avg = Δp / Δt
where Δt is the time for which the ball is in contact with the wall.
Given:
Δt = 0.180 s (time for which the ball is in contact with the wall)
Substituting the values:
F_avg = 0 kg·m/s / 0.180 s
F_avg = 0 N
Therefore, the average force exerted by the wall on the ball is 0 Newtons.
To find the average force exerted by the wall on the ball, we can use the impulse-momentum theorem.
The impulse-momentum theorem states that the change in momentum of an object is equal to the force applied to it multiplied by the time interval over which the force is applied.
The change in momentum is given by:
Δp = m(vf - vi)
Where:
Δp is the change in momentum
m is the mass of the object (3.56 kg in this case)
vf is the final velocity of the object (after bouncing off the wall)
vi is the initial velocity of the object (before hitting the wall)
In this case, the ball bounces off the wall with the same speed and angle, so the final velocity (vf) has the same magnitude (10.0 m/s) and is in the opposite direction of the initial velocity (vi).
So, vf = -10.0 m/s
Now, let's calculate the initial velocity (vi). Since the ball strikes the wall at an angle of 60.0° with the plane of the wall, we need to find the horizontal and vertical components of the initial velocity.
The horizontal component of the initial velocity (vix) can be found using the equation:
vix = vi * cos(θ)
Where:
θ is the angle with the plane of the wall (60.0°)
So, vix = vi * cos(60.0°)
The vertical component of the initial velocity (viy) can be found using the equation:
viy = vi * sin(θ)
Where:
θ is the angle with the plane of the wall (60.0°)
So, viy = vi * sin(60.0°)
Since the ball hits the wall, the vertical component of the initial velocity is zero (viy = 0).
Using the above equation, we can find the initial velocity (vi):
viy = vi * sin(60.0°)
0 = vi * sin(60.0°)
vi = 0 / sin(60.0°)
vi = 0
Therefore, the initial velocity of the ball (vi) is zero.
Now, let's substitute the values into the formula for the change in momentum:
Δp = m(vf - vi)
Δp = 3.56 kg * (-10.0 m/s - 0)
Δp = 3.56 kg * (-10.0 m/s)
Now, we need to determine the time interval over which the force is applied. In this case, it is given as 0.180 s.
Let's calculate the average force (F) using the impulse-momentum theorem:
F = Δp / Δt
Where:
Δt is the time interval (0.180 s)
Substituting the values:
F = (3.56 kg * (-10.0 m/s)) / 0.180 s
Now we can solve for F:
F = (-35.6 kg·m/s) / 0.180 s
F ≈ -198.9 N
Therefore, the average force exerted by the wall on the ball is approximately -198.9 N. The negative sign indicates that the force is directed opposite to the motion of the ball.