How Is Maximum Force Calculated from Impulse in a Bouncing Ball Scenario?

[merzperson]

Homework Statement

A 200 g ball is dropped from a height of 1.9 m, bounces on a hard floor, and rebounds to a height of 1.5 m. The figure shows the impulse received from the floor.
What maximum force does the floor exert on the ball?

Homework Equations

Kinetics:
x-x_o = v_ot + 0.5at²
v₂ = v₀² + 2a(x - x_o)
v = v_o + at

Momentum and Force:
SF*dt = dP = m*dv

The Attempt at a Solution

What I did was calculate the velocity of the ball just as it hit the ground (-6.10m/s) and just as it rebounded (5.42m/s).
Then, using these I found the dv (5.42 + 6.10 = 11.52m/s) and plugged that into the momentum equation:
dP = m * dv
dP = 0.2 * 11.52 = 2.304 = SFdt

From here I don't know how to calculate the maximum force, as I know virtually nothing about integrals. Also, my work so far is probably wrong somewhere so some help would be greatly appreciated! :)

 

[ahmadmz]

I'm in mechanics and i have not seen a question about max force. But i'll try this. From the graph we can see the Impulse which is the area under the curve is (1/2)(0.005s)F_max
This is equal to the change in momentum which we can calculate.

So F_max = 2*(p2-p1)/0.005

Where p2-p1 = m(v2-v1)

 

[kerimek]

I would like to clarify that impulse and force are closely related but not the same. Impulse is the change in momentum of an object, while force is the cause of that change in momentum. In this situation, the floor exerts a force on the ball, causing it to change its momentum.

To calculate the maximum force exerted by the floor on the ball, we can use the impulse-momentum theorem: impulse = change in momentum. In this case, the impulse is equal to the change in momentum of the ball as it hits the floor and rebounds.

First, we need to calculate the initial momentum of the ball before it hits the floor. This can be done using the equation p = mv, where p is momentum, m is mass, and v is velocity. Since the ball is dropped from a height of 1.9 m, its initial velocity is 0 m/s. Therefore, the initial momentum is 0 kg*m/s.

Next, we need to calculate the final momentum of the ball after it rebounds. This can be done using the same equation, but with the velocity of the ball after rebounding. From the given information, we know that the ball rebounds to a height of 1.5 m with a velocity of 5.42 m/s. Therefore, the final momentum is 0.2 kg * 5.42 m/s = 1.084 kg*m/s.

Now, we can calculate the impulse using the equation I = Δp = pf - pi, where I is impulse, pf is final momentum, and pi is initial momentum. Therefore, the impulse is 1.084 kg*m/s.

Finally, we can calculate the maximum force exerted by the floor using the equation F = I/Δt, where F is force and Δt is the time interval during which the impulse occurs. Since the ball is in contact with the floor for a very short time, we can assume that Δt is very small, almost instantaneous. Therefore, the maximum force exerted by the floor on the ball is approximately equal to the impulse, which is 1.084 kg*m/s.

In conclusion, the maximum force exerted by the floor on the ball is approximately 1.084 kg*m/s. It is important to note that this is an idealized calculation and may not accurately represent the actual force exerted due to factors such as air resistance and the elasticity of the ball and floor.