How Do You Calculate the Impulse Exerted by the Ground on a Bouncing Ball?

[NasuSama]

Homework Statement

A rubber ball of mass m = 0.225 kg is dropped from a height ho = 5.92 m, and it rebounds to a height hf = 4.58 m. Assume there is no air friction.

Find J, the magnitude of the impulse exerted by the ground on the ball.

Homework Equations

I = Ft
Energy conservations formula: mgh or ½mv²?

The Attempt at a Solution

Ugh... Don't think I have the right type of approach. -__-

I first found that the energy lost is 2.96 J, which is right. I tried to use this form:

mgh = ½mv² + ΔE

But seems like there is no other way around.

 

[lewando]

Impulse is also defined as the change of momentum of a body.

 

[NasuSama]

Impulse is also defined as the change of momentum of a body.

Unclear of what you are trying to say here. Yes, I do get that impulse is same as the change of momentum, but how can I approach this problem? Is it just by using the impulse and momentum form or solve that problem using energy laws and then, the momentum rules? Still confused. Not helpful.

 

[lep11]

Hi,
On the other hand I=Δv*m so you need to calculate the speeds of the ball after dropping and after bouncing using the equation mgh=½mv^2. When the ball is dropped its potential energy changes into kinetic energy and after the bounce vice versa.
I hope this helps.

 

[NasuSama]

Still stumped. Still not sure how to approach this problem. You might want to give the reasons for such explanation. -__-

 

[Nytik]

Impulse is change in momentum. You can figure out the velocities from immediately before and after the ball bounces using suvat equations; from these you know the momentum before and after the ball bounces.
Difference is the impulse.

 

[lep11]

Still stumped. Still not sure how to approach this problem. You might want to give the reasons for such explanation. -__-

I suppose this was your homework but...
I=Δv*m=m(v_b-v_a), where v_b=velocity before the bounce
v_a=velocity after the bounce
h_before=5.92m
m=0.225kg h_after=4.58m

Let's solve for v
mgh=½mv^2 |:m |*2
2gh=v^2
v=√(2gh)
v_b=√(2gh_b)
... and v_a=√(2gh_a)

I=m*((√(2gh_b))-(√(2gh_a)))...

 

[NasuSama]

I suppose this was your homework but...
I=Δv*m=m(v_b-v_a), where v_b=velocity before the bounce
v_a=velocity after the bounce
h_before=5.92m
m=0.225kg h_after=4.58m

Let's solve for v
mgh=½mv^2 |:m |*2
2gh=v^2
v=√(2gh)
v_b=√(2gh_b)
... and v_a=√(2gh_a)

I=m*((√(2gh_b))-(√(2gh_a)))...

Still not working. Not sure how you prove this.

 

[lep11]

Still not working. Not sure how you prove this.

Why not working?

 

[NasuSama]

Why not working?

Even though your calculation is right, it seems that I obtain the wrong value. Yes, I did 0.292 J*s, but it's incorrect. Don't get quite why. Maybe, that is because you are determining the impulse of the ball.

 

[lewando]

Show your work.

 

[NasuSama]

I believe that is...

mgh_0 = ½mv0²
v0 = √(2gh_0)

½mvf² = mgh_f
vf = √(2gh_f)

That is by some kind of energy conservation.

Then, I guess that...

P = m√(2gh_f) - m√(2gh_0)

So there is negative impulse.

 

[bdh2991]

for right before v = -10.772, for right after v = +9.47...

change in momentum is (9.47)(.225) - (-10.772)(.225) - = 4.55 = impulse

 

[NasuSama]

bdh, you gave away the answer. Anywho, thanks.

 

[bdh2991]

I'm sorry you were on the right track and i wasn't patient enough but just try to remember momentum is a vector so the velocities will have a direction attached to them...