Average force during a ball's elastic bounce off of a plate

[Apashanka]

Homework Statement

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Homework Equations

For this problem I have calculated the average force as ∫ Fdt/ ∫dt
And using a(acceleration)=2h(height)/t² and calculating the numerator it comes out -2m(mass)h/t ,but what will be the limit of the intergral in that case ??
Of course the denominator will be ∫dt =0.5 sec.
Thank you

The Attempt at a Solution

 

[BvU]

To answer your question: from before the collision until after the collision -- same as for the denominator.
But actually you calculated (I hope) $\int F\; dt$ using $F = {dp\over dt}$ as $\int dp = \Delta p$

 

[Apashanka]

To answer your question: from before the collision until after the collision -- same as for the denominator.
But actually you calculated (I hope) $\int F\; dt$ using $F = {dp\over dt}$ as $\int dp = \Delta p$

I have obtained F(force)=f(t) and then intergrate it as ∫ f(t) dt ,but does not know how to use the limit .

 

[jbriggs444]

I have obtained F(force)=f(t) and then intergrate it as ∫ f(t) dt ,but does not know how to use the limit .

As @BvU points out, you do not need to integrate the the force over the interval of interest to obtain the change in momentum over the interval of interest. You can determine the change in momentum directly: $\Delta p = p_f - p_i$.

You say that:

a(acceleration)=2h(height)/t2 and calculating the numerator it comes out -2m(mass)h/t

Can you provide details? What is "a" here, what is "h" here, what is "t" here, what are you calculating and how are you justifying it?

 

[Apashanka]

As @BvU points out, you do not need to integrate the the force over the interval of interest to obtain the change in momentum over the interval of interest. You can determine the change in momentum directly: $\Delta p = p_f - p_i$.

You say that:

Can you provide details? What is "a" here, what is "h" here, what is "t" here, what are you calculating and how are you justifying it?

I have taken the acceleration from Newton's equation to be 2h(t)/t² since initial velocity is 0 where h(t) is distance traveled from the intial point of dropping after time t.
So in calculating the intergral ∫F(t)dt ,Previously I have taken the term h(t) outside the intergral which couldn't be.
So calculating ∫2mh(t)/t² over dt is now become more complicated .
But Δp can be calculated but if given after bouncing off elastically how much height it gains ,then the final momentum can be calculated
For which the avg force becomes Δp/.5sec

 

[jbriggs444]

I have taken the acceleration from Newton's equation to be 2h(t)/t² since initial velocity is 0 where h(t) is distance traveled from the intial point of dropping after time t.

So "h" is the distance below the drop point and "t" is the time since the drop. And you are computing acceleration as 2h(t)/t.

That is incorrect. The units do not even work out correctly. If you divide a distance by a time you get a result with units of a velocity.

You seem to be using $s=tv_{avg}=t\frac{h(t)}{2}$ and, hence, $v_{avg}=\frac{2h(t)}{t}$

 

[Apashanka]

I got it now it is 2p_i where p_i initial momentum imparted on the plate and calculating it 2p_i/.5sec the ans is 4N

h" is the distance below the drop point and "t" is the time since the drop. And you are computing acceleration as 2h(t)/t.

It's t² in the denominator from Newton's 3rd equation of motion

 

[jbriggs444]

Don't know how I missed seeing the square. So yes, dimensionally, $2\frac{h(t)}{t^2}$ comes out right. But without knowing t, that does not allow you to calculate a momentum.

 

[Apashanka]

To answer your question: from before the collision until after the collision -- same as for the denominator.
But actually you calculated (I hope) $\int F\; dt$ using $F = {dp\over dt}$ as $\int dp = \Delta p$

Thanks sir .
I am initially going through so much of complicated works .

 

[Apashanka]

Don't know how I missed seeing the square. So yes, dimensionally, $2\frac{h(t)}{t^2}$ comes out right. But without knowing t, that does not allow you to calculate a momentum.

Don't know how I missed seeing the square. So yes, dimensionally, $2\frac{h(t)}{t^2}$ comes out right. But without knowing t, that does not allow you to calculate a momentum.

The final velocity before hitting the plate can be calculated now and hence momentum ,for elastic collision momentum conservation also gives the final momentum after hitting .

 

[jbriggs444]

The final velocity before hitting the plate can be calculated now and hence momentum ,for elastic collision momentum conservation also gives the final momentum after hitting .

Yes, of course. PE at start => KE at impact => velocity at impact => momentum at impact. KE at impact => KE after rebound => velocity after rebound => momentum after rebound. Delta p divided by delta t to get average acceleration. Multiply by mass to get average net force. Then just one more tweak.

 

[neilparker62]

Alternate method:

Collision impulse for perfectly elastic collision: $$Δp = 2mΔv$$ with $$Δv=\sqrt{2gh}$$ Just divide by Δt for the force.