Linear Impulse - Jackhamer Rebounds

[joemama69]

Homework Statement

During Operation the jack hammer develops on the concrete surface develops a force @ t=0 f=0, @ t=.2, F=90, @ t=.4, F=0. Graph looks like a triangle
To achieve this the 2lb spike S, is fired from rest into the surface @ 200ft/s. Determine the speed of the spike just after rebounding

Homework Equations

mv + E Fdt=mv

The Attempt at a Solution

Well it seems to me that the initial velocity is 0, and the final is 200ft. How do i know the rebound velocity.

 

[aimslin22]

impulse = F *(change in T)

impulse = mv

mv=F *(change in T)

 

[LowlyPion]

Homework Statement

During Operation the jack hammer develops on the concrete surface develops a force @ t=0 f=0, @ t=.2, F=90, @ t=.4, F=0. Graph looks like a triangle
To achieve this the 2lb spike S, is fired from rest into the surface @ 200ft/s. Determine the speed of the spike just after rebounding

Homework Equations

mv + E Fdt=mv

The Attempt at a Solution

Well it seems to me that the initial velocity is 0, and the final is 200ft. How do i know the rebound velocity.

First of all convert to SI units before things get out of whack.
200 f/s = 60.96 m/s
90 lbs force = 400.5 N
2 lbs = .907 kg

Now what is your impulse?

Δmv = F*Δt

But your time varies according to the graph. So your total impulse is the area under the function of Force_(t) that they give you.

With total Impulse then you have your Δmv.

 

[joemama69]

It is still not clicking

mass .907

is my initial velocity from rest, or is 60.96

My force is 400500 N at its peak,

i do not understand which t's to use.

 

[LowlyPion]

Just curious how is your force so high? Are there units to your problem that you haven't supplied? Is the peak force 90lbs or some other value? Your original statement doesn't really say.

 

[joemama69]

90 lbs force = 400.5 N

the Forve is 10^3 N = 400500 N

 

[LowlyPion]

90 lbs force = 400.5 N

the Forve is 10^3 N = 400500 N

And where does the multiplier come in?

Don't they give you the force as 90 already at t = .2 sec?

Because it looks to me like the area under the Force as a function of time that they give, is the area of a triangle, over the range of t = 0 to t = .4.
So ...

I = 1/2*(400.5 N) * .4sec = 80.1 N-s

That impulse is available to change the momentum of the .907 kg hammer.

Figure then the original momentum and subtract the impulse, and that should yield the return momentum and hence the return velocity shouldn't it?

 

[joemama69]

Momentum = P

P=mv = .907*60.96 = 55.301

P-I = -24.799 <-- Pretty sure I'm wrong but I am going to keep going

Return Momentum = RP

RP = mv = -24.799 = .907v therefore v = -27.341 m/s

 

[LowlyPion]

Momentum = P

P=mv = .907*60.96 = 55.301

P-I = -24.799 <-- Pretty sure I'm wrong but I am going to keep going

Return Momentum = RP

RP = mv = -24.799 = .907v therefore v = -27.341 m/s

You probably want to multiply by 3.281 to get back to feet / sec. don't you?

 

[joemama69]

yes, i would, so i take it i am correct

 

[joemama69]

Ok, one more question

For the graph, the units of force is 10^3 N

Which makes F = 400500 N

Do I use this number, it give a totaly different answer

I = 80100 N-s
P=55.301
V=-88234.17601 m/s

 

[LowlyPion]

Ok, one more question

For the graph, the units of force is 10^3 N

Which makes F = 400500 N

Do I use this number, it give a totaly different answer

I = 80100 N-s
P=55.301
V=-88234.17601 m/s

Is there more to this problem than you have provided? Is there a mass to the jackhammer that is more than just 2 lb? Is there a cross section of the tip to consider? These numbers don't seem reasonable to me.

 

[joemama69]

nope that's it

on the graph the yaxis is the force and the x-axis is t(ms)

am i correct based on the info i have provided

 

[LowlyPion]

nope that's it

on the graph the yaxis is the force and the x-axis is t(ms)

am i correct based on the info i have provided

Oh and time is ms. Well that makes the difference. Your F at 10³ and your time at 10^-3 then effectively cancel in magnitude and yield the same thing as though you were calculating with 90 lbs and .4 sec. The earlier answer is what I got.

Is it correct? I guess your grader is the judge.