# Linear Impulse - Jackhamer Rebounds

1. Apr 9, 2009

### joemama69

1. The problem statement, all variables and given/known data
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

2. Relevant equations

mv + E Fdt=mv

3. 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.

2. Apr 9, 2009

### aimslin22

impulse = F *(change in T)

impulse = mv

mv=F *(change in T)

3. Apr 9, 2009

### LowlyPion

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

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.

4. Apr 13, 2009

### 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.

5. Apr 13, 2009

### 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.

6. Apr 13, 2009

### joemama69

90 lbs force = 400.5 N

the Forve is 10^3 N = 400500 N

7. Apr 13, 2009

### LowlyPion

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?

8. Apr 15, 2009

### joemama69

Momentum = P

P=mv = .907*60.96 = 55.301

P-I = -24.799 <-- Pretty sure i'm wrong but im gonna keep going

Return Momentum = RP

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

9. Apr 15, 2009

### LowlyPion

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

10. Apr 15, 2009

### joemama69

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

11. Apr 15, 2009

### 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

12. Apr 15, 2009

### LowlyPion

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.

13. Apr 15, 2009

### joemama69

nope thats 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

14. Apr 15, 2009

### LowlyPion

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.