# Finding Impulse

1. Nov 18, 2008

### swede5670

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

A 110 kg fullback is running at 4.4 m/s to the east and is stopped in 0.75 s by a head-on tackle by a tackler running due west.
A. Calculate the impulse exerted on the fullback.
B. Calculate the impulse exerted on the tackler.
C. Calculate the average force exerted on the tackler.

2. Relevant equations
Fnet*delta time = change in momentum

3. The attempt at a solution
To solve A I set up the equation like this
.75X= 110 * 0 + 110 * 4.14
.75x=484
divide by .75 and you get
645.33 Ns

This is not right but I'm not sure why. Did I calculate the change in momentum incorrectly?

I'm not sure how to find the impulse on the tackler because I don't have his mass given to me and I don't have his velocity .

2. Nov 18, 2008

### Staff: Mentor

What you need is the impulse, not the force. Impulse = FΔt = Δ(mv).

Δ(mv) = mv(final) - mv(initial).

Note that momentum and impulse are vectors, so direction counts.

Consider Newton's 3rd law.

3. Nov 19, 2008

### swede5670

Impulse and Conservation of Momentum

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

A 110 kg fullback is running at 4.4 m/s to the east and is stopped in 0.75 s by a head-on tackle by a tackler running due west.
A. Calculate the impulse exerted on the fullback.
B. Calculate the impulse exerted on the tackler.
C. Calculate the average force exerted on the tackler.

2. Relevant equations
Fnet*delta time = change in momentum

3. The attempt at a solution
To solve A I set up the equation like this
.75X= 110 * 0 + 110 * 4.14
.75x=484
divide by .75 and you get
645.33 Ns

This is not right but I'm not sure why.

I got this help, but it didn't clarify what my problem was

I know that I need impulse and not force, do I have to use an equation like this before I can find the impulse?
(Change in P of fullback before) + (Change in P of cornerback) = (Change in P after the event of fullback) + (change in P after the event cornerback)

I really need someone to tell me what I'm doing wrong, I just don't see whats incorrect.

4. Nov 19, 2008

### swede5670

Right but I thought that this: .75X= 110 * 0 + 110 * 4.14=Impulse = FΔt = Δ(mv)
The .75 is delta t
the X is the force
the change in momentum is (110 *0) + (110*4.14)

I don't see what I'm doing wrong, that is the whole problem

5. Nov 19, 2008

### Staff: Mentor

Re: Impulse and Conservation of Momentum

The problem is that you are solving for X for some reason. X would be the average force, not the impulse.

Explain what that equation is that you set up and what X stands for. Compare it to the equation that I gave you for impulse.

(Also: Do not create multiple threads for the same problem.)

6. Nov 19, 2008

### Staff: Mentor

This is almost exactly right, the only change I would make is to subtract--not add--the initial and final momentums to find the change.
You are solving for force, but all you need for A and B is the impulse, not the force.

7. Nov 19, 2008

### swede5670

Sorry, I was worried that the thread was too far down for it to be seen again.

I don't seem to quite get it, I know that X is equal to force but isn't the force equal to impulse? Impulse = FΔt = Δ(mv).
Do I need to multiply the Force that I find by the change in time?
.75X = -484

8. Nov 19, 2008

### Staff: Mentor

No. Force and impulse are two different things.
That's what impulse equals.
No, don't do that. (It's a waste of effort.)

In general, there are two ways to find the impulse, both of which are contained in the equation above:

(1) You can take the force and multiply it by the time, using the definition of impulse. But in this problem you are given the time, but not the force. So this approach won't help.

(2) You can find the change in momentum. That's the one you need here.

9. Nov 20, 2008

### swede5670

This is fairly frustrating, I have no idea what to do.

Well this is what I think the change in momentum is (the problem is that I have no idea if this is right):

(M1*V1) + (M2*V2)
(110 x 4.4) + (M2*V2) = change in P
so I have 484 + M2V2 = delta P
how do I find the tacklers mass or velocity? Do I need the conservation problem?

(M1*V1) + (M2*V2) = V(m1+m2)
110*4.4 + M2 * V2 = 0 ( 110 + m2)

I'm reasoning that since after the tackle the speed is zero so the final velocity of the system will be zero
Well then I'm back where I started
M2V2 = -484

I really need this for tomorrow, and I have no clue what I'm doing. How do I incorporate the time hes stopped in, into the problem? I feel pretty confused, I'm not sure where I need to go next.

10. Nov 20, 2008

### swede5670

Wow that was an infuriating struggle for such a simple answer. I figured out that the only thing that was wrong with my answer was the negative sign. I tried -484 instead of 484, I'm not sure why it has to be positive though. Could it be negative?

11. Nov 20, 2008

### Staff: Mentor

Here's how I would do part A: To find the impulse exerted on the fullback, just find his change in momentum. Let's call the east direction to be positive.
His initial momentum = (m)(Vi) = (110)(4.4) = 484 kg-m/s.
His final momentum = (m)(Vf) = (110)(0) = 0

The change in momentum is final minus initial: 0 - 484 = -484 kg-m/s. That just means the impulse on him is 484 kg-m/s going west (since a minus means the direction is opposite east). For the answer, they probably just want the magnitude of the impulse, which is just 484 kg-m/s without a minus sign.

So it's much easier than you first thought.

Hint for part B: Don't do any calculations at all to solve this. Consider Newton's 3rd law and the definition of impulse = FΔt.

Hint for part C: Now it's time to make use of the time that they gave you and calculate the average force F. You already know the impulse, just use impulse = FΔt to solve for F.

12. Nov 21, 2008

### swede5670

Thanks a lot for the help, I've got it now.