# Newton's First Law of Motion

1. Sep 20, 2004

### BlackMamba

I don't understand how to even begin with this problem.

Here it is: An arrow, starting from rest, leaves the bow with a speed of 26.0 m/s. If the average force exerted on the arrow by the bow were doubled, all else remaining the same, with what speed would the arrow leave the bow?

I do know:

Vo = 0
V = 26.0

That's pretty much it. How do I move on from here?

2. Sep 20, 2004

### Tide

The force time the distance over which the bow acts on the arrow is the kinetic energy of the arrow when it is released.

3. Sep 20, 2004

### arildno

You should use:
$$\int_{t_{0}}^{t_{1}}Fdt=\bigtriangleup{mv}$$

4. Sep 20, 2004

### Tide

How would you determine the time interval over which to integrate? If you change the force then the time interval of the acceleration necessarily changes too.

5. Sep 20, 2004

### arildno

You've been given that ALL ELSE is unchanged!
(This would be the time&space intervals)
Now, let's look at the kinetic energy equation:
$$\int_{t_{0}}^{t_{1}}Fvdt=\frac{1}{2}v_{f}^{2}$$
($$v_{f}$$ final velocity)
By the mean value-theorem, we may rewrite this as:
$$v(t*)\int_{t_{0}}^{t_{1}}Fdt=\frac{1}{2}v_{f}^{2}$$
($$t_{0}\leq{t*}\leq{t}_{1}$$)

Rewritting in terms of average force, we get:
$$v(t*)\hat{F}(t_{1}-t_{0})=\frac{1}{2}v_{f}^{2}$$

Now, you cannot conclude that t* is the same value in the two cases.
(Or for that matter v(t*))

6. Sep 21, 2004

### arildno

BTW, you criticized my use of the same time interval; I might criticize your approach for using the same space interval.
As I look more at it, I think the exercise is decidedly unclear.

Possibly, they think of modeling the instantaneous force as kx (or 2kx in the second case); if that is true your approach is possibly what they're after.

7. Sep 21, 2004

### Tide

I didn't criticize. I asked a question and noted a property of bows and arrows which, presumably, one would have to know in order to address a problem relating to bows and arrows. Namely, the distance over which the bow accelerates the arrow is fixed. The time is not.

I agree the statement of the problem is somewhat ambiguous but, more than that, it is in error since it is impossible for ALL ELSE to remain unchanged. If both the distance and time interval over which acceleration occurs remain the same then the force cannot be doubled.

8. Sep 21, 2004

### arildno

I agree with you, Tide.
I'd like to apologize for using the word "criticize"; the Norwegian "equivalent" is almost neutral, it seems that it is rather more negatively charged in English than I thought.

As for the problem:
As far as I can see, the only way to put some sense in the wording, is that the bowstring is replaced with a stiffer one

For example, with the same displacement of the string, the potential energy would increase.
However, pulling the (new) string further back might keep the actual time interval constant.

In other words, a poorly stated problem..

9. Sep 21, 2004

### Tide

You're right. In English the verb "to criticize" has a slightly different meaning than the more suitable "to question." I wasn't offended by your use of the the word but I just wanted to clarify.

In any case, we do agree on the ambiguity of the problem! Perhaps the original poster can lend some insight based on his or her familiarity with the textbook of professor.

10. Sep 21, 2004

### Leong

$$\begin{multline*} \begin{split} &Newton's\ 2nd\ Law\\ &\sum \vec{F}=m\vec{a}\\ &F=m*(\frac{v-u}{t})\\ &F=m*(\frac{26}{t})\\ &F'=m*(\frac{v'-u'}{t})\\ &F'=2F\\ &m*(\frac{v'-u'}{t})=2*m*(\frac{26}{t})\\ &(\frac{v'-0}{t})=2*(\frac{26}{t})\\ &v'=2*26=52\ m/s \\ \end{split} \end{multline*}$$

11. Sep 21, 2004

### Tide

Leong,

I don't think you know the times you put in for F and F' are the same.

12. Sep 22, 2004

### Leong

i thought the question said 'all remaining the same' ? can i put this the question this way : if the force is doubled, what is the final velocity of the bow after time t which is the same as the first case ?

13. Sep 22, 2004

### Tide

Leong,

All else cannot remain the same - either the acceleration time changes or how far back you pull the string changes or a combination of both if the average force is to double. That's why Arildno and I concluded it the problem is flawed.

14. Sep 22, 2004

### Leong

opps, i am wrong !