Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have the following problem:

A point with mass $m$ and speed $v$ collides with a fixed obstacle and penetrates it, stopping in a space $\Delta x$. Calculate $\Delta t $.

I understand the way my book solves it: let $F_m$ be the mean value of the force during the collision. We have:

$$\frac{1}{2} mv^2 = F_m \Delta x \qquad mv =F_m \Delta t $$

which is easily solved for $\Delta t$. Although it's pretty obvious to intuition that $F_m$ must be the same in both equations, I was wondering if this is mathematically (and generally) correct. From the equations:

$$F_m=\dfrac{W}{\Delta x}=\dfrac{\int_{x_0}^{x_1}Fdx}{x_1-x_0}$$

On the other hand:

$$F_m=\dfrac {\Delta p}{\Delta t}=\dfrac {\int _{t_0}^{t_1}Fdt}{t_1-t_0}$$

So the mathematical statement behind this is:

$$\dfrac {\int _{t_0}^{t_1}Fdt}{t_1-t_0}=\dfrac{\int _{x_0}^{x_1} Fdx}{x_1-x_0}$$

(that one may prove to be generally false, see accepted answer) and also, if $v_1 \neq v_0 $:

$$\dfrac {W}{\Delta p}=\dfrac {\Delta x}{\Delta t}=v_m \iff v_m = \dfrac {v_0+v_1}{2} $$

Am I confused? any help is appreciated.

share|improve this question
1  
Hi Kazz8, and welcome to Physics Stack Exchange! Good question, but it would be better if you incorporate your edit into the main body of the question - don't add on an "edit" block at the end. –  David Z Feb 17 '13 at 12:51
add comment

1 Answer

up vote 0 down vote accepted

You're dealing with two different kinds of average force here: there's an average over time, and an average over position. In general, these are not going to be the same.

$$\begin{align}\frac{1}{x_1 - x_0}\int_{x_0}^{x_1} F(x)\mathrm{d}x &= \frac{1}{x_1 - x_0}\int_{x=x_0}^{x=x_1} F(x(t))\frac{\mathrm{d}x}{\mathrm{d}t}\mathrm{d}t\\ &= \frac{1}{v_m(t_1 - t_0)}\int_{t_0}^{t_1} F(t)v(t)\mathrm{d}t\end{align}$$

where $v_m = \frac{x_1 - x_0}{t_1 - t_0}$.

There are, however, two general circumstances in which the time average and the position average of $F$ are equal. One is if the velocity is constant; that's easy to prove by just plugging in $v(t) = v_m$. The other is if the force is constant, $F(t) = F_m$. In that case you have

$$\begin{align}\frac{1}{x_1 - x_0}\int_{x_0}^{x_1} F(x)\mathrm{d}x &= \frac{F_m}{v_m(t_1 - t_0)}\int_{t_0}^{t_1} v(t)\mathrm{d}t \\ \frac{1}{t_1 - t_0}\int_{t_0}^{t_1} F(t)\mathrm{d}t &= F_m\end{align}$$

showing that the averages will be equal if $v_m(t_1 - t_0) = \int_{t_0}^{t_1}v(t)\mathrm{d}t$. Both sides of this are equal to $x_1 - x_0$ for any (integrable) function $v(t)$.

Besides those two cases, it's possible that you could "accidentally" get the two sides to equal each other even if neither force nor velocity is constant during the process. But that would just be a numerical coincidence, and wouldn't have any physical significance. (At least, I don't think there are any other general cases, but I can't be sure right now.)

For a collision, your book is presumably taking the force to be constant (obviously the velocity can't be constant because the object stops). The reason this works is not because the collision takes a short time, but because the force is constant to a decent approximation. The equations I used here don't make any reference to a particular time scale, so the requirement for either $F$ or $v$ to be constant can't be bypassed by assuming a long or short time.

share|improve this answer
    
Thank you for your answer, made a lot of things clear. I also noted that in a similar problem from the same book, the author states exactly that the force is constant (to a decent approximation). Also thank you for your advice about editing. –  pppqqq Feb 17 '13 at 13:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.