# Using work on particle system

If $\vec{a}$ and $\vec{v}$ is the acceleration and velocity of a particle system, and the external forces $\vec{F}_i$ are acting on the system, then:

$$\int_{\vec{r}_1}^{\vec{r}_2} \sum_{i}\vec{F_i} \cdot d\vec{r} = \frac{m}{2}\left(\vec{v}_2^2-\vec{v}_1^2\right)$$

(Where $\vec{v}_1$ is the velocity at position $\vec{r}_1$, etc.)

To the question, my book asks about the velocity of the boxes when they have moved a certain distance (in this case $1\text{ m}$):

You can get that velocity in several ways. I solved it by just examining one of the blocks, but the problem with this approach is that you need to find an expression for the tension force, which involves solving a system of equations. Nevertheless, it worked.

However, in a PDF, they simply solved the problem using:

$$(20\text{ kg})(9.81\text{ m/s}^2)(1 \text{ m}) = \frac{1}{2}(24\text{ kg})v^2$$

This wasn't very well-motivated, but I assume that they're letting both boxes be the particle system, which means that the tension-forces are internal. This gives that $\sum_{i}\vec{F_i} = (20\text{ kg})(9.81\text{ m/s}^2)$, thus giving the equation above.

Now to my question: Judging by the initial equation I posted, it seems to me that this only is right if they assume that the velocity of the CM has the same magnitude as the speed of each individual box. I basically began examining this, assuming that $\vec{v} \equiv \dot{\vec{r}}_{CM}$:

$$\vec{r}_{CM} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}$$ $$\dot{\vec{r}}_{CM} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$$ $$v = \left| \dot{\vec{r}}_{CM} \right| = \left|\frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}\right|$$

To me, it doesn't seem like this formula simplifies to $v = |\vec{v}_1| = |\vec{v}_2|$ (even if you let $|\vec{v}_1| = |\vec{v}_2|$) which is why I'm rather confused by the whole thing.

Questions:

1. If the above formula doesn't simplify correctly, why does assuming that the center of mass moves at the same speed as the individual boxes work? What have I done wrong to reach my "contradicting" conclusion?
2. If the above formula does simplify correctly (in case I just don't see how), are there any general guidelines as to when you can make these simplifications?

• Would that not imply however that the center of mass is equal to the speeds of the individual masses? Even though it shouldn't be. The R.H.S. has a $\vec{v}^2 = |\vec{v}|^2$, which should be referring to the CM, so would expressing that in terms of $|\vec{v}_1|$ and $|\vec{v}_2|$ yield a R.H.S. that is $\frac{1}{2}m\vec{v}_{1,2}^2$? To me it seems impossible, which is what I cannot get my head around. Your interpretation of the equation makes more sense, but to me it seems to mathematically be saying another thing. – Max Feb 6 '17 at 19:14
• If the answer had been written as $(20\text{ kg})(9.81\text{ m/s}^2)(1 \text{ m}) = \frac{1}{2}(4\text{ kg})v^2+\frac {1}{2}(20\text{ kg})v^2$ would you have thought about centre of mass? – Farcher Feb 6 '17 at 19:20