Question

Ten particles are moving with the following speeds: four at$200m/s$, two at$500m/s$, and four at$600m/s$. Calculate their

a) Average speed

b) Rms speed

c) Is$v_{rms}>v_{avg}$?

Short answer

.

Step-by-step solution

Given data

• Four particles have speed$200\text{\hspace{0.17em}m}/\text{s}$
• Two particles have speed$500m/s$
• Four particles have speed$600\text{\hspace{0.17em}m}/\text{s}$

Understanding the concept

Average speed is calculated by adding the speed of each molecules and then by dividing the total number of molecules.

The RMS speed of molecules is given by,

${\text{v}}_{rms}=\sqrt{\frac{{\displaystyle {\sum }_{}^{}{\text{v}}^2}}{\text{N}}}$

(a) Calculate the average speed

Average speed is given by

${\text{V}}_{avg}=\frac{1}{\text{N}}\underset{i=1}{\overset{N}{}}{\text{V}}_i$

We have total of$10$particles, so$\text{N}=10$

Four particles have speed of $200\text{\hspace{0.17em}m}/\text{s}.$

So,${\text{V}}_1=4\times 200=800$

Two particles have speed of .$400\text{\hspace{0.17em}m}/\text{s}$

So,${\text{V}}_2=2\times 500=1000$

Four particles have speed of.$600\text{\hspace{0.17em}m}/\text{s}$

So,${\text{V}}_3=4\times 600=2400$

$V_{avg}=\frac{1}{N}(V_1+V_2+V_3)$

$\begin{array}{c}{\text{V}}_{avg}=\frac{1}{10}(800+1000+2400)\\ =420\text{\hspace{0.17em}m}/\text{s}\end{array}$

Therefore the average speed is .$420\text{\hspace{0.17em}m}/\text{s}$

(b) Calculate the RMS speed Vrms

The rms speed is given by

${\text{V}}_{rms}=\sqrt{\frac{1}{\text{N}}\underset{i=1}{\overset{N}{}}{\text{V}}_i^2}$

$\begin{array}{c}{\text{V}}_{rms}=\sqrt{\frac{1}{10}[4{\left(200\frac{\text{m}}{\text{s}}\right)}^2+2{\left(500\frac{\text{m}}{\text{s}}\right)}^2+4{\left(600\frac{\text{m}}{\text{s}}\right)}^2}\\ =458\text{\hspace{0.17em}m}/\text{s}\end{array}$

Therefore the RMS speed is $458\text{\hspace{0.17em}m}/\text{s}$.

(c) Figure out if Vrms>Vavg

Yes; ${\text{V}}_{rms}>{\text{V}}_{avg}$