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Q11P

Expert-verifiedFound in: Page 15

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Kinetic energy K (Chapter 7) has dimensions kg.m^{2}/s^{2} . It can be written in terms of the momentum p (Chapter 9) and mass m as**

**${\mathit{K}}{\mathbf{=}}\frac{{\mathbf{p}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}$**

**(a) Determine the proper units for momentum using dimensional analysis. **

**(b) The unit of force is the newton N, where 1 N = 1 kg.m/s^{2} . What are the units of momentum p in terms of a newton and another fundamental SI unit?**

- The unit of momentum is
**$p=\frac{kg.m}{s}$**. - The unit of momentum in terms of
*N*is N.s .

The formula of kinetic energy is $K=\frac{{\mathrm{p}}^{2}}{2\mathrm{m}}$ where *K* is the kinetic energy, *p* is the momentum, and *m* is the momentum.

It has given that:

The dimension of kinetic energy is kg.m^{2}/s^{2} .

**The analysis of a relationship between different physical quantities by using the units of measurements and dimensions is called dimensional analysis. It is used to examine the correctness of an equation.**

Consider the above equation of kinetic energy.

$K=\frac{{p}^{2}}{2m}$

Rearrange the above equation for** ***p* :

$\begin{array}{rcl}{p}^{2}& =& 2Km\\ p& =& \sqrt{2Km}\end{array}$

Replace** **kg.m^{2}/s^{2} for *K* and kg for *m* .

$\begin{array}{rcl}p& =& \sqrt{\frac{kg.{m}^{2}}{{s}^{2}}\times kg}\\ & =& \sqrt{\frac{{\left(kg\right)}^{2}{m}^{2}}{{s}^{2}}}\\ & =& \frac{kg.m}{s}\end{array}$

Hence, the unit of momentum is $p=\frac{kg.m}{s}$

The momentum is given by the product of force in *L* and some unknown quantity *X* .

p = N . X

Replace kg.m/s^{2} for *N* and kg.m/s for *p* .

$\frac{kg.m}{s}=\frac{kg.m}{{s}^{2}}.X$

Find the unit for X

*X* = *s*

Now, replace *s* for *X* to get the units of *p* in terms of *N* .

$\begin{array}{rcl}p& =& N.X\\ & =& N.s.\end{array}$

Hence, the unit of momentum in terms of *N* is N.s .

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