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SS1: PHYSICS – 1ST TERM

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  1. Introduction to Physics | Week 1
    4 Topics
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    1 Quiz
  2. Measurement I | Week 2
    3 Topics
    |
    1 Quiz
  3. Measurement II | Week 3
    6 Topics
    |
    1 Quiz
  4. Motion | Week 4
    5 Topics
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    1 Quiz
  5. Velocity-Time Graph | Week 5
    4 Topics
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    1 Quiz
  6. Causes of Motion | Week 6
    5 Topics
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    1 Quiz
  7. Work, Energy & Power | Week 7
    3 Topics
  8. Energy Transformation / Power | Week 8
    3 Topics
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    1 Quiz
  9. Heat Energy | Week 9
    5 Topics
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    1 Quiz
  10. Linear Expansion | Week 10
    7 Topics
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    1 Quiz

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SS1: PHYSICS – 1ST TERM Measurement II | Week 3 Dimensions of Physical Quantities
Lesson 3, Topic 6
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Dimensions of Physical Quantities

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  • Dimensions of Physical Quantities

Any quantity that can be measured in units also has dimension, hence the dimension for length, mass, and time is as follows:

QuantitySymbol
MassM
LengthL
TimeT

Some examples of dimension for derived quantities are:

Velocity = \( \frac {length}{time} \\ = \frac {L}{T} \)

Dimension of Velocity is \(\scriptsize [L][T^{-1}] \)

Momentum = Mass × Velocity

= \(\scriptsize M \times LT^{-1} \\ \scriptsize = M LT^{-1} \)

 

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2022 Physics WAEC Theory Past Question 3

The effective potential energy, E, of a lunar satellite of mass, m1, moving in an. elliptical orbit around the moon of mass, m2, is given by

E = \(\frac{K^2}{2m_1 r^2} = \frac{Gm_1m_2}{r}\)

where r is the distance of the satellite from the moon and G is the universal gravitational constant of dimensions, \( \scriptsize M^{-1}L^3 T^{-2} \)

Determine the dimensions of the angular momentum, K, of the satellite using dimensional analysis.

Solution:

E = \(\frac{K^2}{2m_1 r^2} = \frac{Gm_1m_2}{r}\)

∴ \(\frac{K^2}{2m_1 r^2} = \frac{Gm_1m_2}{r}\)

⇒ \( \scriptsize k^2 = \normalsize\frac{Gm_1m_2 \: \times \: 2m_1 \not{r^2}}{\not{r}}\)

⇒ \( \scriptsize k^2 = 2Gm_1m_2 m_1 \: \times \: r\)

(Note: we can ignore 2 in front of G since we are working in dimensions)

Insert Dimensions

Dimension for:
G = \( \scriptsize M^{-1}L^3 T^{-2} \)
m = M
r = L

⇒ \( \scriptsize k^2 =M^{-1}L^3 T^{-2} \: \times \: M \: \times \: M \: \times \: M \: \times \: L \)

⇒ \( \scriptsize k^2 =M^{-1} \: \times \: M^3 \: \times \: L^3 \: \times \: L \: \times \: T^{-2} \)

⇒ \( \scriptsize k^2 =M^{(-1 + 3)} \: \times \: L^{(3+1)} \: \times \: T^{-2} \)

⇒ \( \scriptsize k^2 = M^2 \: \times \: L^4  \: \times \: T^{-2} \)

⇒ \( \scriptsize k^2 = M^2  L^4  T^{-2} \)

multiply both sides by \( \frac{1}{2} \) or find the square root of both sides.

⇒ \( \scriptsize k^{2\: \times \: \frac{1}{2}} = \left(M^2  L^4  T^{-2}\right)^{\frac{1}{2}} \)

⇒ \( \scriptsize k = M^{2\: \times \: \frac{1}{2}}  L^{4\: \times \: \frac{1}{2}}  T^{-2\: \times \: \frac{1}{2}} \)

⇒ \( \scriptsize k = M  L^{2}  T^{-1} \)

∴ dimensions of the angular momentum, K, ⇒ \( \scriptsize  M  L^{2}  T^{-1} \)

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