| Prefix | Symbol | Factor |
|---|---|---|
| kilo | k | 10³ |
| centi | c | 10⁻² |
| milli | m | 10⁻³ |
| micro | µ | 10⁻⁶ |
| nano | n | 10⁻⁹ |
Dimensional analysis: before substituting numbers, check that each term in an equation has the same dimensions (e.g. [MLT⁻²] for force). This guards against algebraic errors and helps identify the form of a required relationship.
| Operation | Rule for uncertainties |
|---|---|
| Addition / subtraction (z = a ± b) | Absolute uncertainties add: \(\displaystyle \Delta z = \Delta a + \Delta b\) |
| Multiplication / division (z = a × b or z = a ⁄ b) | Percentage uncertainties add: \(\displaystyle \frac{\Delta z}{z}\times100\% = \frac{\Delta a}{a}\times100\% + \frac{\Delta b}{b}\times100\%\) |
| Power or root (z = aⁿ) | Relative uncertainty multiplied by \(|n|\): \(\displaystyle \frac{\Delta z}{z}\times100\% = |n|\,\frac{\Delta a}{a}\times100\%\) |
For a function \(z = f(a,b,\dots )\) the combined uncertainty is
\[
\Delta z = \sqrt{\left(\frac{\partial f}{\partial a}\Delta a\right)^{2}
+\left(\frac{\partial f}{\partial b}\Delta b\right)^{2}
+\dots }.
\]
Measurements:
\[
\Delta L = \Delta L{1} + \Delta L{2}=0.2+0.1=0.3\ \text{cm}
\]
\[
L = 12.3 + 8.7 = 21.0\ \text{cm}
\]
\[
\boxed{L = 21.0\ \text{cm} \pm 0.3\ \text{cm}}
\]
Given:
Density \(\rho = m/V\).
Percentage uncertainties add:
\[
\frac{\Delta\rho}{\rho}\times100\% = 0.4\% + 2.5\% = 2.9\%
\]
\[
\rho = \frac{50.0}{20.0}=2.50\ \text{g cm}^{-3}
\]
\[
\Delta\rho = 2.9\%\times2.50 = 0.0725\ \text{g cm}^{-3}\approx0.07\ \text{g cm}^{-3}
\]
\[
\boxed{\rho = 2.50\ \text{g cm}^{-3} \pm 0.07\ \text{g cm}^{-3}}
\]
Speed from a free‑fall experiment: \(v = \sqrt{2gh}\).
Since \(v \propto h^{1/2}\):
\[
\frac{\Delta v}{v}\times100\% = \tfrac12 \times 1.7\% = 0.85\%
\]
\[
v = \sqrt{2\times9.81\times1.20}=4.85\ \text{m s}^{-1}
\]
\[
\Delta v = 0.85\%\times4.85 = 0.041\ \text{m s}^{-1}\approx0.04\ \text{m s}^{-1}
\]
\[
\boxed{v = 4.85\ \text{m s}^{-1} \pm 0.04\ \text{m s}^{-1}}
\]
Work done: \(W = Fd\cos\theta\).
\[
\Delta W = \sqrt{(d\cos\theta\,\Delta F)^{2}
+(F\cos\theta\,\Delta d)^{2}
+(Fd\sin\theta\,\Delta\theta)^{2}}
= 0.38\ \text{J}\;(≈0.4\ \text{J})
\]
\[
W = 12.0\times0.45\times\cos30^{\circ}=9.8\ \text{J}
\]
\[
\boxed{W = 9.8\ \text{J} \pm 0.4\ \text{J}}
\]
| Quantity | Equation |
|---|---|
| Displacement | \(s = ut + \tfrac12 at^{2}\) |
| Final speed | \(v = u + at\) |
| Speed–displacement | \(v^{2}=u^{2}+2as\) |
| Average speed | \(\bar v = \dfrac{\text{total distance}}{\text{total time}}\) |
A cart travels \(s = 1.20\ \text{m} \pm 0.02\ \text{m}\) in \(t = 0.50\ \text{s} \pm 0.01\ \text{s}\). Find the speed and its uncertainty.
\[
v = \frac{s}{t}= \frac{1.20}{0.50}=2.40\ \text{m s}^{-1}
\]
Percentage uncertainties:
\[
\frac{\Delta s}{s}\times100\% = \frac{0.02}{1.20}\times100\% = 1.7\%
\qquad
\frac{\Delta t}{t}\times100\% = \frac{0.01}{0.50}\times100\% = 2.0\%
\]
For division, add percentages:
\[
\frac{\Delta v}{v}\times100\% = 1.7\% + 2.0\% = 3.7\%
\]
\[
\Delta v = 3.7\%\times2.40 = 0.089\ \text{m s}^{-1}\approx0.09\ \text{m s}^{-1}
\]
\[
\boxed{v = 2.40\ \text{m s}^{-1} \pm 0.09\ \text{m s}^{-1}}
\]
| Situation | Equation |
|---|---|
| Horizontal motion with friction | \(F{\text{net}} = ma = F{\text{applied}} - \mu N\) |
| Vertical motion (free fall) | \(v = u + gt,\; s = ut + \tfrac12gt^{2}\) |
| Circular motion (uniform) | \(F_{c}=mv^{2}/r\) |
A block of mass \(m=0.500\ \text{kg}\pm0.005\ \text{kg}\) is pulled horizontally with a force \(F=3.00\ \text{N}\pm0.02\ \text{N}\). The coefficient of kinetic friction is \(\mu_k=0.20\) (no uncertainty). Find the acceleration and its uncertainty.
\[
N = mg = 0.500\times9.81 = 4.905\ \text{N}
\]
\[
F{\text{fr}} = \muk N = 0.20\times4.905 = 0.981\ \text{N}
\]
\[
F{\text{net}} = F - F{\text{fr}} = 3.00 - 0.981 = 2.019\ \text{N}
\]
\[
a = \frac{F_{\text{net}}}{m}= \frac{2.019}{0.500}=4.04\ \text{m s}^{-2}
\]
Uncertainties (multiplication/division → add percentages):
\[
\frac{\Delta F}{F}= \frac{0.02}{3.00}=0.67\%
\quad
\frac{\Delta m}{m}= \frac{0.005}{0.500}=1.0\%
\]
\[
\frac{\Delta a}{a}=0.67\%+1.0\% = 1.67\%
\quad\Rightarrow\quad
\Delta a = 1.67\%\times4.04 = 0.07\ \text{m s}^{-2}
\]
\[
\boxed{a = 4.04\ \text{m s}^{-2} \pm 0.07\ \text{m s}^{-2}}
\]
| Concept | Formula |
|---|---|
| Work | \(W = \mathbf{F}\!\cdot\!\mathbf{s}=Fs\cos\theta\) |
| Kinetic energy | \(K = \tfrac12 mv^{2}\) |
| Gravitational potential energy | \(U = mgh\) |
| Mechanical energy (conserved) | \(E = K+U\) |
| Power | \(P = \dfrac{W}{t}=Fv\) |
A cart of mass \(m=0.250\ \text{kg}\pm0.002\ \text{kg}\) moves at \(v=2.00\ \text{m s}^{-1}\pm0.03\ \text{m s}^{-1}\).
\[
K = \tfrac12 mv^{2}=0.5\times0.250\times(2.00)^{2}=0.500\ \text{J}
\]
Percentage uncertainties:
\[
\frac{\Delta m}{m}=0.8\%
\qquad
\frac{\Delta v}{v}= \frac{0.03}{2.00}=1.5\%
\]
Since \(v^{2}\) appears, double the velocity percentage:
\[
\frac{\Delta K}{K}=0.8\%+2(1.5\%)=3.8\%
\]
\[
\Delta K = 3.8\%\times0.500 = 0.019\ \text{J}\approx0.02\ \text{J}
\]
\[
\boxed{K = 0.50\ \text{J} \pm 0.02\ \text{J}}
\]
| Quantity | Formula |
|---|---|
| Wave speed | \(v = f\lambda\) |
| Period | \(T = 1/f\) |
| Fundamental frequency (string, fixed‑fixed) | \(f_{1}= \dfrac{v}{2L}\) |
| Fundamental frequency (open‑closed tube) | \(f_{1}= \dfrac{v}{4L}\) |
| Interference condition (constructive) | \(\Delta r = n\lambda\) |
| Interference condition (destructive) | \(\Delta r = (n+\tfrac12)\lambda\) |
Measured wavelength \(\lambda = 0.500\ \text{m}\pm0.005\ \text{m}\) and frequency \(f = 250\ \text{Hz}\pm2\ \text{Hz}\).
\[
v = f\lambda = 250\times0.500 = 125\ \text{m s}^{-1}
\]
Percentage uncertainties:
\[
\frac{\Delta \lambda}{\lambda}=1.0\%,\qquad
\frac{\Delta f}{f}= \frac{2}{250}=0.8\%
\]
Add percentages (multiplication):
\[
\frac{\Delta v}{v}=1.0\%+0.8\%=1.8\%
\]
\[
\Delta v = 1.8\%\times125 = 2.3\ \text{m s}^{-1}
\]
\[
\boxed{v = 125\ \text{m s}^{-1} \pm 2\ \text{m s}^{-1}}
\]
Two resistors: \(R{1}=100.0\ \Omega\pm0.5\ \Omega\), \(R{2}=220.0\ \Omega\pm1.0\ \Omega\). A supply voltage \(V=12.0\ \text{V}\pm0.1\ \text{V}\). Find the total current and its uncertainty.
\[
R{\text{eq}} = R{1}+R_{2}=320.0\ \Omega
\]
\[
\frac{\Delta R{\text{eq}}}{R{\text{eq}}}= \frac{0.5+1.0}{320}=0.47\%
\]
\[
I = \frac{V}{R_{\text{eq}}}= \frac{12.0}{320.0}=0.0375\ \text{A}
\]
Percentage uncertainties (division → add):
\[
\frac{\Delta I}{I}= \frac{\Delta V}{V}+ \frac{\Delta R{\text{eq}}}{R{\text{eq}}}
= \frac{0.1}{12.0}+0.0047 = 0.0083+0.0047 = 0.0130\;(1.3\%)
\]
\[
\Delta I = 1.3\%\times0.0375 = 0.00049\ \text{A}\approx0.0005\ \text{A}
\]
\[
\boxed{I = 0.0375\ \text{A} \pm 0.0005\ \text{A}}
\]
For two series resistors \(R{1}\) and \(R{2}\) across a supply \(V\), the voltage across \(R_{2}\) is
\[
V{2}=V\frac{R{2}}{R{1}+R{2}}.
\]
Uncertainty (division & multiplication) → add percentage uncertainties of \(V\), \(R{1}\) and \(R{2}\).
Supply \(V=5.00\ \text{V}\pm0.02\ \text{V}\). Resistors: \(R{1}=1.00\ \text{k}\Omega\pm10\ \Omega\), \(R{2}=2.00\ \text{k}\Omega\pm20\ \Omega\).
\[
V_{2}=5.00\times\frac{2000}{1000+2000}=5.00\times\frac{2}{3}=3.33\ \text{V}
\]
Percentage uncertainties:
\[
\frac{\Delta V}{V}=0.4\%,\;
\frac{\Delta R{1}}{R{1}}=1.0\%,\;
\frac{\Delta R{2}}{R{2}}=1.0\%
\]
For the ratio \(R{2}/(R{1}+R{2})\) the relative uncertainty is the sum of the uncertainties of \(R{2}\) and \((R{1}+R{2})\):
\[
\frac{\Delta (R{1}+R{2})}{R{1}+R{2}} = \frac{10+20}{3000}=1.0\%
\]
Thus
\[
\frac{\Delta V{2}}{V{2}} = 0.4\% + 1.0\% + 1.0\% = 2.4\%
\]
\[
\Delta V_{2}=2.4\%\times3.33=0.08\ \text{V}
\]
\[
\boxed{V_{2}=3.33\ \text{V} \pm 0.08\ \text{V}}
\]
| Quantity | Formula |
|---|---|
| Decay constant | \(\lambda = \dfrac{\ln 2}{t_{1/2}}\) |
| Remaining nuclei | \(N = N_{0}e^{-\lambda t}\) |
| Activity | A = \(\lambda N\) |
| Mass defect | \(\Delta m = Zm{p} + Nm{n} - m_{\text{nuclide}}\) |
| Binding energy | \(E_{b}= \Delta m\,c^{2}\) |
In a counting experiment the activity after 10 min is \(A = 2500\ \text{counts}\pm50\). The initial activity was \(A_{0}=5000\ \text{counts}\pm30\). Determine the half‑life assuming first‑order decay.
\[
\frac{A}{A_{0}} = e^{-\lambda t}\;\Rightarrow\;
\lambda = -\frac{1}{t}\ln
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