Dimensional Analysis

Question

If energy (E), velocity (v), and force (F) are taken as fundamental quantities, what is the dimensional formula for momentum (p)?

RankGuru · Accepted Answer

1. **System Equations**: 
   - Let $p \propto E^a v^b F^c$.
   - Dimensions of $p$: $[MLT^{-1}]$.
   - $[MLT^{-1}] = [ML^2T^{-2}]^a [LT^{-1}]^b [MLT^{-2}]^c$.
2. **Power Matching**:
   - Mass ($M$): $1 = a + c$.
   - Length ($L$): $1 = 2a + b + c$.
   - Time ($T$): $-1 = -2a - b - 2c$.
3. **Solve Equations**:
   - From (1), $c = 1-a$.
   - Insert into (2): $1 = 2a + b + 1 - a \implies a + b = 0 \implies b = -a$.
   - Insert into (3): $-1 = -2a - (-a) - 2(1-a) = -2a + a - 2 + 2a = a-2 \implies a = 1$.
   - Thus, $b = -1$ and $c = 0$.
4. **Result**: Dimension of $p$ is $[E^1 v^{-1} F^0]$, or **$[E v^{-1}]$**.
5. **Concept Check**: $E = Force 	imes Distance$, $p = Mass 	imes Velocity$. $E/v$ has dimensions of momentum ($mv^2 / v = mv$).

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