## Question 5:

### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[v^2] = [pu^\alpha]$ | M1 | For the idea that dimensions of every term must be the same; must relate LHS to RHS; must explicitly state $p$ is dimensionless |
| $[v] = [u]$ and $p$ is dimensionless, or $L^2T^{-2} = L^\alpha T^{-\alpha} \Rightarrow \alpha = 2$ | A1 | |

### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $L^2T^{-2} = (LT^{-2})^\beta L^\gamma$ | M1 | Equating dimensions of other term with $[v^2]$ with $q$ gone and $[a]$ and $[s]$ used; ignore term in $[u^2]$ if present |
| $-2 = -2\beta \Rightarrow \beta = 1$ | A1 | |
| $\beta + \gamma = 2$ | M1 | Equating powers of L |
| $\gamma = 1$ | A1 | SC2 for both correct |

### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $s = 0$ then $v = u$ so $u^2 = pu^2 + 0 \Rightarrow p = 1$ | B1 | Details must be shown; do not allow use of prior knowledge of $v^2 = u^2 + 2as$ |

### Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = \frac{1}{2}qmas = \frac{1}{2}qFs \equiv \frac{1}{2}qW$ | M1 | Where $F$ must be the force acting and $W$ the work done by this force; do not allow use of prior knowledge of $v^2 = u^2 + 2as$ |
| So we can see that this equation is a statement of the Work-Energy principle so $\frac{1}{2}q = 1$ so $q = 2$ | A1 | |