Easy -1.8 This is a straightforward algebraic manipulation requiring only two steps: multiply both sides by 2, then divide by m, then take the square root. It's simpler than a typical quadratic equation as there's no rearrangement into standard form or use of the quadratic formula—just basic inverse operations on a single term.
- M2 for \(v^2 = \dfrac{2E}{m}\) or for \([v =][\pm]\sqrt{\dfrac{E}{\frac{1}{2}m}}\)
M2
- M1 for a correct constructive first step
M1
- M1 for \(v = [\pm]\sqrt{k}\) ft their \(v^2 = k\)
M1
- If M0 then SC1 for \(\sqrt{E/\frac{1}{2}m}\) or \(\sqrt{2E/m}\) etc
SC1
## Question 13:
$[v =][\pm]\sqrt{\dfrac{2E}{m}}$ (www) | 3 marks |
- M2 for $v^2 = \dfrac{2E}{m}$ or for $[v =][\pm]\sqrt{\dfrac{E}{\frac{1}{2}m}}$ | M2 |
- M1 for a correct constructive first step | M1 |
- M1 for $v = [\pm]\sqrt{k}$ ft their $v^2 = k$ | M1 |
- If M0 then SC1 for $\sqrt{E/\frac{1}{2}m}$ or $\sqrt{2E/m}$ etc | SC1 |
---