| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - inclined plane with resistance (no driving force) |
| Difficulty | Standard +0.3 This is a straightforward application of the work-energy principle with clearly given values. Students must identify forces doing work (gravity component and resistance), calculate work done over 8m, and apply ΔKE = work done. The arithmetic is simple (sin α = 3/5 given directly) and requires only substitution into a standard formula with no conceptual obstacles. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Work done \(= \frac{1}{5}mg \times 8 \quad (15.68m)\) | B1 | Work done against friction seen or implied |
| PE Loss \(= 8mg\sin\alpha \quad (47.04m)\) | B1 | PE loss seen or implied. NB: B1B1 for \(\left(\frac{3}{5}mg - \frac{1}{5}mg\right)\times 8 \left(=\frac{16}{5}mg\right)\) |
| KE Gain \(=\) difference of two KE terms | M1 | Difference in two KE terms seen or implied (allow KE loss) |
| \(= \frac{1}{2}mv^2 - \frac{1}{2}m(5)^2\) | A1 | Correct unsimplified expression. Allow \(\pm\) |
| Work done against friction \(=\) PE Loss \(-\) KE Gain | M1 | Work-energy equation with all terms. Must be dimensionally correct but condone sign errors. Note: suvat methods will not score this M1 |
| \(\frac{1}{5}mg \times 8 = 8mg\sin\alpha - \left(\frac{1}{2}mv^2 - \frac{1}{2}m \cdot 5^2\right)\) | A1 | Correct unsimplified equation |
| \(v = 9.4\) or \(9.37\ \text{m s}^{-1}\) | A1 | 2 sf or 3 sf (after use of \(g = 9.8\)) |
## Question 3:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Work done $= \frac{1}{5}mg \times 8 \quad (15.68m)$ | B1 | Work done against friction seen or implied |
| PE Loss $= 8mg\sin\alpha \quad (47.04m)$ | B1 | PE loss seen or implied. NB: B1B1 for $\left(\frac{3}{5}mg - \frac{1}{5}mg\right)\times 8 \left(=\frac{16}{5}mg\right)$ |
| KE Gain $=$ difference of two KE terms | M1 | Difference in two KE terms seen or implied (allow KE loss) |
| $= \frac{1}{2}mv^2 - \frac{1}{2}m(5)^2$ | A1 | Correct unsimplified expression. Allow $\pm$ |
| Work done against friction $=$ PE Loss $-$ KE Gain | M1 | Work-energy equation with all terms. Must be dimensionally correct but condone sign errors. Note: suvat methods will not score this M1 |
| $\frac{1}{5}mg \times 8 = 8mg\sin\alpha - \left(\frac{1}{2}mv^2 - \frac{1}{2}m \cdot 5^2\right)$ | A1 | Correct unsimplified equation |
| $v = 9.4$ or $9.37\ \text{m s}^{-1}$ | A1 | 2 sf or 3 sf (after use of $g = 9.8$) |
**(7 marks)**\begin{enumerate}
\item A particle, $P$, of mass $m \mathrm {~kg}$ is projected with speed $5 \mathrm {~ms} ^ { - 1 }$ down a line of greatest slope of a rough plane. The plane is inclined to the horizontal at an angle $\alpha$, where $\sin \alpha = \frac { 3 } { 5 }$ The total resistance to the motion of $P$ is a force of magnitude $\frac { 1 } { 5 } m g$\\
Use the work-energy principle to find the speed of $P$ at the instant when it has moved a distance 8 m down the plane from the point of projection.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM1 AS 2019 Q3 [7]}}