| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force on horizontal road |
| Difficulty | Standard +0.3 This is a straightforward work-energy problem requiring standard applications of energy conservation and work-energy principles. Part (a) involves calculating kinetic energy and work done (routine calculations), part (b) requires stating that the transition is smooth (standard modeling assumption) and applying energy conservation on a smooth slope. All steps are textbook-standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02c Work by variable force: using integration6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Gain in KE \(= \frac{1}{2} \times 4.2 \times 4.5^2\) (J) | M1 | Correct formula for KE used. Can be implied by awrt 42.5 |
| Work done by force \(= 35 \times 2.4\) (J) | M1 | Correct formula for WD by force used. Can be implied by awrt 84.0. Do not allow assumption that resistance is constant, e.g. by use of suvat, also in part (ii) |
| Energy lost \(= 84.0 - 42.5 =\) awrt 41.5 J | A1 | SC2 if using suvat to find correct average resistance and hence total energy lost |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = 41.5 / 2.4\) | M1 | Their energy loss divided by 2.4 |
| So average resistive force is awrt 17.3 N | A1 | SC1 only for 17.3N, if using suvat/N2L |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Other resistive forces (e.g. air resistance) can be ignored | B1 | "No friction" is not a valid answer here |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Need \(\frac{1}{2} \times 4.2 \times 4.5^2 = 4.2gh\) | M1 | Equating KE with PE (4.2 may be missing on both sides). If "resistive force" term included then M0 unless recovered |
| \(h = 1.033...\) | A1 | |
| Distance \(= 1.033 / \sin 20° =\) awrt 3.02 m | A1 | |
| Alternative method: | ||
| \(a = -g\sin 20°\) | M1 | Correctly deducing the acceleration up the slope |
| \(0^2 = 4.5^2 + 2 \times {-g\sin 20°} \times s\) | M1 | Using a suvat equation, or equations, which lead(s) to \(s\) from \(a\) and \(u\) given with \(v=0\) and consistent signs |
| Distance \(=\) awrt 3.02 m | A1 |
# Question 4:
## Part (a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Gain in KE $= \frac{1}{2} \times 4.2 \times 4.5^2$ (J) | M1 | Correct formula for KE used. Can be implied by awrt 42.5 |
| Work done by force $= 35 \times 2.4$ (J) | M1 | Correct formula for WD by force used. Can be implied by awrt 84.0. Do not allow assumption that resistance is constant, e.g. by use of suvat, also in part (ii) |
| Energy lost $= 84.0 - 42.5 =$ awrt 41.5 J | A1 | SC2 if using suvat to find correct average resistance and hence total energy lost |
## Part (a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = 41.5 / 2.4$ | M1 | Their energy loss divided by 2.4 |
| So average resistive force is awrt 17.3 N | A1 | SC1 only for 17.3N, if using suvat/N2L |
## Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Other resistive forces (e.g. air resistance) can be ignored | B1 | "No friction" is not a valid answer here |
## Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Need $\frac{1}{2} \times 4.2 \times 4.5^2 = 4.2gh$ | M1 | Equating KE with PE (4.2 may be missing on both sides). If "resistive force" term included then M0 unless recovered |
| $h = 1.033...$ | A1 | |
| Distance $= 1.033 / \sin 20° =$ awrt 3.02 m | A1 | |
| **Alternative method:** | | |
| $a = -g\sin 20°$ | M1 | Correctly deducing the acceleration up the slope |
| $0^2 = 4.5^2 + 2 \times {-g\sin 20°} \times s$ | M1 | Using a suvat equation, or equations, which lead(s) to $s$ from $a$ and $u$ given with $v=0$ and consistent signs |
| Distance $=$ awrt 3.02 m | A1 | |
---4 A small box $B$ of mass 4.2 kg is initially at rest at a point $O$ on rough horizontal ground. A horizontal force of magnitude 35 N is applied to $B$.\\
$B$ moves in a straight line until it reaches the point $S$ which is 2.4 m from $O$. At the instant that $B$ reaches $S$ its speed is $4.5 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the energy lost due to the resistive forces acting on $B$ as it moves from $O$ to $S$.
\item Deduce the magnitude of the average resistive force acting on $B$ as it moves from $O$ to $S$.
When $B$ reaches $S$, the force is no longer applied. $B$ continues to move directly up a smooth slope which is inclined at $20 ^ { \circ }$ above the horizontal (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{a65c4b75-b8b4-4a51-8abb-f857dc278271-3_275_1027_1866_244}
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item State an assumption required to model the motion of $B$ up the slope with only the information given.
\item Using the assumption made in part (b)(i), determine the distance travelled by $B$ up the slope until the instant when it comes to rest.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2021 Q4 [9]}}