| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Work done by vector force displacement |
| Difficulty | Standard +0.3 This is a straightforward work-energy problem requiring standard mechanics techniques: resolving forces, calculating work done (F·s with angle), applying work-energy theorem, and finding power and impulse. While it involves multiple parts and vector resolution, each step follows directly from standard formulas without requiring novel insight or complex problem-solving. Slightly easier than average due to clear structure and routine application of mechanics principles. |
| Spec | 6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02k Power: rate of doing work6.02l Power and velocity: P = Fv6.03e Impulse: by a force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F = 80\cos 40° = 45a \Rightarrow a = 80\cos 40° / 45\) | M1 | Using \(F = ma\) with a resolved component of the pulling force and \(m\) (not \(mg\)) to find \(a\). \(1.36185\ldots\) or \(mv = Ft\ (= 306.4)\) |
| \(s = 0.8 \times 5 + \frac{1}{2}(1.36185\ldots) \times 5^2\) | M1 | Using suvat equation(s) with their value of \(a\) to find distance travelled (\(21.0232\ldots\) m). Or final velocity \((v = 0.8 + \frac{1}{2} \times 1.36185\ldots \times 5 = 7.609283\ldots)\); or \(v = \frac{Ft}{m} + 0.8\ \left(= \frac{306.4}{45} + 0.8\right)\) |
| \(\text{WD by pull} = 80\cos 40° \times 21.0232\ldots\) | M1 | Their component of force \(\times\) their distance. If \(a = 0\) used (e.g. 245 Nm) then only this mark can be awarded. Or change in \(\text{KE} = \frac{1}{2} \times 15 \times (7.609283\ldots^2 - 0.8^2)\) |
| \(= \text{awrt}\ 1290\ \text{(J)}\) | A1 | \(1288.377\ldots\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 0.8 + 1.36185\ldots \times 5 = 7.609283\ldots\) and use in \(\text{KE} = \frac{1}{2}mv^2\) | M1 | Attempt to use suvat equation with their \(a\) to find velocity after 5 seconds (with no vertical component) and using this to attempt to find KE (may be seen in (a)). Or \(\text{WD} + \text{initial KE}\ (= 14.4)\) must be initial and not final KE, combined with WD from part (a) |
| So \(\text{KE} = \frac{1}{2} \times 45 \times 7.609283\ldots^2 = \text{awrt}\ 1300\ \text{(J)}\) | A1 | \(Or\ 14.4 + 1288\). \(1302.777\ldots\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Work done should equal the *increase* in (kinetic) energy so no account has been made of the fact that the crate has some initial energy. | B1 | Ignore reference to any kind of resistive force |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Av Power} = 80\cos 40° \times 21.0232\ldots / 5 = \text{awrt}\ 258\ \text{(W)}\) | B1FT | Their \((1290)/5\). Or average velocity \(\times\) resolved force \(= 0.5(0.8 + 7.609283\ldots) \times 80\cos 40°\). B0 if \(a = 0\) used. \(257.675\ldots\) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Impulse \(=\) Their (horizontal) Force \(\times\) time \(= 80\cos 40° \times 5\) | M1 | Or \(=\) change in crate's momentum \(= 45(7.609283\ldots - 0.8)\). Must lead to an answer \(> 0\) |
| \(= \text{awrt}\ 306\ \text{(Ns)}\) | A1 | |
| [2] |
## Question 3:
**Part (a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F = 80\cos 40° = 45a \Rightarrow a = 80\cos 40° / 45$ | M1 | Using $F = ma$ with a resolved component of the pulling force and $m$ (not $mg$) to find $a$. $1.36185\ldots$ or $mv = Ft\ (= 306.4)$ |
| $s = 0.8 \times 5 + \frac{1}{2}(1.36185\ldots) \times 5^2$ | M1 | Using suvat equation(s) with their value of $a$ to find distance travelled ($21.0232\ldots$ m). Or final velocity $(v = 0.8 + \frac{1}{2} \times 1.36185\ldots \times 5 = 7.609283\ldots)$; or $v = \frac{Ft}{m} + 0.8\ \left(= \frac{306.4}{45} + 0.8\right)$ |
| $\text{WD by pull} = 80\cos 40° \times 21.0232\ldots$ | M1 | Their component of force $\times$ their distance. If $a = 0$ used (e.g. 245 Nm) then only this mark can be awarded. Or change in $\text{KE} = \frac{1}{2} \times 15 \times (7.609283\ldots^2 - 0.8^2)$ |
| $= \text{awrt}\ 1290\ \text{(J)}$ | A1 | $1288.377\ldots$ |
| **[4]** | | |
**Part (b)(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 0.8 + 1.36185\ldots \times 5 = 7.609283\ldots$ and use in $\text{KE} = \frac{1}{2}mv^2$ | M1 | Attempt to use suvat equation with their $a$ to find velocity after 5 seconds (with no vertical component) and using this to attempt to find KE (may be seen in (a)). Or $\text{WD} + \text{initial KE}\ (= 14.4)$ must be initial and not final KE, combined with WD from part (a) |
| So $\text{KE} = \frac{1}{2} \times 45 \times 7.609283\ldots^2 = \text{awrt}\ 1300\ \text{(J)}$ | A1 | $Or\ 14.4 + 1288$. $1302.777\ldots$ |
| **[2]** | | |
**Part (b)(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Work done should equal the *increase* in (kinetic) energy so no account has been made of the fact that the crate has some initial energy. | B1 | Ignore reference to any kind of resistive force |
| **[1]** | | |
**Part (c)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Av Power} = 80\cos 40° \times 21.0232\ldots / 5 = \text{awrt}\ 258\ \text{(W)}$ | B1FT | Their $(1290)/5$. Or average velocity $\times$ resolved force $= 0.5(0.8 + 7.609283\ldots) \times 80\cos 40°$. B0 if $a = 0$ used. $257.675\ldots$ |
| **[1]** | | |
**Part (d)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Impulse $=$ Their (horizontal) Force $\times$ time $= 80\cos 40° \times 5$ | M1 | Or $=$ change in crate's momentum $= 45(7.609283\ldots - 0.8)$. Must lead to an answer $> 0$ |
| $= \text{awrt}\ 306\ \text{(Ns)}$ | A1 | |
| **[2]** | | |3 A crate of mass 45 kg is sliding with a speed of $0.8 \mathrm {~ms} ^ { - 1 }$ in a straight line across a smooth horizontal floor. One end of a light inextensible rope is attached to the crate. At a certain instant a builder takes the other end of the rope and starts to pull, applying a constant force of 80 N for 5 seconds.
While the builder is pulling the crate, the rope makes a constant angle of $40 ^ { \circ }$ above the horizontal. Both the rope and the velocity of the crate lie in the same vertical plane (see diagram).\\
\begin{tikzpicture}[
>= Stealth,
line cap = round,
line join = round
]
% Ground line
\draw[thick] (0.8, 0.74) -- (12.2, 0.74);
% Left gray block
\filldraw[fill=gray!45, draw=black, line width=0.6pt]
(1.70, 0.74) rectangle (3.00, 1.54);
% Right dark block
\filldraw[fill=black!80, draw=black, line width=0.6pt]
(6.00, 0.74) rectangle (7.56, 1.54);
% Velocity arrow (gray) and label
\draw[->, gray, line width=1.5pt] (0.60, 1.24) -- (1.64, 1.24);
\node[gray, font=\small] at (1.12, 1.80) {$0.8\,\mathrm{ms}^{-1}$};
% Push arrow (black, left of dark block)
\draw[->, line width=2pt] (5.20, 1.14) -- (5.92, 1.14);
% 80 N force line at 40 degrees
\draw[->] (7.56, 1.54) -- (9.38, 3.08);
\node[above right, font=\small] at (9.38, 3.08) {$80\,\mathrm{N}$};
% Dashed horizontal reference line
\draw[dashed] (7.56, 1.54) -- (10.10, 1.54);
% Arc showing 40-degree angle (centre at force-line origin, radius 1.0)
\draw (8.56, 1.54) arc (0:40:1.0);
% 40-degree label
\node[font=\small] at (8.90, 1.38) {$40^\circ$};
\end{tikzpicture}
It may be assumed that there is no resistance to the motion of the crate.
\begin{enumerate}[label=(\alph*)]
\item Determine the work done by the builder in pulling the crate.
\item \begin{enumerate}[label=(\roman*)]
\item Find the kinetic energy of the crate at the instant when the builder stops pulling the crate.
\item Explain why the answers to part (a) and part (b)(i) are not equal.
\end{enumerate}\item Find the average power developed by the builder in pulling the crate.
\item Calculate the total impulse exerted on the crate by the builder.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2023 Q3 [10]}}