| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find total distance |
| Difficulty | Moderate -0.8 This is a straightforward M1 SUVAT question with clearly defined stages and all necessary information provided. Students must calculate areas under a trapezoid-shaped speed-time graph using basic formulas (area of triangle + rectangle). The multi-part structure guides students through the solution methodically, requiring only routine application of kinematics principles with no problem-solving insight needed. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T\cos\alpha - F = 2g\cos 60°\) | M1 A1 | M1 for resolving parallel to plane; correct no. of terms; both \(T\) and \(2g\) terms resolved. A1 for correct equation (use of \(\alpha\) instead of \(30°\) or \(60°\) is A error not M error; use of \(\sin(3/5)\) or \(\cos(4/5)\) scores M1A0) |
| \(T\sin\alpha + R = 2g\cos 30°\) | M1 A1 | M1 for resolving perpendicular to plane; correct no. of terms; both \(T\) and \(2g\) terms resolved. Same A1 conditions as above |
| \(F = \frac{1}{3}R\) | B1 | B1 for \(F = \frac{1}{3}R\) seen or implied |
| Eliminating \(F\) and \(R\) | DM1 | Dependent on first two M marks and appropriate angles used in both equations |
| \(T = g\!\left(1 + \dfrac{1}{\sqrt{3}}\right)\), \(1.6g\) (or better), \(15.5\), \(15\) (N) | DM1 A1 | Fourth M1 dependent on third M1, for solving for \(T\); A1 for \(15\)(N) or \(15.5\)(N) |
| Total | (8) | N.B. First two M marks can be for two resolutions in any directions. Use of \(\tan\alpha = 4/3\) leads to \(17.83\ldots\) and can score max 7/8 |
# Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T\cos\alpha - F = 2g\cos 60°$ | M1 A1 | M1 for resolving parallel to plane; correct no. of terms; both $T$ and $2g$ terms resolved. A1 for correct equation (use of $\alpha$ instead of $30°$ or $60°$ is A error not M error; use of $\sin(3/5)$ or $\cos(4/5)$ scores M1A0) |
| $T\sin\alpha + R = 2g\cos 30°$ | M1 A1 | M1 for resolving perpendicular to plane; correct no. of terms; both $T$ and $2g$ terms resolved. Same A1 conditions as above |
| $F = \frac{1}{3}R$ | B1 | B1 for $F = \frac{1}{3}R$ seen or implied |
| Eliminating $F$ and $R$ | DM1 | Dependent on first two M marks and appropriate angles used in both equations |
| $T = g\!\left(1 + \dfrac{1}{\sqrt{3}}\right)$, $1.6g$ (or better), $15.5$, $15$ (N) | DM1 A1 | Fourth M1 dependent on third M1, for solving for $T$; A1 for $15$(N) or $15.5$(N) |
| **Total** | **(8)** | N.B. First two M marks can be for two resolutions in any directions. Use of $\tan\alpha = 4/3$ leads to $17.83\ldots$ and can score max 7/8 |3.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{94d9432d-1723-4549-ad5e-d4be0f5fd083-005_851_1073_312_456}
\end{center}
\end{figure}
A sprinter runs a race of 200 m . Her total time for running the race is 25 s . Figure 2 is a sketch of the speed-time graph for the motion of the sprinter. She starts from rest and accelerates uniformly to a speed of $9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in 4 s . The speed of $9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is maintained for 16 s and she then decelerates uniformly to a speed of $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the end of the race. Calculate
\begin{enumerate}[label=(\alph*)]
\item the distance covered by the sprinter in the first 20 s of the race,
\item the value of $u$,
\item the deceleration of the sprinter in the last 5 s of the race.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q3}}