| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Constant acceleration vector (i and j) |
| Difficulty | Moderate -0.3 This is a straightforward M1 vector kinematics question requiring standard techniques: Pythagoras for speed, position vector formula r = r₀ + vt, and solving simultaneous equations for interception. All parts follow routine procedures with no novel problem-solving required, though the multi-part structure and vector notation place it slightly below average difficulty for A-level overall. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For \(A\): \(T = 2ma\) | B1 | |
| For \(B\): \(3mg - T = 3ma\) | M1 A1 | First M1 for resolving vertically for \(B\) with correct number of terms (allow omission of \(m\) provided 3 is there). First A1 for correct equation. |
| \(3mg = 5ma\) | DM1 | Second M1, dependent on first M1, for eliminating \(T\) to give equation in \(a\) only |
| \(\frac{3g}{5} = a\) (5.9 or 5.88 m s\(^{-2}\)) | A1 | Second A1 for 0.6g, 5.88 or 5.9. N.B. 'Whole system' equation \(3mg = 5ma\) earns first 4 marks but any error loses all 4. |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = \frac{6mg}{5}\); \(12m\); \(11.8m\) | B1 | B1 for \(\frac{6mg}{5}\), \(11.8m\), \(12m\) |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = \sqrt{T^2 + T^2}\) | M1 A1 ft | M1 for \(\sqrt{(T^2+T^2)}\) or \(\frac{T}{\sin 45°}\) or \(\frac{T}{\cos 45°}\) or \(2T\cos 45°\) or \(2T\sin 45°\) (allow if \(m\) omitted). M0 for \(T\sin 45°\). First A1 ft on their \(T\). |
| \(F = \frac{6mg\sqrt{2}}{5}\); \(1.7mg\) (or better); \(16.6m\); \(17m\) | A1 | Second A1 cao for \(\frac{6mg\sqrt{2}}{5}\) oe, \(1.7mg\) (or better), \(16.6m\), \(17m\) |
| Direction clearly marked on diagram with arrow and \(45°\) (oe) marked | B1 | B1 for direction clearly shown on diagram with arrow and \(45°\) marked |
| (4) | ||
| [10] |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For $A$: $T = 2ma$ | B1 | |
| For $B$: $3mg - T = 3ma$ | M1 A1 | First M1 for resolving vertically for $B$ with correct number of terms (allow omission of $m$ provided 3 is there). First A1 for correct equation. |
| $3mg = 5ma$ | DM1 | Second M1, dependent on first M1, for eliminating $T$ to give equation in $a$ only |
| $\frac{3g}{5} = a$ (5.9 or 5.88 m s$^{-2}$) | A1 | Second A1 for 0.6g, 5.88 or 5.9. N.B. 'Whole system' equation $3mg = 5ma$ earns first 4 marks but any error loses all 4. |
| | **(5)** | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = \frac{6mg}{5}$; $12m$; $11.8m$ | B1 | B1 for $\frac{6mg}{5}$, $11.8m$, $12m$ |
| | **(1)** | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \sqrt{T^2 + T^2}$ | M1 A1 ft | M1 for $\sqrt{(T^2+T^2)}$ or $\frac{T}{\sin 45°}$ or $\frac{T}{\cos 45°}$ or $2T\cos 45°$ or $2T\sin 45°$ (allow if $m$ omitted). M0 for $T\sin 45°$. First A1 ft on their $T$. |
| $F = \frac{6mg\sqrt{2}}{5}$; $1.7mg$ (or better); $16.6m$; $17m$ | A1 | Second A1 cao for $\frac{6mg\sqrt{2}}{5}$ oe, $1.7mg$ (or better), $16.6m$, $17m$ |
| Direction clearly marked on diagram with arrow and $45°$ (oe) marked | B1 | B1 for direction clearly shown on diagram with arrow and $45°$ marked |
| | **(4)** | |
| | **[10]** | |\begin{enumerate}
\item \hspace{0pt} [In this question, the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are horizontal vectors due east and north respectively.]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q8}}