| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Find acceleration from distances/times |
| Difficulty | Standard +0.3 This is a standard SUVAT problem requiring systematic application of kinematic equations across two intervals with a constraint relating the distances. While it involves algebraic manipulation and simultaneous equations, the approach is methodical and well-practiced in M1 courses, making it slightly easier than average but still requiring careful setup. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use constant acceleration equations to obtain an expression for either \(s_{AB}\) or \(s_{BC}\) in terms of \(a\) | M1 | |
| \(s_{AB} = 2 \times 4.5 - \frac{1}{2} \times a \times 2^2\) | A1 | or \(s_{AB} = \frac{1}{2}(v_A + v_B) \times 2 = 9 - 2a\) |
| \(s_{BC} = 2 \times 4.5 + \frac{1}{2} \times a \times 2^2\) | A1 | or \(s_{BC} = \frac{1}{2}(v_B + v_C) \times 2 = 9 + 2a\) |
| \(2 \times 4.5 - \frac{1}{2}a \times 2^2 = \frac{4}{5}(2 \times 4.5 + \frac{1}{2}a \times 2^2)\) | M1 | Use the given information to find a valid equation for \(a\) |
| \(a = 0.5\ \text{ms}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4.5 = u + 2a,\ s_{AC} = 4u + 8a,\ s_{AB} = 2u + 2a\) | M1 | Any two relevant equations in \(u\), \(a\), \(s_{AB}\) and \(s_{AC}\) where \(u\) is velocity at \(A\) |
| Two correct equations | A1 | |
| Three correct equations | A1 | |
| \(2(4.5 - 2a) + 6a = \frac{5}{4}\{2(4.5 - 2a) + 2a\}\) | M1 | Use the given information that \(BC = \frac{5}{4}AB\) to find valid equation involving \(a\) only |
| \(a = 0.5\ \text{ms}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(AC = 4.5 \times 4\) | M1 | Using \(AC = v_B \times 4\) since \(v_B\) is the average velocity over \(AC\) |
| \(BC = \frac{5}{9} \times AC\) or \(AB = \frac{4}{9} \times AC\) | M1 | |
| \(BC = 10\) or \(AB = 8\) | A1 | |
| \(10 = 4.5 \times 2 + 2a\) or \(8 = 4.5 \times 2 - 2a\) | M1 | Using \(s = ut + \frac{1}{2}at^2\) for \(BC\) or \(s = vt - \frac{1}{2}at^2\) for \(AB\) |
| \(a = 0.5\ \text{ms}^{-2}\) | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s_{AB} = 2 \times 4.5 - \frac{1}{2} \times 0.5 \times 2^2 = 8\) OR \(s_{BC} = 2 \times 4.5 + \frac{1}{2} \times 0.5 \times 2^2 = 10\) | M1 | Attempt to find \(s_{AB}\) or \(s_{BC}\); OR attempt to find \(s_{AB}\) directly as \(s_{AC} = 3.5\times4 + \frac{1}{2}\times a\times4^2\) or \(\frac{1}{2}(4.5-2a+4.5+2a)\times4\); or add the 2 expressions found in 4(a) for \(s_{AB}\) and \(s_{BC}\) |
| \(s_{AC} = 8 + \frac{5}{4} \times 8 = 18\) m OR \(s_{AC} = 10 + \frac{4}{5} \times 10 = 18\) m | A1 | |
| Total: 2 |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use constant acceleration equations to obtain an expression for either $s_{AB}$ or $s_{BC}$ in terms of $a$ | M1 | |
| $s_{AB} = 2 \times 4.5 - \frac{1}{2} \times a \times 2^2$ | A1 | or $s_{AB} = \frac{1}{2}(v_A + v_B) \times 2 = 9 - 2a$ |
| $s_{BC} = 2 \times 4.5 + \frac{1}{2} \times a \times 2^2$ | A1 | or $s_{BC} = \frac{1}{2}(v_B + v_C) \times 2 = 9 + 2a$ |
| $2 \times 4.5 - \frac{1}{2}a \times 2^2 = \frac{4}{5}(2 \times 4.5 + \frac{1}{2}a \times 2^2)$ | M1 | Use the given information to find a valid equation for $a$ |
| $a = 0.5\ \text{ms}^{-2}$ | A1 | |
**Alternative method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $4.5 = u + 2a,\ s_{AC} = 4u + 8a,\ s_{AB} = 2u + 2a$ | M1 | Any two relevant equations in $u$, $a$, $s_{AB}$ and $s_{AC}$ where $u$ is velocity at $A$ |
| Two correct equations | A1 | |
| Three correct equations | A1 | |
| $2(4.5 - 2a) + 6a = \frac{5}{4}\{2(4.5 - 2a) + 2a\}$ | M1 | Use the given information that $BC = \frac{5}{4}AB$ to find valid equation involving $a$ only |
| $a = 0.5\ \text{ms}^{-2}$ | A1 | |
**Alternative method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $AC = 4.5 \times 4$ | M1 | Using $AC = v_B \times 4$ since $v_B$ is the average velocity over $AC$ |
| $BC = \frac{5}{9} \times AC$ **or** $AB = \frac{4}{9} \times AC$ | M1 | |
| $BC = 10$ or $AB = 8$ | A1 | |
| $10 = 4.5 \times 2 + 2a$ **or** $8 = 4.5 \times 2 - 2a$ | M1 | Using $s = ut + \frac{1}{2}at^2$ for $BC$ or $s = vt - \frac{1}{2}at^2$ for $AB$ |
| $a = 0.5\ \text{ms}^{-2}$ | A1 | |
| **Total: 5** | | |
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $s_{AB} = 2 \times 4.5 - \frac{1}{2} \times 0.5 \times 2^2 = 8$ **OR** $s_{BC} = 2 \times 4.5 + \frac{1}{2} \times 0.5 \times 2^2 = 10$ | M1 | Attempt to find $s_{AB}$ or $s_{BC}$; **OR** attempt to find $s_{AB}$ directly as $s_{AC} = 3.5\times4 + \frac{1}{2}\times a\times4^2$ or $\frac{1}{2}(4.5-2a+4.5+2a)\times4$; **or** add the 2 expressions found in **4(a)** for $s_{AB}$ and $s_{BC}$ |
| $s_{AC} = 8 + \frac{5}{4} \times 8 = 18$ m **OR** $s_{AC} = 10 + \frac{4}{5} \times 10 = 18$ m | A1 | |
| **Total: 2** | | |4 A cyclist travels along a straight road with constant acceleration. He passes through points $A , B$ and $C$. The cyclist takes 2 seconds to travel along each of the sections $A B$ and $B C$ and passes through $B$ with speed $4.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The distance $A B$ is $\frac { 4 } { 5 }$ of the distance $B C$.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the cyclist.
\item Find $A C$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q4 [7]}}