| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| 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 with algebraic unknowns. While it involves multiple stages and algebraic manipulation, the approach is straightforward: use s=ut+½at² for both segments with the constraint that BC=2AB. This is slightly above average difficulty due to the algebraic nature and two-part structure, but follows a well-practiced method without requiring novel insight. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(AB: [s =] u\times8 + \frac{1}{2}a\times8^2\) \([= 8u + 32a]\) or \(\dfrac{u + u + 8a}{2}\times8\) | B1B1 | For use of \(s = ut + \frac{1}{2}at^2\) or \(s = \frac{u+v}{2}t\). B1 for any one correct expression, B2 for two correct expressions. |
| \(BC: [2s =](u+8a)\times10 + \frac{1}{2}a\times10^2\) \([= 10u + 130a]\) or \(\dfrac{u+8a+u+8a+10a}{2}\times10\) | ||
| \(AC: [3s =] u\times18 + \frac{1}{2}a\times18^2\) \([= 18u + 162a]\) or \(\dfrac{u + u + 18a}{2}\times18\) | ||
| Attempt to solve simultaneously: \((u+8a)\times10 + \frac{1}{2}a\times10^2 = 2\!\left(u\times8 + \frac{1}{2}a\times8^2\right)\) \([\Rightarrow 10u + 130a = 2(8u+32a)]\) OR \(u\times18 + \frac{1}{2}a\times18^2 = 3\!\left(u\times8 + \frac{1}{2}a\times8^2\right)\) \([\Rightarrow 18u+162a = 3(8u+32a)]\) | M1 | To obtain an equation in \(u\) and \(a\) only. Must have come from correct expressions but allow \(\times\frac{1}{3}\) instead of \(\times3\) or \(\times\frac{1}{2}\) instead of \(\times2\) or \(\times\frac{2}{3}\) instead of \(\times\frac{3}{2}\). Note: M0 for \(u\times10 + \frac{1}{2}a\times10^2 = 2\!\left(u\times8+\frac{1}{2}a\times8^2\right)\) leading to \(u = -\frac{7}{3}a\). Note: M0 for distance \(AC = 2AB\) leading to \(u = -49a\). |
| \(u = 11a\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([s =] \dfrac{(u+8a)^2 - u^2}{2a}\) or \([2s =] \dfrac{(u+18a)^2-(u+8a)^2}{2a}\) or \([3s =] \dfrac{(u+18a)^2 - u^2}{2a}\) | (B1B1) | B1 for any one correct expression, B2 for two correct. |
| \(\dfrac{(u+18a)^2 - u^2}{2a} = 3\!\left(\dfrac{(u+8a)^2-u^2}{2a}\right)\) or \(\dfrac{(u+18a)^2-(u+8a)^2}{2a} = 2\!\left(\dfrac{(u+8a)^2-u^2}{2a}\right)\) or \(\dfrac{(u+18a)^2-u^2}{2a} = \dfrac{3}{2}\!\left(\dfrac{(u+18a)^2-(u+8a)^2}{2a}\right)\) | (M1) | To obtain an equation in \(u\) and \(a\) only. |
| \(u = 11a\) | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 11a + a\times18\) | M1 | For use of \(v = u + at\) or other complete suvat method using their \(u\) in terms of \(a\), e.g. \(v^2 = (11a)^2 + 2a(18\times11a + 162a)\ [= 841a^2]\) |
| Speed \(= 29a\) | A1FT | FT their expression for \(v\) so their \(u + 18a\). Note: If answer to part (a) is \(u = -\frac{7}{3}a\), then speed \(= \frac{47}{3}a\). |
| 2 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $AB: [s =] u\times8 + \frac{1}{2}a\times8^2$ $[= 8u + 32a]$ or $\dfrac{u + u + 8a}{2}\times8$ | B1B1 | For use of $s = ut + \frac{1}{2}at^2$ or $s = \frac{u+v}{2}t$. B1 for any one correct expression, B2 for two correct expressions. |
| $BC: [2s =](u+8a)\times10 + \frac{1}{2}a\times10^2$ $[= 10u + 130a]$ or $\dfrac{u+8a+u+8a+10a}{2}\times10$ | | |
| $AC: [3s =] u\times18 + \frac{1}{2}a\times18^2$ $[= 18u + 162a]$ or $\dfrac{u + u + 18a}{2}\times18$ | | |
| Attempt to solve simultaneously: $(u+8a)\times10 + \frac{1}{2}a\times10^2 = 2\!\left(u\times8 + \frac{1}{2}a\times8^2\right)$ $[\Rightarrow 10u + 130a = 2(8u+32a)]$ OR $u\times18 + \frac{1}{2}a\times18^2 = 3\!\left(u\times8 + \frac{1}{2}a\times8^2\right)$ $[\Rightarrow 18u+162a = 3(8u+32a)]$ | M1 | To obtain an equation in $u$ and $a$ only. Must have come from correct expressions but allow $\times\frac{1}{3}$ instead of $\times3$ or $\times\frac{1}{2}$ instead of $\times2$ or $\times\frac{2}{3}$ instead of $\times\frac{3}{2}$. Note: M0 for $u\times10 + \frac{1}{2}a\times10^2 = 2\!\left(u\times8+\frac{1}{2}a\times8^2\right)$ leading to $u = -\frac{7}{3}a$. Note: M0 for distance $AC = 2AB$ leading to $u = -49a$. |
| $u = 11a$ | A1 | |
| | **4** | |
**Alternative: Using $v^2 = u^2 + 2as$**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[s =] \dfrac{(u+8a)^2 - u^2}{2a}$ or $[2s =] \dfrac{(u+18a)^2-(u+8a)^2}{2a}$ or $[3s =] \dfrac{(u+18a)^2 - u^2}{2a}$ | (B1B1) | B1 for any one correct expression, B2 for two correct. |
| $\dfrac{(u+18a)^2 - u^2}{2a} = 3\!\left(\dfrac{(u+8a)^2-u^2}{2a}\right)$ or $\dfrac{(u+18a)^2-(u+8a)^2}{2a} = 2\!\left(\dfrac{(u+8a)^2-u^2}{2a}\right)$ or $\dfrac{(u+18a)^2-u^2}{2a} = \dfrac{3}{2}\!\left(\dfrac{(u+18a)^2-(u+8a)^2}{2a}\right)$ | (M1) | To obtain an equation in $u$ and $a$ only. |
| $u = 11a$ | (A1) | |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 11a + a\times18$ | M1 | For use of $v = u + at$ or other complete suvat method using their $u$ in terms of $a$, e.g. $v^2 = (11a)^2 + 2a(18\times11a + 162a)\ [= 841a^2]$ |
| Speed $= 29a$ | A1FT | FT their expression for $v$ so their $u + 18a$. Note: If answer to part (a) is $u = -\frac{7}{3}a$, then speed $= \frac{47}{3}a$. |
| | **2** | |
---3 A car travels along a straight road with constant acceleration $a \mathrm {~ms} ^ { - 2 }$, where $a > 0$. The car passes through points $A , B$ and $C$ in that order. The speed of the car at $A$ is $u \mathrm {~ms} ^ { - 1 }$ in the direction $A B$. The distance $B C$ is twice the distance $A B$. The car takes 8 seconds to travel from $A$ to $B$ and 10 seconds to travel from $B$ to $C$.
\begin{enumerate}[label=(\alph*)]
\item Find $u$ in terms of $a$.
\item Find the speed of the car at $C$ in terms of $a$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2024 Q3 [6]}}