| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Collision or meeting problems |
| Difficulty | Standard +0.8 This question requires integrating non-constant accelerations twice to find positions, setting up equations for collision conditions (equal velocities in part a, equal positions in part b), and solving the resulting system. While the integration itself is straightforward, coordinating multiple conditions and managing the algebra across two particles with different acceleration functions elevates this above a standard mechanics problem. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v = \int 12t + 12\,dt\), \(v_X = 6t^2 + 12t\) or \(v = \int 24t - 8\,dt\), \(v_Y = 12t^2 - 8t\) | M1 | Uses \(v = \int a\,dt\) |
| \(v_X = 6t^2 + 12t\), \(v_Y = 12t^2 - 8t\) | A1 | All correct |
| \(6t^2 + 12t = 12t^2 - 8t\) leading to \(t = \dfrac{10}{3}\) | B1 | Solves \(v_X = v_Y\) |
| \(s_X = \int 6t^2 + 12t\,dt\), \(s_X = 2t^3 + 6t^2\); \(s_Y = \int 12t^2 - 8t\,dt\), \(s_Y = 4t^3 - 4t^2\) | *M1 | Uses \(s = \int v\,dt\) |
| \(s_X = 2\!\left(\dfrac{10}{3}\right)^3 + 6\!\left(\dfrac{10}{3}\right)^2\), \(s_Y = 4\!\left(\dfrac{10}{3}\right)^3 - 4\!\left(\dfrac{10}{3}\right)^2\) | DM1 | Evaluates \(s\!\left(\dfrac{10}{3}\right)\) for each particle |
| \(AB = \dfrac{3800}{27} - \dfrac{2800}{27} = \dfrac{1000}{27}\), Distance \(AB = 37\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(AB = -2t^3 + 10t^2 = -2\times3^3 + 10\times3^2\) | M1 | Calculates distance \(AB\) |
| \([t=3]\), \(AB = 36\) m | A1 | AG |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $v = \int 12t + 12\,dt$, $v_X = 6t^2 + 12t$ or $v = \int 24t - 8\,dt$, $v_Y = 12t^2 - 8t$ | M1 | Uses $v = \int a\,dt$ |
| $v_X = 6t^2 + 12t$, $v_Y = 12t^2 - 8t$ | A1 | All correct |
| $6t^2 + 12t = 12t^2 - 8t$ leading to $t = \dfrac{10}{3}$ | B1 | Solves $v_X = v_Y$ |
| $s_X = \int 6t^2 + 12t\,dt$, $s_X = 2t^3 + 6t^2$; $s_Y = \int 12t^2 - 8t\,dt$, $s_Y = 4t^3 - 4t^2$ | *M1 | Uses $s = \int v\,dt$ |
| $s_X = 2\!\left(\dfrac{10}{3}\right)^3 + 6\!\left(\dfrac{10}{3}\right)^2$, $s_Y = 4\!\left(\dfrac{10}{3}\right)^3 - 4\!\left(\dfrac{10}{3}\right)^2$ | DM1 | Evaluates $s\!\left(\dfrac{10}{3}\right)$ for each particle |
| $AB = \dfrac{3800}{27} - \dfrac{2800}{27} = \dfrac{1000}{27}$, Distance $AB = 37$ m | A1 | |
---
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $AB = -2t^3 + 10t^2 = -2\times3^3 + 10\times3^2$ | M1 | Calculates distance $AB$ |
| $[t=3]$, $AB = 36$ m | A1 | AG |
---5 Particles $X$ and $Y$ move in a straight line through points $A$ and $B$. Particle $X$ starts from rest at $A$ and moves towards $B$. At the same instant, $Y$ starts from rest at $B$.
At time $t$ seconds after the particles start moving
\begin{itemize}
\item the acceleration of $X$ in the direction $A B$ is given by $( 12 t + 12 ) \mathrm { m } \mathrm { s } ^ { - 2 }$,
\item the acceleration of $Y$ in the direction $A B$ is given by $( 24 t - 8 ) \mathrm { m } \mathrm { s } ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item It is given that the velocities of $X$ and $Y$ are equal when they collide.
\end{itemize}
Calculate the distance $A B$.
\item It is given instead that $A B = 36 \mathrm {~m}$.
Verify that $X$ and $Y$ collide after 3 s.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q5 [8]}}