| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Standard +0.3 This is a standard variable acceleration problem requiring integration with initial conditions to find velocity and displacement, then applying standard techniques (setting s=0, finding stationary points). The algebra is straightforward, and all steps follow routine M1 procedures with no novel insight required—slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 16k - kt^2,\quad v = 16kt - \frac{1}{3}kt^3\) | M1 | Uses \(v = \int a\ dt\) |
| \(8 = 16k \times 4 - \frac{1}{3}k \times 4^3\) leading to \(k = \ldots\) | M1 | Substitutes \(t = 4,\ v = 8\) |
| \(v = 16kt - \frac{kt^3}{3}\) and \(k = \frac{3}{16}\) | A1 | OE |
| \(s = 8kt^2 - \frac{1}{12}kt^4\) leading to \(s = \frac{24}{16}t^2 - \frac{3}{192}t^4\) | M1 | Uses \(s = \int v\ dt\) and attempts to find \(s\) in terms of \(t\) only. May be using \(v = 3t - \frac{1}{16}t^3\) |
| \(s = \frac{1}{64}t^2(96 - t^2)\) | A1 | AG, no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = 0,\ t^2 = 96,\ t = 4\sqrt{6}\) | M1 | Attempt to find \(t\) when \(s = 0\) |
| \(v = 16 \times \frac{3}{16} \times \sqrt{96} - \frac{3}{16} \times \frac{1}{3} \times \sqrt{96}^3\) | M1 | Attempt to find \(v\) at this \(t\) value |
| Speed is \(29.4\ \text{ms}^{-1}\) | A1 | Do not condone \(v = -29.4\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = 0,\ t^2 = 48,\ t = 4\sqrt{3}\) | M1 | Determine the time \(t\) (or \(t^2\)) at which \(v = 0\) |
| \(s = \frac{1}{64} \times 48 \times (96 - 48)\) | M1 | Use substitution of the \(t\) or \(t^2\) value to find \(s\) |
| \(s = 36\) m | A1 |
## Question 5:
### Part (a):
| $a = 16k - kt^2,\quad v = 16kt - \frac{1}{3}kt^3$ | M1 | Uses $v = \int a\ dt$ |
| $8 = 16k \times 4 - \frac{1}{3}k \times 4^3$ leading to $k = \ldots$ | M1 | Substitutes $t = 4,\ v = 8$ |
| $v = 16kt - \frac{kt^3}{3}$ and $k = \frac{3}{16}$ | A1 | OE |
| $s = 8kt^2 - \frac{1}{12}kt^4$ leading to $s = \frac{24}{16}t^2 - \frac{3}{192}t^4$ | M1 | Uses $s = \int v\ dt$ and attempts to find $s$ in terms of $t$ only. May be using $v = 3t - \frac{1}{16}t^3$ |
| $s = \frac{1}{64}t^2(96 - t^2)$ | A1 | AG, no errors seen |
### Part (b):
| $s = 0,\ t^2 = 96,\ t = 4\sqrt{6}$ | M1 | Attempt to find $t$ when $s = 0$ |
| $v = 16 \times \frac{3}{16} \times \sqrt{96} - \frac{3}{16} \times \frac{1}{3} \times \sqrt{96}^3$ | M1 | Attempt to find $v$ at this $t$ value |
| Speed is $29.4\ \text{ms}^{-1}$ | A1 | Do not condone $v = -29.4$ |
### Part (c):
| $v = 0,\ t^2 = 48,\ t = 4\sqrt{3}$ | M1 | Determine the time $t$ (or $t^2$) at which $v = 0$ |
| $s = \frac{1}{64} \times 48 \times (96 - 48)$ | M1 | Use substitution of the $t$ or $t^2$ value to find $s$ |
| $s = 36$ m | A1 | |
---5 A particle $P$ moves in a straight line, starting from rest at a point $O$ on the line. At time $t \mathrm {~s}$ after leaving $O$ the acceleration of $P$ is $k \left( 16 - t ^ { 2 } \right) \mathrm { m } \mathrm { s } ^ { - 2 }$, where $k$ is a positive constant, and the displacement from $O$ is $s \mathrm {~m}$. The velocity of $P$ is $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when $t = 4$.
\begin{enumerate}[label=(\alph*)]
\item Show that $s = \frac { 1 } { 64 } t ^ { 2 } \left( 96 - t ^ { 2 } \right)$.
\item Find the speed of $P$ at the instant that it returns to $O$.
\item Find the maximum displacement of the particle from $O$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q5 [11]}}