| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Challenging +1.2 This is a multi-part mechanics question requiring integration of piecewise acceleration functions with fractional powers, followed by a connected particles problem. Part (a) requires straightforward integration of t^(1/2), part (b) uses continuity conditions, and part (c) involves finding displacement through integration. The fractional powers and piecewise nature add moderate complexity beyond standard constant acceleration, but the techniques are systematic M1 content without requiring novel insight. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt to integrate \(a\) for \(0 \leq t \leq 1\) | M1 | Increase power by 1 and change in coefficient in at least one term (same term); \(v = at\) is M0. Expect \((v=) -t^{\frac{3}{2}}[+c]\) |
| \((v=) -\frac{1.5}{\phantom{x}}t^{\frac{3}{2}} + 6 = -t^{\frac{3}{2}} + 6\) | A1 | Allow for \(-t^{\frac{3}{2}} + c\) and \(c = 6\) seen from CWO. Allow unsimplified. |
| Velocity at \(t=1\) is \(5\ \text{ms}^{-1}\) | A1 | CWO. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt to integrate \(a\) for \(t > 1\): \((v=)\frac{1.5}{\frac{3}{2}}t^{\frac{3}{2}} - \frac{3}{\frac{1}{2}}t^{\frac{1}{2}}[+c] = t^{\frac{3}{2}} - 6t^{\frac{1}{2}}[+c]\) | \*M1 | Increase power by 1 and change in coefficient in at least one term (same term); \(v = at\) is M0. |
| Use \(v = their\ 5\) when \(t=1\) to find \(c\): \(\left[5 = 1 - 6 + c\right]\) | DM1 | Must get numerical expression for \(c\); if no substitution seen, \(c\) must be correct for their expression for \(v\). Their 5 must not be a made up value. |
| \((v=) t^{\frac{3}{2}} - 6t^{\frac{1}{2}} + 10\) | A1 | OE, complete (possibly unsimplified) expression. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt to integrate their \(v\) for either section | \*M1 | Increase power by 1 and change in coefficient in at least one term (same term); \(s = vt\) is M0. Their \(v\)'s must come from integration. |
| \((s_1=) -\frac{1}{\frac{5}{2}}t^{\frac{5}{2}} + 6t[+c] = -\frac{2}{5}t^{\frac{5}{2}} + 6t[+c]\) \((s_2=) \frac{1}{\frac{5}{2}}t^{\frac{5}{2}} - \frac{6}{\frac{3}{2}}t^{\frac{3}{2}} + 10t[+c] = \frac{2}{5}t^{\frac{5}{2}} - 4t^{\frac{3}{2}} + 10t[+c]\) | A1 | For either correct. Allow unsimplified. |
| Use limits 0 and 1 for \(s_1\) and 1 and 4 for \(s_2\) | DM1 | Using correct limits correctly in expressions from integration; would lead to 2 positive values. \(s_1\) and \(s_2\) must have correct number of non-constant terms and include a linear term. |
| Total distance \(= 20\ \text{m}\) | A1 | \(s_1 = \left(-\frac{2}{5}\times1 + 6\times1\right) - 0 = \frac{28}{5} = 5.6\); \(s_2 = \left(\frac{2}{5}\times4^{\frac{5}{2}} - 4\times4^{\frac{3}{2}} + 10\times4\right) - \left(\frac{2}{5}\times1^{\frac{5}{2}} - 4\times1^{\frac{3}{2}} + 10\times1\right) = \frac{72}{5} = 14.4\) |
| Answer | Marks |
|---|---|
| Answer | Mark |
| \(s_1 = \frac{28}{5}\) or \(s_2 = \frac{72}{5}\) | B1 |
| Total distance \(= 20\ \text{m}\) | B1 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to integrate $a$ for $0 \leq t \leq 1$ | **M1** | Increase power by 1 and change in coefficient in at least one term (same term); $v = at$ is M0. Expect $(v=) -t^{\frac{3}{2}}[+c]$ |
| $(v=) -\frac{1.5}{\phantom{x}}t^{\frac{3}{2}} + 6 = -t^{\frac{3}{2}} + 6$ | **A1** | Allow for $-t^{\frac{3}{2}} + c$ and $c = 6$ seen from CWO. Allow unsimplified. |
| Velocity at $t=1$ is $5\ \text{ms}^{-1}$ | **A1** | CWO. |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to integrate $a$ for $t > 1$: $(v=)\frac{1.5}{\frac{3}{2}}t^{\frac{3}{2}} - \frac{3}{\frac{1}{2}}t^{\frac{1}{2}}[+c] = t^{\frac{3}{2}} - 6t^{\frac{1}{2}}[+c]$ | **\*M1** | Increase power by 1 and change in coefficient in at least one term (same term); $v = at$ is M0. |
| Use $v = their\ 5$ when $t=1$ to find $c$: $\left[5 = 1 - 6 + c\right]$ | **DM1** | Must get numerical expression for $c$; if no substitution seen, $c$ must be correct for their expression for $v$. Their 5 must not be a made up value. |
| $(v=) t^{\frac{3}{2}} - 6t^{\frac{1}{2}} + 10$ | **A1** | OE, complete (possibly unsimplified) expression. |
---
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to integrate their $v$ for either section | **\*M1** | Increase power by 1 and change in coefficient in at least one term (same term); $s = vt$ is M0. Their $v$'s must come from integration. |
| $(s_1=) -\frac{1}{\frac{5}{2}}t^{\frac{5}{2}} + 6t[+c] = -\frac{2}{5}t^{\frac{5}{2}} + 6t[+c]$ $(s_2=) \frac{1}{\frac{5}{2}}t^{\frac{5}{2}} - \frac{6}{\frac{3}{2}}t^{\frac{3}{2}} + 10t[+c] = \frac{2}{5}t^{\frac{5}{2}} - 4t^{\frac{3}{2}} + 10t[+c]$ | **A1** | For either correct. Allow unsimplified. |
| Use limits 0 and 1 for $s_1$ **and** 1 and 4 for $s_2$ | **DM1** | Using correct limits correctly in expressions from integration; would lead to 2 positive values. $s_1$ and $s_2$ must have correct number of non-constant terms and include a linear term. |
| Total distance $= 20\ \text{m}$ | **A1** | $s_1 = \left(-\frac{2}{5}\times1 + 6\times1\right) - 0 = \frac{28}{5} = 5.6$; $s_2 = \left(\frac{2}{5}\times4^{\frac{5}{2}} - 4\times4^{\frac{3}{2}} + 10\times4\right) - \left(\frac{2}{5}\times1^{\frac{5}{2}} - 4\times1^{\frac{3}{2}} + 10\times1\right) = \frac{72}{5} = 14.4$ |
**SC for integration not seen:**
| Answer | Mark |
|--------|------|
| $s_1 = \frac{28}{5}$ or $s_2 = \frac{72}{5}$ | **B1** |
| Total distance $= 20\ \text{m}$ | **B1** |
---6 A particle, $P$, travels in a straight line, starting from a point $O$ with velocity $6 \mathrm {~ms} ^ { - 1 }$. The acceleration of $P$ at time $t \mathrm {~s}$ after leaving $O$ is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$, where
$$\begin{array} { l l }
a = - 1.5 t ^ { \frac { 1 } { 2 } } & \text { for } 0 \leqslant t \leqslant 1 , \\
a = 1.5 t ^ { \frac { 1 } { 2 } } - 3 t ^ { - \frac { 1 } { 2 } } & \text { for } t > 1 .
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of $P$ at $t = 1$.
\item Given that there is no change in the velocity of $P$ when $t = 1$, find an expression for the velocity of $P$ for $t > 1$.\\
\includegraphics[max width=\textwidth, alt={}, center]{145d93bd-7f56-4e8c-a646-938330511347-11_2725_35_99_20}
\item Given that the velocity of $P$ is positive for $t \leqslant 4$, find the total distance travelled between $t = 0$ and $t = 4$.\\
\includegraphics[max width=\textwidth, alt={}, center]{145d93bd-7f56-4e8c-a646-938330511347-12_723_762_248_653}
Two particles, $A$ and $B$, of masses 0.2 kg and 0.3 kg respectively, are attached to the ends of a light inextensible string. The string passes over a small fixed smooth pulley which is attached to the bottom of a rough plane inclined at an angle $\theta$ to the horizontal where $\sin \theta = 0.6$. Particle $A$ lies on the plane, and particle $B$ hangs vertically below the pulley, 0.25 m above horizontal ground. The string between $A$ and the pulley is parallel to a line of greatest slope of the plane (see diagram). The coefficient of friction between $A$ and the plane is 1.125 . Particle $A$ is released from rest.\\
(a) Find the tension in the string and the magnitude of the acceleration of the particles.\\
\includegraphics[max width=\textwidth, alt={}, center]{145d93bd-7f56-4e8c-a646-938330511347-12_2716_38_109_2012}\\
(b) When $B$ reaches the ground, it comes to rest.
Find the total distance that $A$ travels down the plane from when it is released until it comes to rest. You may assume that $A$ does not reach the pulley.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.\\
\includegraphics[max width=\textwidth, alt={}, center]{145d93bd-7f56-4e8c-a646-938330511347-14_2715_31_106_2016}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2024 Q6 [10]}}