| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on smooth incline, particle hanging |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question combining kinematics (finding minimum velocity and distance from a quadratic velocity function) with a two-part pulley system problem involving equilibrium and motion on an incline. All parts use routine techniques: completing the square/differentiation for part (i), integration with attention to direction changes for part (ii), and standard F=ma with resolved forces for the pulley system. The energy method in the final part is a direct application of conservation principles. While multi-step, it requires no novel insight beyond textbook methods. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration3.03n Equilibrium in 2D: particle under forces3.03o Advanced connected particles: and pulleys6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = 2t - 8\) | M1 | Differentiate to find \(a\) |
| \(a = 0 \rightarrow t = 4\) | M1 | Set \(a = 0\) and solve for \(t\) |
| Minimum \(v = -4\) ms\(^{-1}\) | A1 | Full marks available for correct use of a \(v\)-\(t\) graph or correct use of "\(t = -b/2a\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v = (t-4)^2 - 4\) | M1 | Attempt to complete the square for \(v\) |
| \([t = 4]\) | M1 | Choose the \(t\) value which gives minimum \(v\) |
| Minimum \(v = -4\) ms\(^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v = 0\) when \((t-2)(t-6) = 0\) | M1 | Find values of \(t\) when \(v = 0\), factorise or formula |
| \(t = 2\) or \(t = 6\) | A1 | |
| \([s = \frac{1}{3}t^3 - 4t^2 + 12t\ (+c)]\) | M1 | Integrate \(v\) to find \(s\) |
| A1 | Correct integration | |
| \(0 \leq t \leq 2\): \(s_1 = 8/3 - 16 + 24 = 32/3\) | M1 | Attempt to find \(s_1\), \(s_2\) and \(s_3\); look for consideration of the need for 3 intervals; allow use of symmetry when finding \(s_1\) and \(s_3\) |
| \(2 \leq t \leq 6\): \(s_2 = (216/3 - 144 + 72) - (8/3 - 16 + 24) = -32/3\) | ||
| \(6 \leq t \leq 8\): \(s_3 = (512/3 - 4 \times 8^2 + 12 \times 8) - (216/3 - 144 + 72) = 32/3\) | ||
| A1 | 2 correct values of displacement | |
| Total distance \(= 32\) m | A1 | All correct |
## Question 5(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = 2t - 8$ | M1 | Differentiate to find $a$ |
| $a = 0 \rightarrow t = 4$ | M1 | Set $a = 0$ and solve for $t$ |
| Minimum $v = -4$ ms$^{-1}$ | A1 | Full marks available for correct use of a $v$-$t$ graph or correct use of "$t = -b/2a$" |
**Alternative method for question 5(i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $v = (t-4)^2 - 4$ | M1 | Attempt to complete the square for $v$ |
| $[t = 4]$ | M1 | Choose the $t$ value which gives minimum $v$ |
| Minimum $v = -4$ ms$^{-1}$ | A1 | |
## Question 5(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $v = 0$ when $(t-2)(t-6) = 0$ | M1 | Find values of $t$ when $v = 0$, factorise or formula |
| $t = 2$ or $t = 6$ | A1 | |
| $[s = \frac{1}{3}t^3 - 4t^2 + 12t\ (+c)]$ | M1 | Integrate $v$ to find $s$ |
| | A1 | Correct integration |
| $0 \leq t \leq 2$: $s_1 = 8/3 - 16 + 24 = 32/3$ | M1 | Attempt to find $s_1$, $s_2$ and $s_3$; look for consideration of the need for 3 intervals; allow use of symmetry when finding $s_1$ and $s_3$ |
| $2 \leq t \leq 6$: $s_2 = (216/3 - 144 + 72) - (8/3 - 16 + 24) = -32/3$ | | |
| $6 \leq t \leq 8$: $s_3 = (512/3 - 4 \times 8^2 + 12 \times 8) - (216/3 - 144 + 72) = 32/3$ | | |
| | A1 | 2 correct values of displacement |
| Total distance $= 32$ m | A1 | All correct |5 A particle $P$ moves in a straight line from a fixed point $O$. The velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ of $P$ at time $t \mathrm {~s}$ is given by
$$v = t ^ { 2 } - 8 t + 12 \quad \text { for } 0 \leqslant t \leqslant 8$$
(i) Find the minimum velocity of $P$.\\
(ii) Find the total distance travelled by $P$ in the interval $0 \leqslant t \leqslant 8$.\\
\includegraphics[max width=\textwidth, alt={}, center]{555678d3-f37d-4822-a005-de8c6094dc50-12_401_1102_260_520}
Two particles $A$ and $B$, of masses 0.4 kg and 0.2 kg respectively, are connected by a light inextensible string. Particle $A$ is held on a smooth plane inclined at an angle of $\theta ^ { \circ }$ to the horizontal. The string passes over a small smooth pulley $P$ fixed at the top of the plane, and $B$ hangs freely 0.5 m above horizontal ground (see diagram). The particles are released from rest with both sections of the string taut.\\
(i) Given that the system is in equilibrium, find $\theta$.\\
(ii) It is given instead that $\theta = 20$. In the subsequent motion particle $A$ does not reach $P$ and $B$ remains at rest after reaching the ground.
\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string and the acceleration of the system.
\item Find the speed of $A$ at the instant $B$ reaches the ground.
\item Use an energy method to find the total distance $A$ moves up the plane before coming to instantaneous rest.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2019 Q5 [10]}}