| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Collision on slope |
| Difficulty | Challenging +1.2 This is a multi-part mechanics problem requiring resolution of forces on a slope, limiting equilibrium, kinematics with constant acceleration, collision/coalescence, and energy calculations. While it involves several connected steps and careful bookkeeping across three parts, each individual technique is standard M1 material (resolving forces, SUVAT equations, conservation of momentum, work-energy). The 'show that' part (a) is routine, and parts (b)-(c) require systematic application of standard methods rather than novel insight, making this moderately above average difficulty for A-level. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = 3g\cos\alpha = 3\times10\times0.8\) | B1 | |
| \(F = 3g\sin\alpha = 3\times10\times0.6\) | M1 | Resolving parallel to plane |
| \(\mu = \dfrac{18}{24} = 0.75\) or \(\mu = \dfrac{3g\sin\alpha}{3g\cos\alpha} = \tan\alpha = 0.75\) | A1 | Uses \(\mu = \dfrac{F}{R}\), AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = g\sin\alpha\) or PE loss \(= 1.5gx\sin\alpha\) for \(AB\) and \(a = 0\) for \(BC\); \(4.5g\times\sin\alpha - 0.75\times4.5g\cos\alpha = 4.5a\) leading to \(a = 0\) | B1 | Accelerations for \(AB\) and \(BC\) |
| \(v_1^2 = 2\times g\sin\alpha\times x\) or \([1.5g\times x\sin\alpha = 0.5\times1.5\times v_1^2]\) | M1 | Uses 'suvat' or PE loss = KE gain for \(AB\) |
| \(v_1^2 = 20x\sin\alpha = 12x\) leading to \(v_1 = \sqrt{12x}\) | A1 | |
| \(1.5\times\sqrt{12x} + 0 = 4.5\times v_2\) leading to \(v_2 = \dfrac{1}{3}\sqrt{12x}\) | M1 | Conservation of momentum |
| \(4 = \dfrac{2}{3}\times\sqrt{12x}\) | M1 | Use of \(s = vt\) on \(BC\) since \(a = 0\) |
| \(x = 3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = g\sin\alpha\) or PE loss \(= 1.5gx\sin\alpha\) for \(AB\) and \(a = 0\) for \(BC\); \(4.5g \times \sin\alpha - 0.75 \times 4.5g\cos\alpha = 4.5a\) leading to \(a = 0\) | B1 | Accelerations for \(AB\) and \(BC\) |
| \(4 = 2v_2\) leading to \(v_2 = 2\) | M1 | Uses \(s = vt\) on \(BC\) since \(a = 0\) |
| \(1.5 \times v_1 + 0 = 4.5 \times 2\) | M1 | Conservation of momentum |
| \(v_1 = 6\) | A1 | Velocity before collision |
| \(6^2 = 2 \times g\sin\alpha \times x\) or \(1.5g \times x\sin\alpha = 0.5 \times 1.5 \times 6^2\) | M1 | Uses *suvat* or PE loss \(=\) KE gain for \(AB\) |
| \(x = 3\) | A1 | |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(KE = 0.5 \times 4.5 \times 2^2 = 9\text{ J}\) | B1 | KE gain for \(AC\) |
| PE loss \(= 15 \times (4+3) \times \dfrac{3}{5} + 30 \times 4 \times \dfrac{3}{5} = 135\text{ J}\) | M1 | Evaluates PE loss for \(AC\) |
| Loss of energy \(= 126\text{ J}\) | A1 | |
| 3 |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = 3g\cos\alpha = 3\times10\times0.8$ | B1 | |
| $F = 3g\sin\alpha = 3\times10\times0.6$ | M1 | Resolving parallel to plane |
| $\mu = \dfrac{18}{24} = 0.75$ or $\mu = \dfrac{3g\sin\alpha}{3g\cos\alpha} = \tan\alpha = 0.75$ | A1 | Uses $\mu = \dfrac{F}{R}$, AG |
---
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = g\sin\alpha$ or PE loss $= 1.5gx\sin\alpha$ for $AB$ and $a = 0$ for $BC$; $4.5g\times\sin\alpha - 0.75\times4.5g\cos\alpha = 4.5a$ leading to $a = 0$ | B1 | Accelerations for $AB$ and $BC$ |
| $v_1^2 = 2\times g\sin\alpha\times x$ or $[1.5g\times x\sin\alpha = 0.5\times1.5\times v_1^2]$ | M1 | Uses 'suvat' or PE loss = KE gain for $AB$ |
| $v_1^2 = 20x\sin\alpha = 12x$ leading to $v_1 = \sqrt{12x}$ | A1 | |
| $1.5\times\sqrt{12x} + 0 = 4.5\times v_2$ leading to $v_2 = \dfrac{1}{3}\sqrt{12x}$ | M1 | Conservation of momentum |
| $4 = \dfrac{2}{3}\times\sqrt{12x}$ | M1 | Use of $s = vt$ on $BC$ since $a = 0$ |
| $x = 3$ | A1 | |
## Question 7(b) [Alternative Method]:
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = g\sin\alpha$ or PE loss $= 1.5gx\sin\alpha$ for $AB$ **and** $a = 0$ for $BC$; $4.5g \times \sin\alpha - 0.75 \times 4.5g\cos\alpha = 4.5a$ leading to $a = 0$ | **B1** | Accelerations for $AB$ and $BC$ |
| $4 = 2v_2$ leading to $v_2 = 2$ | **M1** | Uses $s = vt$ on $BC$ since $a = 0$ |
| $1.5 \times v_1 + 0 = 4.5 \times 2$ | **M1** | Conservation of momentum |
| $v_1 = 6$ | **A1** | Velocity before collision |
| $6^2 = 2 \times g\sin\alpha \times x$ **or** $1.5g \times x\sin\alpha = 0.5 \times 1.5 \times 6^2$ | **M1** | Uses *suvat* or PE loss $=$ KE gain for $AB$ |
| $x = 3$ | **A1** | |
| | **6** | |
---
## Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $KE = 0.5 \times 4.5 \times 2^2 = 9\text{ J}$ | **B1** | KE gain for $AC$ |
| PE loss $= 15 \times (4+3) \times \dfrac{3}{5} + 30 \times 4 \times \dfrac{3}{5} = 135\text{ J}$ | **M1** | Evaluates PE loss for $AC$ |
| Loss of energy $= 126\text{ J}$ | **A1** | |
| | **3** | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{4a2bad7c-6720-414c-b336-060afb2255e9-12_560_716_258_712}
Particles of masses 1.5 kg and 3 kg lie on a plane which is inclined at an angle of $\alpha$ to the horizontal, where $\tan \alpha = \frac { 3 } { 4 }$. The section of the plane from $A$ to $B$ is smooth and the section of the plane from $B$ to $C$ is rough. The 1.5 kg particle is held at rest at $A$ and the 3 kg particle is in limiting equilibrium at $B$. The distance $A B$ is $x \mathrm {~m}$ and the distance $B C$ is 4 m (see diagram).
\begin{enumerate}[label=(\alph*)]
\item Show that the coefficient of friction between the particle at $B$ and the plane is 0.75 .\\
The 1.5 kg particle is released from rest. In the subsequent motion the two particles collide and coalesce. The time taken for the combined particle to travel from $B$ to $C$ is 2 s . The coefficient of friction between the combined particle and the plane is still 0.75 .
\item Find $x$.
\item Find the total loss of energy of the particles from the time the 1.5 kg particle is released until the combined particle reaches $C$.\\
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 2022 Q7 [12]}}