| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Motion on rough inclined plane |
| Difficulty | Standard +0.3 This is a multi-part mechanics problem involving inclined planes with changing friction conditions and energy methods. While it requires careful bookkeeping across multiple parts and application of SUVAT equations plus energy principles, the individual steps are standard M1 techniques (resolving forces, calculating work/energy, applying kinematics). The problem is slightly above average due to its length and the need to track information across parts, but doesn't require novel insight or particularly sophisticated problem-solving. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Acceleration is \(2.8 \text{ ms}^{-2}\) | B1 | Using acceleration \(= g\sin\alpha\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([mg \times 0.28 - 0.5mg \times 0.96 = ma]\) | M1 | For using Newton's 2nd law |
| Acceleration is \(-2 \text{ ms}^{-2}\) | A1 | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| M1 | For using \(v^2 = u^2 + 2as\) for \(AB\) and for \(BC\) and using \(AB + BC = 5\) | |
| \(v_B^2 = 2 \times 2.8(AB)\) and \(2^2 = 5.6(AB) - 2 \times 2(5 - AB)\) | A1\(\checkmark\) | ft incorrect answers in (i) |
| Distance is \(2.5 \text{ m}\) | A1 | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([mg \times 5 \times 0.28 = \frac{1}{2}m \cdot 2^2 + \mu \times mg \times 0.96 \times BC]\) | M1 | For using Loss in PE \(=\) Gain in KE \(+\) WD against Friction for motion \(A\) to \(C\) |
| \(14 = 2 + 4.8 \times BC\) | A1 | Correct equation |
| \(BC = 12/4.8 = 2.5 \text{ m}\) | A1 | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| M1 | For using \(t = 2s \div (u + v)\) for \(AB\) and \(BC\) | |
| \(T = 2 \times 2.5 \div (0 + \sqrt{14}) + 2 \times 2.5 \div (\sqrt{14} + 2)\) | A1 | |
| Time taken is \(2.21 \text{ s}\) | A1 | 3 marks total |
## Question 6:
### Part (i)(a):
| Working | Mark | Guidance |
|---------|------|----------|
| Acceleration is $2.8 \text{ ms}^{-2}$ | B1 | Using acceleration $= g\sin\alpha$ |
### Part (i)(b):
| Working | Mark | Guidance |
|---------|------|----------|
| $[mg \times 0.28 - 0.5mg \times 0.96 = ma]$ | M1 | For using Newton's 2nd law |
| Acceleration is $-2 \text{ ms}^{-2}$ | A1 | 3 marks total |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| | M1 | For using $v^2 = u^2 + 2as$ for $AB$ and for $BC$ and using $AB + BC = 5$ |
| $v_B^2 = 2 \times 2.8(AB)$ and $2^2 = 5.6(AB) - 2 \times 2(5 - AB)$ | A1$\checkmark$ | ft incorrect answers in (i) |
| Distance is $2.5 \text{ m}$ | A1 | 3 marks total |
*Alternative method for (ii):*
| Working | Mark | Guidance |
|---------|------|----------|
| $[mg \times 5 \times 0.28 = \frac{1}{2}m \cdot 2^2 + \mu \times mg \times 0.96 \times BC]$ | M1 | For using Loss in PE $=$ Gain in KE $+$ WD against Friction for motion $A$ to $C$ |
| $14 = 2 + 4.8 \times BC$ | A1 | Correct equation |
| $BC = 12/4.8 = 2.5 \text{ m}$ | A1 | 3 marks total |
### Part (iii):
| Working | Mark | Guidance |
|---------|------|----------|
| | M1 | For using $t = 2s \div (u + v)$ for $AB$ and $BC$ |
| $T = 2 \times 2.5 \div (0 + \sqrt{14}) + 2 \times 2.5 \div (\sqrt{14} + 2)$ | A1 | |
| Time taken is $2.21 \text{ s}$ | A1 | 3 marks total |
---\begin{enumerate}[label=(\alph*)]
\item the work done against the frictional force acting on $B$,
\item the loss of potential energy of the system,
\item the gain in kinetic energy of the system.
At the instant when $B$ has moved 0.9 m the string breaks. $A$ is at a height of 0.54 m above a horizontal floor at this instant.\\
(ii) Find the speed with which $A$ reaches the floor.\\
$6 \quad A B C$ is a line of greatest slope of a plane inclined at angle $\alpha$ to the horizontal, where $\sin \alpha = 0.28$ and $\cos \alpha = 0.96$. The point $A$ is at the top of the plane, the point $C$ is at the bottom of the plane and the length of $A C$ is 5 m . The part of the plane above the level of $B$ is smooth and the part below the level of $B$ is rough. A particle $P$ is released from rest at $A$ and reaches $C$ with a speed of $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The coefficient of friction between $P$ and the part of the plane below $B$ is 0.5 . Find
\begin{enumerate}[label=(\roman*)]
\item the acceleration of $P$ while moving\\
(a) from $A$ to $B$,\\
(b) from $B$ to $C$,
\item the distance $A B$,
\item the time taken for $P$ to move from $A$ to $C$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2014 Q6 [9]}}