| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion down rough slope |
| Difficulty | Moderate -0.3 This is a straightforward two-part mechanics problem applying standard SUVAT equations and Newton's second law on an inclined plane. Part (a) requires energy conservation or kinematics with constant acceleration (a = g sin 12°). Part (b) adds friction (μ = 0.03) requiring resolution of forces and calculation of net acceleration, then using SUVAT to find time. Both parts follow standard textbook procedures with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = 2.08\) \([2.07911...]\) | B1 | From \(mg\sin 12 = ma\). Allow exact (e.g. \(a = g\sin 12\)) |
| \(v^2 = 5^2 + 2a \times 60\) | M1 | For use of \(v^2 = u^2 + 2as\) with \(u = 5\) and \(s = 60\). Allow sign errors but \(a\) must be either \(g\sin 12\) or \(g\cos 12\) only |
| Speed \(= 16.6\text{ ms}^{-1}\) \([16.567861...]\) | A1 | AWRT 16.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For attempt at work energy equation | (M1) | 3 terms, dimensionally correct. Allow sign errors; allow sin/cos mix on PE term – condone \(m\) missing from all terms. Must be a weight component |
| \(\frac{1}{2}mv^2 = \frac{1}{2}m \times 5^2 + mg \times 60\sin 12\) | (A1) | Correct equation. (for reference: \(60\sin 12 = 12.4747...\)) |
| Speed \(= 16.6\text{ ms}^{-1}\) \([16.567861...]\) | (A1) | AWRT 16.6 |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = mg\cos 12\) | B1 | Resolving correctly perpendicular to the plane |
| \(mg\sin 12 - F = ma\) | *M1 | Use of Newton's second law, correct number of terms; allow sign errors; allow sin/cos mix (must be a weight component) |
| For use of \(F = 0.03R\) to get equation in \(a\) (and \(m\)) only | DM1 | Where \(R\) is a component of weight only (dimensionally correct but allow sin/cos mix). \(\left[\Rightarrow a = 1.79\ [1.78567....]\right]\) |
| \(60 = 5t + \frac{1}{2}at^2\) and solve for \(t\) | DM1 | Dependent on previous two M marks. For use of \(s = ut + \frac{1}{2}at^2\) with \(s = 60\), \(u = 5\) and *their* \(a\) or other complete method to find positive value(s) of \(t\) |
| Time \(= 5.86\text{ s}\) \([5.86260...]\) | A1 | AWRT 5.86 (from using \(a = 1.79\) or better). AWRT 5.85 (from using \(a = 1.8\)). AWRT 5.87 (from correct working) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R = mg\cos 12\) | B1 | Resolving correctly perpendicular to the plane. |
| \(\frac{1}{2}mv^2 - \frac{1}{2}m \times 5^2 = 60 \times mg\sin 12 - 60 \times F\) | \*M1 | Use of work-energy principle, correctly number of relevant terms; allow sign errors; allow sin/cos mix on PE term (must be a weight component). |
| For use of \(F = 0.03R\) to get equation in \(v\) (and \(m\)) only | DM1 | Where \(R\) is a component of weight only (dimensionally correct but allow sin/cos mix). \(\left[\Rightarrow v = 15.5\ [15.46870\ldots]\right]\) |
| \(60 = \frac{1}{2}(5+v)t\) and solve for \(t\) | DM1 | Dependent on previous two M marks. Use of \(s = \frac{1}{2}(u+v)t\) with \(s=60\), \(u=5\) and *their* \(v\) or other complete method to find positive value(s) of \(t\). |
| Time \(= 5.86\text{s}\ [5.86260\ldots]\) | A1 | |
| 5 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = 2.08$ $[2.07911...]$ | B1 | From $mg\sin 12 = ma$. Allow exact (e.g. $a = g\sin 12$) |
| $v^2 = 5^2 + 2a \times 60$ | M1 | For use of $v^2 = u^2 + 2as$ with $u = 5$ and $s = 60$. Allow sign errors but $a$ must be either $g\sin 12$ or $g\cos 12$ only |
| Speed $= 16.6\text{ ms}^{-1}$ $[16.567861...]$ | A1 | AWRT 16.6 |
**Alternative Method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| For attempt at work energy equation | (M1) | 3 terms, dimensionally correct. Allow sign errors; allow sin/cos mix on PE term – condone $m$ missing from all terms. Must be a weight component |
| $\frac{1}{2}mv^2 = \frac{1}{2}m \times 5^2 + mg \times 60\sin 12$ | (A1) | Correct equation. (for reference: $60\sin 12 = 12.4747...$) |
| Speed $= 16.6\text{ ms}^{-1}$ $[16.567861...]$ | (A1) | AWRT 16.6 |
| | **3** | |
---
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = mg\cos 12$ | B1 | Resolving correctly perpendicular to the plane |
| $mg\sin 12 - F = ma$ | *M1 | Use of Newton's second law, correct number of terms; allow sign errors; allow sin/cos mix (must be a weight component) |
| For use of $F = 0.03R$ to get equation in $a$ (and $m$) only | DM1 | Where $R$ is a component of weight only (dimensionally correct but allow sin/cos mix). $\left[\Rightarrow a = 1.79\ [1.78567....]\right]$ |
| $60 = 5t + \frac{1}{2}at^2$ and solve for $t$ | DM1 | Dependent on previous two M marks. For use of $s = ut + \frac{1}{2}at^2$ with $s = 60$, $u = 5$ and *their* $a$ or other complete method to find positive value(s) of $t$ |
| Time $= 5.86\text{ s}$ $[5.86260...]$ | A1 | AWRT 5.86 (from using $a = 1.79$ or better). AWRT 5.85 (from using $a = 1.8$). AWRT 5.87 (from correct working) |
## Question 5(b): Alternative method using energy
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = mg\cos 12$ | **B1** | Resolving correctly perpendicular to the plane. |
| $\frac{1}{2}mv^2 - \frac{1}{2}m \times 5^2 = 60 \times mg\sin 12 - 60 \times F$ | **\*M1** | Use of work-energy principle, correctly number of relevant terms; allow sign errors; allow sin/cos mix on PE term (must be a weight component). |
| For use of $F = 0.03R$ to get equation in $v$ (and $m$) only | **DM1** | Where $R$ is a component of weight only (dimensionally correct but allow sin/cos mix). $\left[\Rightarrow v = 15.5\ [15.46870\ldots]\right]$ |
| $60 = \frac{1}{2}(5+v)t$ and solve for $t$ | **DM1** | Dependent on previous two M marks. Use of $s = \frac{1}{2}(u+v)t$ with $s=60$, $u=5$ and *their* $v$ or other complete method to find positive value(s) of $t$. |
| Time $= 5.86\text{s}\ [5.86260\ldots]$ | **A1** | |
| | **5** | |
---5 A straight slope of length 60 m is inclined at an angle of $12 ^ { \circ }$ to the horizontal. A bobsled starts at the top of the slope with a speed of $5 \mathrm {~ms} ^ { - 1 }$. The bobsled slides directly down the slope.
\begin{enumerate}[label=(\alph*)]
\item It is given that there is no resistance to the bobsled's motion.
Find its speed when it reaches the bottom of the slope.
\item It is given instead that the coefficient of friction between the bobsled and the slope is 0.03 .
Find the time that it takes for the bobsled to reach the bottom of the slope.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2024 Q5 [8]}}