| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Find acceleration from distances/times |
| Difficulty | Moderate -0.5 Part (a) is a straightforward SUVAT calculation with u=0, a=2, s=1.44 to find t, requiring only one formula. Part (b) involves resolving forces in two directions and applying F=ma with friction, which is standard M1 content but requires multiple steps. Overall, this is slightly easier than average as the SUVAT part is trivial and the force resolution follows a standard template with given angle components. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03e Resolve forces: two dimensions3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1.44 = 0 + \frac{1}{2} \times 2t^2\) | M1 | For using a complete method which would lead to an equation for finding a value of \(t\) such as \(s = ut + \frac{1}{2}at^2\) with \(u = 0\), \(s = 1.44\) and \(a = 2\) |
| \(t = 1.2\) s | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R = 0.4g - 3 \times \frac{3}{5} = 0.4g - 3\sin 36.9\ [= 2.2]\) | B1 | |
| \(3 \times \frac{4}{5} - F = 3\cos 36.9 - F = 0.4 \times 2\ \ [F = 1.6]\) | M1 | Use Newton's 2nd law, 3 terms, to find \(F\) |
| \(\mu = \dfrac{3\times\frac{4}{5} - 0.4\times 2}{0.4g - 3\times\frac{3}{5}} = \dfrac{1.6}{2.2}\) | M1 | Use of \(\mu = \dfrac{F}{R}\) |
| \(\mu = 0.727\) | A1 | Allow \(\mu = \dfrac{8}{11}\) |
| Total: 4 |
## Question 2(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1.44 = 0 + \frac{1}{2} \times 2t^2$ | M1 | For using a complete method which would lead to an equation for finding a value of $t$ such as $s = ut + \frac{1}{2}at^2$ with $u = 0$, $s = 1.44$ and $a = 2$ |
| $t = 1.2$ s | A1 | |
| **Total: 2** | | |
## Question 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $R = 0.4g - 3 \times \frac{3}{5} = 0.4g - 3\sin 36.9\ [= 2.2]$ | B1 | |
| $3 \times \frac{4}{5} - F = 3\cos 36.9 - F = 0.4 \times 2\ \ [F = 1.6]$ | M1 | Use Newton's 2nd law, 3 terms, to find $F$ |
| $\mu = \dfrac{3\times\frac{4}{5} - 0.4\times 2}{0.4g - 3\times\frac{3}{5}} = \dfrac{1.6}{2.2}$ | M1 | Use of $\mu = \dfrac{F}{R}$ |
| $\mu = 0.727$ | A1 | Allow $\mu = \dfrac{8}{11}$ |
| **Total: 4** | | |2 A particle $P$ of mass 0.4 kg is on a rough horizontal floor. The coefficient of friction between $P$ and the floor is $\mu$. A force of magnitude 3 N is applied to $P$ upwards at an angle $\alpha$ above the horizontal, where $\tan \alpha = \frac { 3 } { 4 }$. The particle is initially at rest and accelerates at $2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the time it takes for $P$ to travel a distance of 1.44 m from its starting point.
\item Find $\mu$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q2 [6]}}