| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Particle on inclined plane |
| Difficulty | Standard +0.3 This is a straightforward SUVAT problem requiring two simultaneous equations from s=ut+½at² applied to two different journeys, followed by standard resolution of forces on an inclined plane (a=g sin θ). All steps are routine applications of standard mechanics formulas with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(1.76 = 0.8u + 0.32a\) | M1 | For using \(s = ut + \frac{1}{2}at^2\) for \(AB\) |
| A1 | ||
| \([1.76 + 2.16 = (0.8 + 0.6)u + \frac{1}{2}(0.8 + 0.6)a\) or \(2.16 = (u + 0.8a)0.6 + \frac{1}{2}0.6^2a]\) | M1 | For using \(s = ut + \frac{1}{2}at^2\) for \(AC\) or \(v = u + at\) for \(AB\) and \(s = ut + \frac{1}{2}at^2\) for \(BC\) |
| \(3.92 = 1.4u + 0.98a\) or \(2.16 = 0.6u + 0.66a\) | A1 | |
| \(u = 1.4\) and \(a = 2\) | M1 | For solving for \(u\) and \(a\) |
| A1 | 6 | |
| (ii) \([2 = 10\sin\theta]\) | M1 | For using \(a = g\sin\theta\) |
| \(\theta = 11.5\) | A1 | 2 |
| **(i)** $1.76 = 0.8u + 0.32a$ | M1 | For using $s = ut + \frac{1}{2}at^2$ for $AB$ |
| | A1 | |
| $[1.76 + 2.16 = (0.8 + 0.6)u + \frac{1}{2}(0.8 + 0.6)a$ or $2.16 = (u + 0.8a)0.6 + \frac{1}{2}0.6^2a]$ | M1 | For using $s = ut + \frac{1}{2}at^2$ for $AC$ or $v = u + at$ for $AB$ and $s = ut + \frac{1}{2}at^2$ for $BC$ |
| $3.92 = 1.4u + 0.98a$ or $2.16 = 0.6u + 0.66a$ | A1 | |
| $u = 1.4$ and $a = 2$ | M1 | For solving for $u$ and $a$ |
| | A1 | 6 |
| **(ii)** $[2 = 10\sin\theta]$ | M1 | For using $a = g\sin\theta$ |
| $\theta = 11.5$ | A1 | 2 |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{28562a1b-ec9a-40d2-bbb3-729770688971-2_449_1273_1829_438}\\
$A , B$ and $C$ are three points on a line of greatest slope of a smooth plane inclined at an angle of $\theta ^ { \circ }$ to the horizontal. $A$ is higher than $B$ and $B$ is higher than $C$, and the distances $A B$ and $B C$ are 1.76 m and 2.16 m respectively. A particle slides down the plane with constant acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The speed of the particle at $A$ is $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see diagram). The particle takes 0.8 s to travel from $A$ to $B$ and takes 1.4 s to travel from $A$ to $C$. Find\\
(i) the values of $u$ and $a$,\\
(ii) the value of $\theta$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{28562a1b-ec9a-40d2-bbb3-729770688971-3_188_510_260_388}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{28562a1b-ec9a-40d2-bbb3-729770688971-3_196_570_255_1187}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A block of mass 2 kg is at rest on a horizontal floor. The coefficient of friction between the block and the floor is $\mu$. A force of magnitude 12 N acts on the block at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 3 } { 4 }$. When the applied force acts downwards as in Fig. 1 the block remains at rest.\\
\hfill \mbox{\textit{CAIE M1 2011 Q4 [8]}}