| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle with string at angle to wall |
| Difficulty | Moderate -0.3 This is a standard M1 equilibrium problem with straightforward geometry (3-4-5 triangle) and resolving forces in two perpendicular directions. Part (i) requires one resolution equation, part (ii) another. The kinematics problem is also routine suvat with constant acceleration phases. Both parts are textbook exercises requiring method application rather than problem-solving insight, making this slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03a Force: vector nature and diagrams3.03b Newton's first law: equilibrium3.03f Weight: W=mg |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sin\theta = \frac{3}{5}\) or \(\cos\theta = \frac{4}{5}\) or \(\tan\theta = \frac{3}{4}\) | B1 | Any correct trig ratio seen |
| \(R(\rightarrow): F\cos\theta = 16\sin\theta\) e.g. \(F = 16\tan\theta\) | M1A1 | 1st equation |
| \(R(\nearrow): F = mg\sin\theta\) | ||
| \(R(\uparrow): mg = 16\cos\theta + F\sin\theta\) | ||
| \(R(\swarrow): 16 = mg\cos\theta\) | M1A1 | 2nd equation |
| \((mg)^2 = F^2 + 16^2\) (Pythagoras from triangle of forces) | ||
| \(F = 12\) | A1 (i) | A0 if 12 from rounding inaccurate answer. If \(F=12\) given without rounding evidence, award BOD A1 |
| \(m = 2.04\) or \(2.0\) | A1 (ii)[7] | A0 for 2 |
# Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sin\theta = \frac{3}{5}$ or $\cos\theta = \frac{4}{5}$ or $\tan\theta = \frac{3}{4}$ | B1 | Any correct trig ratio seen |
| $R(\rightarrow): F\cos\theta = 16\sin\theta$ e.g. $F = 16\tan\theta$ | M1A1 | 1st equation |
| $R(\nearrow): F = mg\sin\theta$ | | |
| $R(\uparrow): mg = 16\cos\theta + F\sin\theta$ | | |
| $R(\swarrow): 16 = mg\cos\theta$ | M1A1 | 2nd equation |
| $(mg)^2 = F^2 + 16^2$ (Pythagoras from triangle of forces) | | |
| $F = 12$ | A1 (i) | A0 if 12 from rounding inaccurate answer. If $F=12$ given without rounding evidence, award BOD A1 |
| $m = 2.04$ or $2.0$ | A1 (ii)[7] | A0 for 2 |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5f2d38d9-b719-4205-8cb0-caa959afc46f-12_540_584_294_680}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A particle $P$ of mass $m \mathrm {~kg}$ is attached to one end of a light inextensible string of length 2.5 m . The other end of the string is attached to a fixed point $A$ on a vertical wall. The tension in the string is 16 N . The particle is held in equilibrium by a force of magnitude $F$ newtons, acting in the vertical plane which is perpendicular to the wall and contains the string. This force acts in a direction perpendicular to the string, as shown in Figure 2.
Given that the horizontal distance of $P$ from the wall is 1.5 m , find\\
(i) the value of $F$,\\
(ii) the value of $m$.
\begin{center}
\end{center}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5f2d38d9-b719-4205-8cb0-caa959afc46f-16_186_830_292_557}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Two posts, $A$ and $B$, are fixed at the side of a straight horizontal road and are 816 m apart, as shown in Figure 3. A car and a van are at rest side by side on the road and level with $A$. The car and the van start to move at the same time in the direction $A B$. The car accelerates from rest with constant acceleration until it reaches a speed of $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car then moves at a constant speed of $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The van accelerates from rest with constant acceleration for 12 s until it reaches a speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The van then moves at a constant speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. When the car has been moving at $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for 30 s , the van draws level with the car at $B$, and each vehicle has then travelled a distance of 816 m .
\begin{enumerate}[label=(\alph*)]
\item Sketch, on the same diagram, a speed-time graph for the motion of each vehicle from $A$ to $B$.
\item Find the time for which the car is accelerating.
\item Find the value of $V$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2018 Q4 [7]}}