| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: equilibrium (find unknowns) |
| Difficulty | Standard +0.3 This is a multi-part mechanics question requiring standard techniques: F=ma, vector resolution, friction in equilibrium, and SUVAT equations. While it has 5 parts and requires careful bookkeeping, each individual step uses routine M1 methods without requiring novel insight or complex problem-solving. The vector notation and multi-stage scenario make it slightly above average difficulty for M1, but it remains a straightforward application question. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03r Friction: concept and vector form3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 0.6 \approx 0.6\text{ m s}^{-2}\) | M1, A1 | Use of N2L. Allow \(F=mga\). Accept no reference to direction. |
| Answer | Marks | Guidance |
|---|---|---|
| \(0\text{ N}\); \(-\)i direction | B1, B1 | Allow correct description or arrow. [Accept '\(-150\) in i direction' for B1 B1] |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 786.4017\ldots \approx 786\textbf{i}\text{ N}\) | M1, B1, A1, M1, A1, A1 | Resolution of both terms attempted. Allow \(\sin\leftrightarrow\cos\) if in both terms. Allow \(250\) or \(250g\) present. \(300\sin40\) or \(450\sin\theta\). Accept \(\pm\). Accept answer rounding to \(25.5\). Allow SC1 if seen in this part. Proper resolution attempted of \(450\) and \(300\). Allow \(\sin\leftrightarrow\cos\) if in both terms. Accept use of their \(\theta\) or just \(\theta\). Either resolution correct. Accept their \(\theta\) or just \(\theta\). Accept \(\sin/\cos\) consistent with use for perpendicular to i. Accept no reference to direction cao. Allow SC1 WW. |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 661.4017\ldots \approx 661\text{ N}\) | M1, A1, M1, A1, E1 | Appropriate (sequence of) suvat. [WW M0 A0]. Use of \(F=ma\) with their \(786.4\) and their \(a\). No extra forces. Allow sign errors. All correct using their \(786.4\) and \(a\). Use of N2L clearly shown. (Accept \(0.5\) used WW) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = 1.5 \approx 1.5\text{ m s}^{-1}\) | M1, F1, M1, F1, A1 | Use of \(F=ma\) with their values. Allow 1 force missing. FT only their \(786\ldots\) and their \(661\ldots\). Appropriate (sequence of) suvat with \(u\neq 0\). Must be 'new' \(a\) obtained by using N2L. Only FT use of \(\pm\) their \(a_1\). cao. |
## Question 5:
### Part (i)
**i** direction: $150 = 250a$
$a = 0.6 \approx 0.6\text{ m s}^{-2}$ | M1, A1 | Use of N2L. Allow $F=mga$. Accept no reference to direction.
### Part (ii)
$0\text{ N}$; $-$**i** direction | B1, B1 | Allow correct description or arrow. [Accept '$-150$ in **i** direction' for B1 B1]
### Part (iii)
For force only in direction perpendicular to **i**:
$300\sin40 = 450\sin\theta$
$\theta = 25.37300\ldots \approx 25.4°$
In **i** direction:
$300\cos40 + 150 + 450\cos\theta$
$= 786.4017\ldots \approx 786\textbf{i}\text{ N}$ | M1, B1, A1, M1, A1, A1 | Resolution of both terms attempted. Allow $\sin\leftrightarrow\cos$ if in both terms. Allow $250$ or $250g$ present. $300\sin40$ or $450\sin\theta$. Accept $\pm$. Accept answer rounding to $25.5$. Allow SC1 if seen in this part. Proper resolution attempted of $450$ and $300$. Allow $\sin\leftrightarrow\cos$ if in both terms. Accept use of their $\theta$ or just $\theta$. Either resolution correct. Accept their $\theta$ or just $\theta$. Accept $\sin/\cos$ consistent with use for perpendicular to **i**. Accept no reference to direction cao. Allow SC1 WW.
### Part (iv)
Using $s = ut + 0.5at^2$:
$1 = 0.5a \times 2^2$
$a = 0.5$
Using N2L in **i** direction:
$786.4017\ldots - F = 250 \times 0.5$
$= 661.4017\ldots \approx 661\text{ N}$ | M1, A1, M1, A1, E1 | Appropriate (sequence of) suvat. [WW M0 A0]. Use of $F=ma$ with their $786.4$ and their $a$. No extra forces. Allow sign errors. All correct using their $786.4$ and $a$. Use of N2L clearly shown. (Accept $0.5$ used WW)
### Part (v)
Using N2L in **i** direction:
**Either:**
$125 - 200 = 250a_1$
**Or** (starting again):
$786.4017\ldots - (200 + 661.4017\ldots) = 250a_1$
so $a_1 = -0.3$
Using $v^2 = u^2 + 2a_1s$:
$v^2 = 1.8^2 + 2\times(-0.3)\times1.65$
$v = 1.5 \approx 1.5\text{ m s}^{-1}$ | M1, F1, M1, F1, A1 | Use of $F=ma$ with their values. Allow 1 force missing. FT only their $786\ldots$ and their $661\ldots$. Appropriate (sequence of) suvat with $u\neq 0$. Must be 'new' $a$ obtained by using N2L. Only FT use of $\pm$ their $a_1$. cao.5 A cylindrical tub of mass 250 kg is on a horizontal floor. Resistance to its motion other than that due to friction is negligible.
The first attempt to move the tub is by pulling it with a force of 150 N in the $\mathbf { i }$ direction, as shown in Fig. 8.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5a1895e1-abe3-4739-876a-f19458f0f6ed-4_310_1349_451_435}
\captionsetup{labelformat=empty}
\caption{Fig. 5.1}
\end{center}
\end{figure}
(i) Calculate the acceleration of the tub if friction is ignored.
In fact, there is friction and the tub does not move.\\
(ii) Write down the magnitude and direction of the frictional force opposing the pull.
Two more forces are now added to the 150 N force in a second attempt to move the tub, as shown in Fig. 8.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5a1895e1-abe3-4739-876a-f19458f0f6ed-4_497_927_1350_646}
\captionsetup{labelformat=empty}
\caption{Fig. 5.2}
\end{center}
\end{figure}
Angle $\theta$ is acute and chosen so that the resultant of the three forces is in the $\mathbf { i }$ direction.\\
(iii) Determine the value of $\theta$ and the resultant of the three forces.
With this resultant force, the tub moves with constant acceleration and travels 1 metre from rest in 2 seconds.\\
(iv) Show that the magnitude of the friction acting on the tub is 661 N , correct to 3 significant figures.
When the speed of the tub is $1.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, it comes to a part of the floor where the friction on the tub is 200 N greater. The pulling forces stay the same.\\
(v) Find the velocity of the tub when it has moved a further 1.65 m .
\hfill \mbox{\textit{OCR MEI M1 Q5 [20]}}