OCR MEI M1 — Question 6 16 marks

Exam BoardOCR MEI
ModuleM1 (Mechanics 1)
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeModelling assumptions and refinements
DifficultyStandard +0.3 This is a straightforward M1 mechanics question involving resolving forces on an inclined plane. Parts (i)-(ii) require simple equilibrium equations with smooth plane assumptions, (iii) is a diagram, and (iv)-(v) add friction and an angled string but still use standard resolution techniques. All steps are routine applications of basic mechanics principles with no novel problem-solving required.
Spec3.03a Force: vector nature and diagrams3.03e Resolve forces: two dimensions3.03f Weight: W=mg3.03i Normal reaction force3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03v Motion on rough surface: including inclined planes
6 An empty open box of mass 4 kg is on a plane that is inclined at \(25 ^ { \circ }\) to the horizontal.
In one model the plane is taken to be smooth. The box is held in equilibrium by a string with tension \(T \mathrm {~N}\) parallel to the plane, as shown in Fig. 6.1. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{5a1895e1-abe3-4739-876a-f19458f0f6ed-5_308_561_559_828} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
\end{figure}
  1. Calculate \(T\). A rock of mass \(m \mathrm {~kg}\) is now put in the box. The system is in equilibrium when the tension in the string, still parallel to the plane, is 50 N .
  2. Find \(m\). In a refined model the plane is rough. The empty box, of mass 4 kg , is in equilibrium when a frictional force of 20 N acts down the plane and the string has a tension of \(P \mathrm {~N}\) inclined at \(15 ^ { \circ }\) to the plane, as shown in Fig. 6.2. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{5a1895e1-abe3-4739-876a-f19458f0f6ed-5_359_559_1599_830} \captionsetup{labelformat=empty} \caption{Fig. 6.2}
    \end{figure}
  3. Draw a diagram showing all the forces acting on the box.
  4. Calculate \(P\).
  5. Calculate the normal reaction of the plane on the box.
Question 6:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Up the plane: \(T - 4g\sin 25 = 0\)M1 Resolving parallel to the plane. If any other direction used, all forces must be present. Accept \(s \leftrightarrow c\). Allow use of \(m\). No extra forces.
\(T = 16.5666...\) so \(T = 16.6\) N (3 s.f.)A1
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Down the plane: \((4+m)g\sin 25 - 50 = 0\)M1 No extra forces. Must attempt resolution in at least 1 term. Accept \(s \leftrightarrow c\). Accept \(Mg\sin 25\). Accept use of mass.
A1Accept \(Mg\sin 25\)
\(m = 8.0724...\) so \(m = 8.07\) (3 s.f.)A1
Part (iii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
DiagramB1 Any 3 of weight, friction, normal reaction and \(P\) present in approx correct directions with arrows.
B1All forces present with suitable directions, labels and arrows. Accept \(W\), \(mg\), \(4g\) and \(39.2\).
Part (iv):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Resolving up the planeM1 Resolving parallel to the plane or all forces must be present. Accept \(s \leftrightarrow c\). Allow use of \(m\). At least one resolution attempted and accept wrong angles. Allow sign errors.
\(P\cos 15\) term correctB1 Allow sign error.
\(P\cos 15 - 20 - 4g\sin 25 = 0\)B1 Both resolutions correct. Weight used. Allow sign errors. FT use of \(P\sin 15\).
A1All correct but FT use of \(P\sin 15\).
\(P = 37.8565...\) so \(P = 37.9\) N (3 s.f.)A1
Part (v):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Resolving perpendicular to the planeM1 May use other directions. All forces present. No extras. Allow \(s \leftrightarrow c\). Weight not mass used. Both resolutions attempted. Allow sign errors.
\(R + P\sin 15 - 4g\cos 25 = 0\)B1 Both resolutions correct. Allow sign errors. Allow use of \(P\cos 15\) if \(P\sin 15\) used in (iv).
F1All correct. Only FT their \(P\) and their use of \(P\cos 15\).
\(R = 25.729...\) so \(R = 25.7\) NA1 cao 
## Question 6:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Up the plane: $T - 4g\sin 25 = 0$ | M1 | Resolving parallel to the plane. If any other direction used, all forces must be present. Accept $s \leftrightarrow c$. Allow use of $m$. No extra forces. |
| $T = 16.5666...$ so $T = 16.6$ N (3 s.f.) | A1 | |

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Down the plane: $(4+m)g\sin 25 - 50 = 0$ | M1 | No extra forces. Must attempt resolution in at least 1 term. Accept $s \leftrightarrow c$. Accept $Mg\sin 25$. Accept use of mass. |
| | A1 | Accept $Mg\sin 25$ |
| $m = 8.0724...$ so $m = 8.07$ (3 s.f.) | A1 | |

### Part (iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Diagram | B1 | Any 3 of weight, friction, normal reaction and $P$ present in approx correct directions with arrows. |
| | B1 | All forces present with suitable directions, labels and arrows. Accept $W$, $mg$, $4g$ and $39.2$. |

### Part (iv):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolving up the plane | M1 | Resolving parallel to the plane or all forces must be present. Accept $s \leftrightarrow c$. Allow use of $m$. At least one resolution attempted and accept wrong angles. Allow sign errors. |
| $P\cos 15$ term correct | B1 | Allow sign error. |
| $P\cos 15 - 20 - 4g\sin 25 = 0$ | B1 | Both resolutions correct. Weight used. Allow sign errors. FT use of $P\sin 15$. |
| | A1 | All correct but FT use of $P\sin 15$. |
| $P = 37.8565...$ so $P = 37.9$ N (3 s.f.) | A1 | |

### Part (v):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolving perpendicular to the plane | M1 | May use other directions. All forces present. No extras. Allow $s \leftrightarrow c$. Weight not mass used. Both resolutions attempted. Allow sign errors. |
| $R + P\sin 15 - 4g\cos 25 = 0$ | B1 | Both resolutions correct. Allow sign errors. Allow use of $P\cos 15$ if $P\sin 15$ used in (iv). |
| | F1 | All correct. Only FT **their** $P$ and their use of $P\cos 15$. |
| $R = 25.729...$ so $R = 25.7$ N | A1 | cao |
6 An empty open box of mass 4 kg is on a plane that is inclined at $25 ^ { \circ }$ to the horizontal.\\
In one model the plane is taken to be smooth.

The box is held in equilibrium by a string with tension $T \mathrm {~N}$ parallel to the plane, as shown in Fig. 6.1.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{5a1895e1-abe3-4739-876a-f19458f0f6ed-5_308_561_559_828}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{center}
\end{figure}

(i) Calculate $T$.

A rock of mass $m \mathrm {~kg}$ is now put in the box. The system is in equilibrium when the tension in the string, still parallel to the plane, is 50 N .\\
(ii) Find $m$.

In a refined model the plane is rough.

The empty box, of mass 4 kg , is in equilibrium when a frictional force of 20 N acts down the plane and the string has a tension of $P \mathrm {~N}$ inclined at $15 ^ { \circ }$ to the plane, as shown in Fig. 6.2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{5a1895e1-abe3-4739-876a-f19458f0f6ed-5_359_559_1599_830}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\end{center}
\end{figure}

(iii) Draw a diagram showing all the forces acting on the box.\\
(iv) Calculate $P$.\\
(v) Calculate the normal reaction of the plane on the box.

\hfill \mbox{\textit{OCR MEI M1  Q6 [16]}}
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