| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Two-part friction scenarios |
| Difficulty | Moderate -0.3 This is a standard two-part friction problem requiring application of F=μR in equilibrium. Part (i) is straightforward substitution. Part (ii) involves resolving forces at an angle and solving simultaneous equations, which is routine M1 content requiring no novel insight, making it slightly easier than average. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(F = 5\) and \(F = \mu R\) | |
| \(R = mg\) | M1 | |
| \(m = 2.55\) | A1 | 3 Accept 2.5 or 2.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P\cos\alpha = 6\) | B1 | |
| M1 | For resolving vertically with 3 distinct forces | |
| \(R = P\sin\alpha + 25\) | A1ft | Or \(P\sin\alpha + (\text{cv}\,m)g\) |
| \(0.2R = 6\) | B1 | For using \(F = 6\) and \(F = \mu R\) |
| \(0.2(P\sin\alpha + 25) = 6\) | M1 | For an equation in \(P\sin\alpha\) (\(=5\)) after elimination of \(R\) |
| \(\alpha = 39.8°\) | A1 | Accept a.r.t \(40°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P^2 = 6^2 + 5^2\) or \(P\cos 39.8° = 6\) or \(P\sin 39.8° = 5\) | M1 | For eliminating or substituting for \(\alpha\) with cv(6). Evidence needed that 5 is value of \(P\sin\alpha\) rather than original frictional force |
| \(P = 7.81\) | A1 | 8 Accept a.r.t 7.8 |
# Question 5:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $F = 5$ and $F = \mu R$ |
| $R = mg$ | M1 | |
| $m = 2.55$ | A1 | **3** Accept 2.5 or 2.6 |
## Part (ii)a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P\cos\alpha = 6$ | B1 | |
| | M1 | For resolving vertically with 3 distinct forces |
| $R = P\sin\alpha + 25$ | A1ft | Or $P\sin\alpha + (\text{cv}\,m)g$ |
| $0.2R = 6$ | B1 | For using $F = 6$ and $F = \mu R$ |
| $0.2(P\sin\alpha + 25) = 6$ | M1 | For an equation in $P\sin\alpha$ ($=5$) after elimination of $R$ |
| $\alpha = 39.8°$ | A1 | Accept a.r.t $40°$ |
## Part (ii)b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P^2 = 6^2 + 5^2$ or $P\cos 39.8° = 6$ or $P\sin 39.8° = 5$ | M1 | For eliminating or substituting for $\alpha$ with cv(6). Evidence needed that 5 is value of $P\sin\alpha$ rather than original frictional force |
| $P = 7.81$ | A1 | **8** Accept a.r.t 7.8 |
---5 A block of mass $m \mathrm {~kg}$ is at rest on a horizontal plane. The coefficient of friction between the block and the plane is 0.2 .\\
(i) When a horizontal force of magnitude 5 N acts on the block, the block is on the point of slipping. Find the value of $m$.\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{8ee41313-b516-48cb-bc87-cd8e54245d28-3_312_711_1244_758}
When a force of magnitude $P \mathrm {~N}$ acts downwards on the block at an angle $\alpha$ to the horizontal, as shown in the diagram, the frictional force on the block has magnitude 6 N and the block is again on the point of slipping. Find
\begin{enumerate}[label=(\alph*)]
\item the value of $\alpha$ in degrees,
\item the value of $P$.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2006 Q5 [11]}}