| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Two-part friction scenarios |
| Difficulty | Standard +0.3 This is a standard M1 friction question with straightforward resolution of forces and application of F=μR. Part (i) is a simple 'show that' for coefficient of friction. Parts (ii) and (iii) require resolving forces at angles and applying Newton's second law, but follow routine procedures without requiring novel insight. Slightly above average difficulty due to the three-part structure and need to consider different force configurations. |
| Spec | 3.03r Friction: concept and vector form3.03s Contact force components: normal and frictional3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu = 1/3\) | AG | Uses \(F = \mu R\); Allow 0.333 or 0.3 recurring |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = 4.08(333...)\) N [or 49/12 N] | M1, A1, A1, M1, A1 | 3 force vertical equation; Accept 12.2 or 12.3; Uses \(F = \mu R\) with new R (may be seen in part b) |
| Answer | Marks |
|---|---|
| \(m = 14.7/9.8 = 1.5\) kg | B1, M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Friction = 2.45 N | A1, A2, [5], B1, B1, M1, A1 | N2L horizontally with 2 relevant forces, including \(4.9\sin\cos30°\); Allow cv(F) \(SR\) Award A1 if m=14.7 used \(SR\) A1 for 0.11, 0.109 or at 0.011 from m = 14.7; 3.49, accept 3.5; 2.45, accept 2.4 or 2.5; Comparing two values; Not 2.4 or 2.5; Explicit (M1 essential) |
### Part i
$4.9 = \mu \times 14.7$
$\mu = 1/3$ | AG | Uses $F = \mu R$; Allow 0.333 or 0.3 recurring
### Part iia
$R + 4.9\sin30° = 14.7$
$R = 12.25$ N
$F = 12.25 \times 1/3$
$F = 4.08(333...)$ N [or 49/12 N] | M1, A1, A1, M1, A1 | 3 force vertical equation; Accept 12.2 or 12.3; Uses $F = \mu R$ with new R (may be seen in part b)
### Part iib
$m = 14.7/9.8 = 1.5$ kg | B1, M1 |
### Part iii
$4.9\cos30° - 4.08(333...) = 1.5a$
$a = 0.107$ ms$^{-2}$
$\mu R = (14.7 - 4.9\cos30°)/3$
Horizontal component of force = $4.9\sin30°$
Horizontal component of force < $\otimes R$
Friction = 2.45 N | A1, A2, [5], B1, B1, M1, A1 | N2L horizontally with 2 relevant forces, including $4.9\sin\cos30°$; Allow cv(F) $SR$ Award A1 if m=14.7 used $SR$ A1 for 0.11, 0.109 or at 0.011 from m = 14.7; 3.49, accept 3.5; 2.45, accept 2.4 or 2.5; Comparing two values; Not 2.4 or 2.5; Explicit (M1 essential)
---6 A block of weight 14.7 N is at rest on a horizontal floor. A force of magnitude 4.9 N is applied to the block.\\
(i) The block is in limiting equilibrium when the 4.9 N force is applied horizontally. Show that the coefficient of friction is $\frac { 1 } { 3 }$.\\
(ii)\\
\begin{tikzpicture}
% Floor (hatched)
\fill[pattern=north east lines] (-1.5,-0.15) rectangle (3.5,0);
\draw[thick] (-1.5,0) -- (3.5,0);
% Block
\draw[thick] (0,0) rectangle (2,1);
% Dashed horizontal line from top-right corner of block
\draw[dashed] (2,1) -- (3.8,1);
% Force arrow at 30 above horizontal from top-right corner
\draw[thick,->] (2,1) -- ++(30:2.2) node[right] {4.9\,N};
% Angle arc
\draw (2.8,1) arc[start angle=0, end angle=30, radius=0.8];
\node at (2.95,1.25) {$30°$};
\end{tikzpicture}
When the force of 4.9 N is applied at an angle of $30 ^ { \circ }$ above the horizontal, as shown in the diagram, the block moves across the floor. Calculate
\begin{enumerate}[label=(\alph*)]
\item the vertical component of the contact force between the floor and the block, and the magnitude of the frictional force,
\item the acceleration of the block.\\
(iii) Calculate the magnitude of the frictional force acting on the block when the 4.9 N force acts at an angle of $30 ^ { \circ }$ to the upward vertical, justifying your answer fully.
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2008 Q6 [16]}}