| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Minimum/maximum force for equilibrium |
| Difficulty | Standard +0.3 This is a standard M1 friction problem requiring resolution of forces in two directions and application of friction laws. Part (a) is routine verification (3 marks), while part (b) requires understanding limiting friction in both directions but follows a well-established method taught in all M1 courses. The 3-4-5 triangle simplifies calculations significantly. |
| Spec | 3.03e Resolve forces: two dimensions3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks |
|---|---|
| 3(a) | R |
| Answer | Marks |
|---|---|
| (i.e. verification that F = 0 => X =14.7) | M1 A1 |
| so F = 0* oe | A1* |
| Answer | Marks |
|---|---|
| 3(a) | M1 Equation in F only, correct no of terms, condone sign errors and |
| Answer | Marks |
|---|---|
| 3(b) | F = 0.5S |
| 1 | B1 |
| Answer | Marks |
|---|---|
| 1 | M1A2 |
| Answer | Marks |
|---|---|
| X = 4g/11, 3.6 or 3.56 or 3.57 | A1 |
| Answer | Marks |
|---|---|
| 3(b) | B1 F = 0.5S seen e.g. on a diagram (even if wrong direction) |
Question 3:
--- 3(a) ---
3(a) | R
F
14.7
2g
( ),14.7cos=2gsin+F (could be – F)
( ) , 1 4 . 7 F c o s R s i n → + =
OR: AND eliminate R to give an
( ) , R c o s F s i n 2 g + =
equation in F only.
Verificaton methods
1 4 . 7 c o s ( 1 1 . 7 6 ) 2 g s i n = =
(i.e. verification that X = 14.7 => F = 0)
OR: X c o s 2 g s i n = => X = 14.7
(i.e. verification that F = 0 => X =14.7) | M1 A1
so F = 0* oe | A1*
(3)
S
F
1
X
2g
3(a) | M1 Equation in F only, correct no of terms, condone sign errors and
sin/cos confusion (M0 if they use F = 0.5R)
N.B. Allow the equation without F
Allow use of m instead of 2 for the Mmark
A1 Correct equation
A1* cao Must state a conclusion or ,
if verifying, must state clearly
X = 14.7 => F = 0 OR F = 0 => X = 14.7
--- 3(b) ---
3(b) | F = 0.5S
1 | B1
Two equations taken from:
( ),X cos+F =2gsin
1
( ),S = X sin+2gcos
(→),X +F cos= Ssin
1
(),Scos+F sin=2g
1 | M1A2
M1A2
N.B. M0 for both equations if they put X = 14.7 anywhere
X = 4g/11, 3.6 or 3.56 or 3.57 | A1
N.B. Enter marks for the equations on ePen in the order in which
they appear above.
(8)
(11)
Notes for question 3
3(b) | B1 F = 0.5S seen e.g. on a diagram (even if wrong direction)
1
M1 A resolution, correct no of terms, condone sign errors and sin/cos
confusion
Allow use of m instead of 2 for the A mark
A2 Correct equation, -1 each error
M1 A resolution, correct no of terms, condone sign errors and sin/cos
confusion
Allow use of m instead of 2 for the A mark
A2 Correct equation, -1 each error
A1 cao\includegraphics{figure_2}
A rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac{3}{4}$
A particle $P$ of mass 2 kg is held in equilibrium on the plane by a horizontal force of magnitude $X$ newtons, as shown in Figure 2. The force acts in a vertical plane which contains a line of greatest slope of the inclined plane.
\begin{enumerate}[label=(\alph*)]
\item Show that when $X = 14.7$ there is no frictional force acting on $P$ [3]
The coefficient of friction between $P$ and the plane is 0.5
\item Find the smallest possible value of $X$. [8]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2022 Q3 [11]}}