| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle suspended by strings |
| Difficulty | Moderate -0.3 This is a standard two-string equilibrium problem requiring resolution of forces in two perpendicular directions and solving simultaneous equations. While it involves multiple steps (resolving horizontally and vertically, handling trigonometry with given angles), it's a textbook M1 exercise with no novel insight required. The 8 marks reflect the working steps rather than conceptual difficulty, making it slightly easier than average for A-level. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks |
|---|---|
| 2 | First M1 for resolving horizontally with correct no. of terms and both T |
| Answer | Marks |
|---|---|
| 2 alt 1 | See Alternative 1 using a Triangle of Forces and the Sine Rule. |
| 2 alt 2 | Alternative 2 is to resolve perpendicular to each string: |
| Answer | Marks |
|---|---|
| 3. | R |
| Answer | Marks |
|---|---|
| 3 | M1A1 |
Question 2:
2 | First M1 for resolving horizontally with correct no. of terms and both T
A
and T terms resolved.
B
First A1 for a correct equation.
Second M1 for resolving vertically with correct no. of terms and both T
A
and T terms resolved.
B
Second A1 for a correct equation.
Third M1, dependent on first two M marks, for eliminating T or T
A B
Third A1 for a correct equation in one unknown
Fourth A1 for T = 8.4 (N) or better.
A
Fifth A1 for T = 7.6 (N) or better.
B
N.B. The first two M marks can be for two resolutions in any two
directions.
N.B. If the two tensions are taken to be equal, can score max M1A0 for
vertical resolution.
2 alt 1 | See Alternative 1 using a Triangle of Forces and the Sine Rule.
2 alt 2 | Alternative 2 is to resolve perpendicular to each string:
The scheme is similar to Alt 1 and gives the same expressions for T and
A
T
B
M1A1 resolving perp to both strings as a complete method.
M1A1A1 for finding T
A
M1A1A1 for finding T
B
3. | R
T
T
A
B
F
4g
2g
30°
Equation of motion of B: 4g−T =4a
Equation of motion of A: T −F −2gsin30=2a
OR: 4g−F −2gsin30=6a
Resolve perpendicular to the plane at A: R=2gcos30
1
Use of F =µR : F = ×2gcos30(= g)
3
T −g−g =T −2g =2a
8g
2T −4g =4g−T , 3T =8g , T = (≈26) 26.1(N)
3 | M1A1
M1A2
B1
M1
DM1A1
(9)
[9]
Notes for Question 3\includegraphics{figure_1}
A particle of weight 8 N is attached at $C$ to the ends of two light inextensible strings $AC$ and $BC$. The other ends, $A$ and $B$, are attached to a fixed horizontal ceiling. The particle hangs at rest in equilibrium, with the strings in a vertical plane. The string $AC$ is inclined at 35° to the horizontal and the string $BC$ is inclined at 25° to the horizontal, as shown in Figure 1. Find
\begin{enumerate}[label=(\roman*)]
\item the tension in the string $AC$,
\item the tension in the string $BC$.
\end{enumerate}
[8]
\hfill \mbox{\textit{Edexcel M1 2013 Q2 [8]}}