| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring resolution of forces in two perpendicular directions to find F and R, followed by a standard application of SUVAT equations or energy methods. The multi-step nature and combination of statics with kinematics places it slightly above average, but all techniques are routine for M1 students. |
| Spec | 3.03b Newton's first law: equilibrium3.03c Newton's second law: F=ma one dimension3.03p Resultant forces: using vectors6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 5(i) | For resolving forces either horizontally or vertically | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| = Rcos15 | 1 | A1 |
| 10sin30 – F sin70 = R sin15 | 1 | A1 |
| 1 | M1 | For solving simultaneously |
| F = 1.90 N and R = 12.4 N | 1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Y = 0.94 F – 5] | 1 | (M1) |
| Answer | Marks | Guidance |
|---|---|---|
| = R2 | 1 | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| = (5 – 0.94F) / (0.342F + 11.34) | 1 | (A1) |
| 1 | (M1) | Solve the tan 15 equation for F and substitute to find R |
| F = 1.90 N and R = 12.4 N | 1 | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | 11.72 = 0 + 2a × 3 | |
| a = 22.815 | 1 | B1 |
| R cos15 = m × 22.815 | 1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Mass of bead = 0.526 kg | 1 | A1 |
Question 5:
--- 5(i) ---
5(i) | For resolving forces either horizontally or vertically | 1 | M1
Fcos70 + 20 – 10 cos 30
= Rcos15 | 1 | A1
10sin30 – F sin70 = R sin15 | 1 | A1
1 | M1 | For solving simultaneously
F = 1.90 N and R = 12.4 N | 1 | A1
Alternative method for 5(i)
[X = 0.342 F + 11.34
Y = 0.94 F – 5] | 1 | (M1) | For finding components of the forces in the x and y
directions
(0.342 F + 11.34)2 + (0.94 F – 5)2
= R2 | 1 | (A1)
tan15
= (5 – 0.94F) / (0.342F + 11.34) | 1 | (A1)
1 | (M1) | Solve the tan 15 equation for F and substitute to find R
F = 1.90 N and R = 12.4 N | 1 | (A1)
5
--- 5(ii) ---
5(ii) | 11.72 = 0 + 2a × 3
a = 22.815 | 1 | B1
R cos15 = m × 22.815 | 1 | M1 | Applying Newton’s second law to the particle in
direction AB
Mass of bead = 0.526 kg | 1 | A1
3
Marks\includegraphics{figure_5}
A small bead $Q$ can move freely along a smooth horizontal straight wire $AB$ of length 3 m. Three horizontal forces of magnitudes $F$ N, 10 N and 20 N act on the bead in the directions shown in the diagram. The magnitude of the resultant of the three forces is $R$ N in the direction shown in the diagram.
\begin{enumerate}[label=(\roman*)]
\item Find the values of $F$ and $R$. [5]
\item Initially the bead is at rest at $A$. It reaches $B$ with a speed of 11.7 m s$^{-1}$. Find the mass of the bead. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 Q5 [8]}}