| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Equilibrium of particle under coplanar forces |
| Difficulty | Moderate -0.3 This is a standard two-part equilibrium and F=ma question requiring resolution of forces in perpendicular directions. Part (i) involves routine application of equilibrium conditions (sum of forces = 0 in two directions) to find two unknowns. Part (ii) is a straightforward application of Newton's second law. The techniques are standard M1 material with no novel problem-solving required, making it slightly easier than average but not trivial due to the multi-step nature and need for careful resolution. |
| Spec | 3.03b Newton's first law: equilibrium3.03c Newton's second law: F=ma one dimension3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| 5(i) | Resolve forces either horizontally or vertically | M1 |
| 7.5cos60+ 4.5cos20 = Fcosθ [=7.97861] | A1 | |
| 7.5sin60– 4.5sin20 = Fsinθ [= 4.95609] | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| F = 7.982 +4.962 | M1 | Use Pythagoras or use the value found for θ to find F |
| Answer | Marks | Guidance |
|---|---|---|
| 7.98 | M1 | Use trigonometry or the value found for F to find θ |
| F=9.39 and θ = 31.8 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| sin80 sin ( 120+θ ) sin ( 160−θ ) | M1 | Attempt to use Lami |
| A1 | One correct pair of terms | |
| A1 | A second correct pair of terms | |
| [4.5sin(160 – θ) = 7.5sin(120+θ)] | M1 | Attempt to solve for θ |
| Use the θ value found by valid trigonometry to find F | M1 | |
| F=9.39 and θ = 31.8 | A1 | |
| Question | Answer | Marks |
| 5(i) | Alternative method for question 5(i) | |
| Forces 4.5, 7.5, F opposite angles 60–θ, θ+20, 100 | M1 | Illustrate a triangle of forces |
| [F2=4.52 + 7.52 – 2 × 4.5×7.5×cos100] | M1 | For application of cosine rule to find F |
| A1 | Correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| sin100 sin ( 60−θ ) sin ( θ+20 ) | M1 | One application of the sine rule to find θ |
| A1 | Correct equation | |
| F=9.39 and θ = 31.8 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | 9.5cos30– 7.5cos60 – 4.5cos20=m×1.5 | M1 |
| m=0.166 kg | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | Resolve forces either horizontally or vertically | M1
7.5cos60+ 4.5cos20 = Fcosθ [=7.97861] | A1
7.5sin60– 4.5sin20 = Fsinθ [= 4.95609] | A1
( )
F = 7.982 +4.962 | M1 | Use Pythagoras or use the value found for θ to find F
θ = tan–1(4.96)
7.98 | M1 | Use trigonometry or the value found for F to find θ
F=9.39 and θ = 31.8 | A1
Alternative method for question 5(i)
F 4.5 7.5
= =
sin80 sin ( 120+θ ) sin ( 160−θ ) | M1 | Attempt to use Lami
A1 | One correct pair of terms
A1 | A second correct pair of terms
[4.5sin(160 – θ) = 7.5sin(120+θ)] | M1 | Attempt to solve for θ
Use the θ value found by valid trigonometry to find F | M1
F=9.39 and θ = 31.8 | A1
Question | Answer | Marks | Guidance
5(i) | Alternative method for question 5(i)
Forces 4.5, 7.5, F opposite angles 60–θ, θ+20, 100 | M1 | Illustrate a triangle of forces
[F2=4.52 + 7.52 – 2 × 4.5×7.5×cos100] | M1 | For application of cosine rule to find F
A1 | Correct equation
9.39 4.5 7.5
= =
sin100 sin ( 60−θ ) sin ( θ+20 ) | M1 | One application of the sine rule to find θ
A1 | Correct equation
F=9.39 and θ = 31.8 | A1
6
--- 5(ii) ---
5(ii) | 9.5cos30– 7.5cos60 – 4.5cos20=m×1.5 | M1 | Apply Newton’s second law to the ring along AB (4 terms)
m=0.166 kg | A1
2
Question | Answer | Marks | Guidance\includegraphics{figure_5}
A small ring $P$ is threaded on a fixed smooth horizontal rod $AB$. Three horizontal forces of magnitudes 4.5 N, 7.5 N and $F$ N act on $P$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Given that these three forces are in equilibrium, find the values of $F$ and $\theta$. [6]
\end{enumerate}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item It is given instead that the values of $F$ and $\theta$ are 9.5 and 30 respectively, and the acceleration of the ring is 1.5 m s$^{-2}$. Find the mass of the ring. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2019 Q5 [8]}}