| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Single particle, Newton's second law – scalar (1D, horizontal or inclined) |
| Difficulty | Moderate -0.3 This is a straightforward two-part mechanics question requiring resolution of forces in two perpendicular directions to find F and R, then applying a standard kinematic equation (v² = u² + 2as) with Newton's second law. All steps are routine M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For resolving forces either horizontally or vertically | |
| \(F\cos 70 + 20 - 10\cos 30 = R\cos 15\) | A1 | |
| \(10\sin 30 - F\sin 70 = R\sin 15\) | A1 | |
| M1 | For solving simultaneously | |
| \(F = 1.90\) N and \(R = 12.4\) N | A1 (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([X = 0.342F + 11.34\), \(Y = 0.94F - 5]\) | M1 | For finding components of the forces in the \(x\) and \(y\) directions |
| \((0.342F + 11.34)^2 + (0.94F - 5)^2 = R^2\) | A1 | |
| \(\tan 15 = (5 - 0.94F)/(0.342F + 11.34)\) | A1 | |
| M1 | Solve the \(\tan 15\) equation for \(F\) and substitute to find \(R\) | |
| \(F = 1.90\) N and \(R = 12.4\) N | A1 (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(11.7^2 = 0 + 2a \times 3\) | ||
| \(a = 22.815\) | B1 | |
| \(R\cos 15 = m \times 22.815\) | M1 | Applying Newton's second law to the particle in direction \(AB\) |
| Mass of bead \(= 0.526\) kg | A1 (3) |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For resolving forces either horizontally or vertically |
| $F\cos 70 + 20 - 10\cos 30 = R\cos 15$ | A1 | |
| $10\sin 30 - F\sin 70 = R\sin 15$ | A1 | |
| | M1 | For solving simultaneously |
| $F = 1.90$ N **and** $R = 12.4$ N | A1 (5) | |
**Alternative method for 5(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[X = 0.342F + 11.34$, $Y = 0.94F - 5]$ | M1 | For finding components of the forces in the $x$ and $y$ directions |
| $(0.342F + 11.34)^2 + (0.94F - 5)^2 = R^2$ | A1 | |
| $\tan 15 = (5 - 0.94F)/(0.342F + 11.34)$ | A1 | |
| | M1 | Solve the $\tan 15$ equation for $F$ and substitute to find $R$ |
| $F = 1.90$ N **and** $R = 12.4$ N | A1 (5) | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $11.7^2 = 0 + 2a \times 3$ | | |
| $a = 22.815$ | B1 | |
| $R\cos 15 = m \times 22.815$ | M1 | Applying Newton's second law to the particle in direction $AB$ |
| Mass of bead $= 0.526$ kg | A1 (3) | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{2a91fb7a-0eaf-4c50-8a2c-4755c0b44c17-3_355_1048_255_552}
A small bead $Q$ can move freely along a smooth horizontal straight wire $A B$ of length 3 m . Three horizontal forces of magnitudes $F \mathrm {~N} , 10 \mathrm {~N}$ and 20 N act on the bead in the directions shown in the diagram. The magnitude of the resultant of the three forces is $R \mathrm {~N}$ in the direction shown in the diagram.\\
(i) Find the values of $F$ and $R$.\\
(ii) Initially the bead is at rest at $A$. It reaches $B$ with a speed of $11.7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the mass of the bead.
\hfill \mbox{\textit{CAIE M1 2015 Q5 [8]}}