| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Stacked boxes, friction between surfaces |
| Difficulty | Standard +0.3 This is a standard two-body friction problem requiring systematic application of Newton's laws and limiting friction conditions. Part (i) uses equilibrium at the threshold of motion (straightforward), part (ii) requires considering friction between the boxes to limit acceleration (routine constraint analysis), and part (iii) applies limiting friction at the floor. While it requires careful bookkeeping of multiple friction forces and masses, the techniques are standard M1 material with no novel insights needed. |
| Spec | 3.03d Newton's second law: 2D vectors3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = 4500\text{ N}\) | B1 | |
| \(3150 = \mu\times4500\) | M1 | For using limiting equilibrium \(\Rightarrow P = \mu R\) |
| Coefficient is 0.7 | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.2\times200g = 200a\) | M1 | For resolving forces horizontally on A when A is about to slide |
| A1 | AG | |
| No sliding \(\Rightarrow a \leq 2\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P - F = 450a\); \(P - F - F_2 = 250a\) | M1 | For applying Newton's second law to A and B combined or to B |
| \(P_{\max} = 3150 + 450\times2\) or \(P_{\max} = 3150 + 0.2\times2000 + 250\times2\) | A1 | |
| \(P_{\max} = 4050\text{ N}\) | A1 | |
| [3] |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = 4500\text{ N}$ | B1 | |
| $3150 = \mu\times4500$ | M1 | For using limiting equilibrium $\Rightarrow P = \mu R$ |
| Coefficient is 0.7 | A1 | |
| **[3]** | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.2\times200g = 200a$ | M1 | For resolving forces horizontally on A when A is about to slide |
| | A1 | AG |
| No sliding $\Rightarrow a \leq 2$ | A1 | |
| **[3]** | | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P - F = 450a$; $P - F - F_2 = 250a$ | M1 | For applying Newton's second law to A and B combined or to B |
| $P_{\max} = 3150 + 450\times2$ or $P_{\max} = 3150 + 0.2\times2000 + 250\times2$ | A1 | |
| $P_{\max} = 4050\text{ N}$ | A1 | |
| **[3]** | | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{dafc271d-a77b-4401-9170-e13e484d6e5f-4_246_665_253_739}
Two rectangular boxes $A$ and $B$ are of identical size. The boxes are at rest on a rough horizontal floor with $A$ on top of $B$. Box $A$ has mass 200 kg and box $B$ has mass 250 kg . A horizontal force of magnitude $P$ N is applied to $B$ (see diagram). The boxes remain at rest if $P \leqslant 3150$ and start to move if $P > 3150$.\\
(i) Find the coefficient of friction between $B$ and the floor.
The coefficient of friction between the two boxes is 0.2 . Given that $P > 3150$ and that no sliding takes place between the boxes,\\
(ii) show that the acceleration of the boxes is not greater than $2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$,\\
(iii) find the maximum possible value of $P$.
\hfill \mbox{\textit{CAIE M1 2010 Q7 [9]}}