| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2004 |
| Session | November |
| 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 friction inequalities. While it involves multiple friction surfaces and requires careful consideration of limiting cases, the approach is methodical: calculate maximum friction forces, apply F=ma to the system, then check the no-slip condition between boxes. The 'show that' format provides guidance, and the techniques are standard M1 fare, making it slightly easier than average. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = 8000 \text{ N}\) | B1 | |
| For obtaining \(P \leq 6000\) | B1 | From \(P = F \leq \mu R = 0.75 \times 8000\) — 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F \leq 0.4 \times 4000\) or \(F_{\max} = 0.4 \times 4000\) | B1 | |
| M1 | For applying Newton's \(2^{\text{nd}}\) law to the upper box and using \(F \leq 1600\) or \(F_{\max} = 1600\) | |
| \(400a \leq 1600\) or \(400a_{\max} = 1600\) | A1 | From \(F = 400a\) |
| \(a \leq 4\) | A1 | |
| M1 | For applying Newton's \(2^{\text{nd}}\) law to the boxes | |
| \(P_{\max} - 6000 = 800 \times 4\) or \(P - 6000 = 800a \leq 800 \times 4\) | A1 | |
| Maximum possible value of \(P\) is \(9200\) | A1 | 7 marks total |
# Question 6:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = 8000 \text{ N}$ | B1 | |
| For obtaining $P \leq 6000$ | B1 | From $P = F \leq \mu R = 0.75 \times 8000$ — **2 marks total** |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F \leq 0.4 \times 4000$ or $F_{\max} = 0.4 \times 4000$ | B1 | |
| | M1 | For applying Newton's $2^{\text{nd}}$ law to the upper box and using $F \leq 1600$ or $F_{\max} = 1600$ |
| $400a \leq 1600$ or $400a_{\max} = 1600$ | A1 | From $F = 400a$ |
| $a \leq 4$ | A1 | |
| | M1 | For applying Newton's $2^{\text{nd}}$ law to the boxes |
| $P_{\max} - 6000 = 800 \times 4$ or $P - 6000 = 800a \leq 800 \times 4$ | A1 | |
| Maximum possible value of $P$ is $9200$ | A1 | **7 marks total** |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{38ece0f6-1c29-4e7a-9d66-16c3e2b695f9-3_330_572_1037_788}
Two identical boxes, each of mass 400 kg , are at rest, with one on top of the other, on horizontal ground. A horizontal force of magnitude $P$ newtons is applied to the lower box (see diagram). The coefficient of friction between the lower box and the ground is 0.75 and the coefficient of friction between the two boxes is 0.4 .\\
(i) Show that the boxes will remain at rest if $P \leqslant 6000$.
The boxes start to move with acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(ii) Given that no sliding takes place between the boxes, show that $a \leqslant 4$ and deduce the maximum possible value of $P$.
\hfill \mbox{\textit{CAIE M1 2004 Q6 [9]}}