| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Block on rough horizontal surface – accelerating (finding acceleration or applied force) |
| Difficulty | Standard +0.3 This is a standard M1 friction problem requiring resolution of forces and Newton's second law. Students must resolve vertically to find the normal reaction (including the vertical component of the applied force), then apply F=μR and Newton's second law horizontally. While it involves multiple steps and careful handling of the angled force, it follows a completely standard method taught in all M1 courses with no novel insight required. |
| Spec | 3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = T \sin 60° + 49\) | M1A1 | M1: Seeing ±49 or 5g or \(mg\) and \(T \sin60°\) or \(T \sin30°\) or \(T \cos60°\) or \(T \cos30°\), with no other terms. A1: Any correct expression for \(R\) in terms of only \(T\), but need not write \(R =\) at the start. Allow 5g but not \(mg\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = 0.2(T \sin 60° + 49)\), \(T \cos 60° - F = 5 \times 0.9\), \(T \cos 60° - 0.2(T \sin 60° + 49) = 4.5\), \(T = \frac{4.5 + 9.8}{\cos 60° - 0.2 \sin 60°} = 43.8 \text{ N}\) | M1A1; M1, A1F; dM1, A1 | M1: Using \(F = 0.2R\) where \(R\) is their answer to (a) and a function of \(T\). A1: Correct expression for \(F\). M1: Resolving horizontally to obtain a three term equation of motion. Must contain \(T \sin60°\) or \(T \sin30°\) or \(T \cos60°\) or \(T \cos30°\) and \(F\) or their \(F\) and \(5 \times 0.9\). A1F: Correct equation. Allow their \(F\). dM1: Substituting expression for \(F\). A1: Correct \(T\) from correct working. Accept AWFW 43.7 to 43.8. Using \(g = 9.81\) still gives 43.8 as the final answer. |
**6. (a)**
$R = T \sin 60° + 49$ | M1A1 | M1: Seeing ±49 or 5g or $mg$ and $T \sin60°$ or $T \sin30°$ or $T \cos60°$ or $T \cos30°$, with no other terms. A1: Any correct expression for $R$ in terms of only $T$, but need not write $R =$ at the start. Allow 5g but not $mg$. | 2 marks
**6. (b)**
$F = 0.2(T \sin 60° + 49)$, $T \cos 60° - F = 5 \times 0.9$, $T \cos 60° - 0.2(T \sin 60° + 49) = 4.5$, $T = \frac{4.5 + 9.8}{\cos 60° - 0.2 \sin 60°} = 43.8 \text{ N}$ | M1A1; M1, A1F; dM1, A1 | M1: Using $F = 0.2R$ where $R$ is their answer to (a) and a function of $T$. A1: Correct expression for $F$. M1: Resolving horizontally to obtain a three term equation of motion. Must contain $T \sin60°$ or $T \sin30°$ or $T \cos60°$ or $T \cos30°$ and $F$ or their $F$ and $5 \times 0.9$. A1F: Correct equation. Allow their $F$. dM1: Substituting expression for $F$. A1: Correct $T$ from correct working. Accept AWFW 43.7 to 43.8. Using $g = 9.81$ still gives 43.8 as the final answer. | 6 marks
**Total for Question 6: 8 marks**
---6 A floor polisher consists of a heavy metal block with a polishing cloth attached to the underside. A light rod is also attached to the block and is used to push the block across the floor that is to be polished. The block has mass 5 kg . Assume that the floor is horizontal. Model the block as a particle.
The coefficient of friction between the cloth and the floor is 0.2 .\\
A person pushes the rod to exert a force on the block. The force is at an angle of $60 ^ { \circ }$ to the horizontal and the block accelerates at $0.9 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The diagram shows the block and the force exerted by the rod in this situation.\\
\includegraphics[max width=\textwidth, alt={}, center]{5dd17095-18a6-470b-a24a-4456c8e3ed31-14_309_205_772_1009}
The rod exerts a force of magnitude $T$ newtons on the block.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $T$, the magnitude of the normal reaction force acting on the block.
\item $\quad$ Find $T$.\\[0pt]
[6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2016 Q6 [6]}}