| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Pulley at edge of table, specific geometry |
| Difficulty | Moderate -0.3 This is a standard AS-level pulley system problem requiring Newton's second law for connected particles, followed by kinematics in two stages. The setup is straightforward with clear given values, and part (a) is a 'show that' which guides students to the expected answer. The two-stage motion in part (b) requires careful thinking but uses routine SUVAT equations. This is slightly easier than average due to the guided nature and standard techniques, though the two-stage aspect prevents it from being trivial. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equation of motion for \(Q\) | M1 | Equation of motion for \(Q\) with correct no. of terms, condone sign errors |
| \(0.6g - T = 0.6a\) | A1 | A correct equation |
| Equation of motion for \(P\) | M1 | Equation of motion for \(P\) with correct no. of terms, condone sign errors |
| \(T = 0.8a\) | A1 | A correct equation |
| \(a = 4.2\) (m s\(^{-2}\)) | A1* | Given acceleration obtained correctly. You must see an equation in \(a\) only before reaching \(a=4.2\). N.B. if just whole system equation \(0.6g = 1.4a\), can only score max M1A1M0A0A0. N.B. Use of \(g = 9.81\) or \(10\) loses final A mark only. Complete verification using both equations can score full marks. |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(0.4 = \frac{1}{2} \times 4.2 \times t_1^2\) (or find \(v\) first then use \(v = 4.2t_1\)) | M1 | Complete method to find time for \(Q\) to hit the floor. M0 if 0.4 not used as distance or 4.2 not used as acceleration |
| \(t_1 = 0.436(4357\ldots)\) Allow 0.43, 0.44, 0.436 or better, or surd form e.g. \(\frac{2}{\sqrt{21}}\) | A1 | See alternatives |
| \(v = 4.2 \times t_1\) or \(v = \sqrt{2 \times 4.2 \times 0.4}\) or \(0.4 = \frac{(0+v)}{2} \times t_1\), \((v = 1.8330\ldots)\) | M1 | Complete method to find speed of \(Q\) as it hits floor. M0 if 0.4 not used as distance or 4.2 not used as acceleration |
| \(t_2 = \frac{1.5 - 0.4}{v}\) | M1 | Uses distance/speed to find time for \(P\) to hit pulley after \(Q\) hits floor. Independent of previous M mark |
| Complete strategy: finding sum of two times \(t_1 + t_2\) | DM1 | Complete method finding and adding two required times. Dependent on previous three M marks |
| \(1.0\) (s) or \(1.04\) (s) | A1 | |
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| e.g. rope being light; rope being inextensible; pulley being smooth; pulley being small; balls being particles | B1 | Clear statement. Allow negatives. Must be a limitation of the model stated in the question. Penalise incorrect or irrelevant extras. B0 for: air resistance, table being smooth |
| (1) |
# Question 2:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Equation of motion for $Q$ | M1 | Equation of motion for $Q$ with correct no. of terms, condone sign errors |
| $0.6g - T = 0.6a$ | A1 | A correct equation |
| Equation of motion for $P$ | M1 | Equation of motion for $P$ with correct no. of terms, condone sign errors |
| $T = 0.8a$ | A1 | A correct equation |
| $a = 4.2$ (m s$^{-2}$) | A1* | Given acceleration obtained correctly. **You must see an equation in $a$ only before reaching $a=4.2$**. N.B. if just whole system equation $0.6g = 1.4a$, can only score max M1A1M0A0A0. N.B. Use of $g = 9.81$ or $10$ loses final A mark only. Complete verification using both equations can score full marks. |
| | **(5)** | |
## Question (b): Pulley/Motion Problem
| Working/Answer | Mark | Guidance |
|---|---|---|
| $0.4 = \frac{1}{2} \times 4.2 \times t_1^2$ (or find $v$ first then use $v = 4.2t_1$) | M1 | Complete method to find time for $Q$ to hit the floor. M0 if 0.4 not used as distance or 4.2 not used as acceleration |
| $t_1 = 0.436(4357\ldots)$ Allow 0.43, 0.44, 0.436 or better, or surd form e.g. $\frac{2}{\sqrt{21}}$ | A1 | See alternatives |
| $v = 4.2 \times t_1$ or $v = \sqrt{2 \times 4.2 \times 0.4}$ or $0.4 = \frac{(0+v)}{2} \times t_1$, $(v = 1.8330\ldots)$ | M1 | Complete method to find speed of $Q$ as it hits floor. M0 if 0.4 not used as distance or 4.2 not used as acceleration |
| $t_2 = \frac{1.5 - 0.4}{v}$ | M1 | Uses distance/speed to find time for $P$ to hit pulley after $Q$ hits floor. Independent of previous M mark |
| Complete strategy: finding sum of two times $t_1 + t_2$ | DM1 | Complete method finding and adding two required times. Dependent on previous three M marks |
| $1.0$ (s) or $1.04$ (s) | A1 | |
| **(6)** | | |
**Part (c):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| e.g. rope being light; rope being inextensible; pulley being smooth; pulley being small; balls being particles | B1 | Clear statement. Allow negatives. Must be a limitation of the model stated in the question. Penalise incorrect or irrelevant extras. B0 for: air resistance, table being smooth |
| **(1)** | | |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ad0eca04-7b0b-4163-b0da-3a6dc85fec22-06_711_1264_251_402}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A small ball, $P$, of mass 0.8 kg , is held at rest on a smooth horizontal table and is attached to one end of a thin rope.
The rope passes over a pulley that is fixed at the edge of the table.\\
The other end of the rope is attached to another small ball, $Q$, of mass 0.6 kg , that hangs freely below the pulley.
Ball $P$ is released from rest, with the rope taut, with $P$ at a distance of 1.5 m from the pulley and with $Q$ at a height of 0.4 m above the horizontal floor, as shown in Figure 1.
Ball $Q$ descends, hits the floor and does not rebound.\\
The balls are modelled as particles, the rope as a light and inextensible string and the pulley as small and smooth.
Using this model,
\begin{enumerate}[label=(\alph*)]
\item show that the acceleration of $Q$, as it falls, is $4.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$
\item find the time taken by $P$ to hit the pulley from the instant when $P$ is released.
\item State one limitation of the model that will affect the accuracy of your answer to part (a).
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 2 2019 Q2 [12]}}