| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on rough horizontal surface, particle hanging |
| Difficulty | Standard +0.3 This is a standard M1 pulley system question with connected particles. Part (a) is a 'show that' using s=ut+½at² (routine kinematics). Parts (b-d) involve standard applications of Newton's second law and friction. Part (e) requires checking if particle A stops before reaching the pulley using energy/kinematics after B hits the floor. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 3.03o Advanced connected particles: and pulleys3.03r Friction: concept and vector form |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1.4 = \frac{1}{2}a \times 2^2\) | M1 | Complete method to obtain equation in \(a\) only; allow verification |
| \(a = 0.7 \ (\text{m s}^{-2})\) GIVEN ANSWER | A1* | Given answer correctly obtained or verification completed correctly |
| Total | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Inextensibility of string | B1 | B0 if any extras given |
| Total | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3g - T = 3\times0.7\) (for \(B\)) | M1 A1 | Equation of motion for \(B\) with usual rules |
| Resultant \(= 2T\cos45°\) OR \(\sqrt{T^2+T^2}\) OR \(\frac{T}{\cos45°}\) | M1 | Correct expression in terms of \(T\) |
| \(= 39\) or \(38.6\) (N) | A1 | |
| Total | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T - F = 4\times0.7\) (for \(A\)) OR \(3g - F = 7\times0.7\) (whole system) | M1 A1 | Equation of motion for \(A\) or whole system |
| \(R = 4g\) | B1 | |
| \(F = \mu \times R\) | B1 | |
| \(27.3 - \mu\times4g = 4\times0.7\) OR \(3g - \mu\times4g = 7\times0.7\) | DM1 | Solving to give equation in \(\mu\) only; dependent on first M1 |
| \(\mu = 0.625\) or \(0.63\) | A1 | \(5/8\) is A0; DM0 if \(T=3g\) used |
| Total | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 0.7\times2\) or \(v = \sqrt{2\times0.7\times1.4}\) | M1 | Finding speed or speed² of either particle when \(B\) hits floor |
| \(-\mu\times4g = 4a\) | M1 | Equation of motion for \(A\); allow without \(-\)ve sign |
| \(0^2 = 1.4^2 - 2\times\frac{5g}{8}s\) | M1 | Complete method to find distance moved by \(A\) until stops; independent M but M0 if no new deceleration found |
| \(s = 0.16\) or \(0.159\) | A1 | |
| \(0.16 + 1.4 < 2 \Rightarrow\) Does not reach pulley | A1 cso | Must see \('<2'\) or use 2 in working |
| Total | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v^2 = 1.4^2 - 2\times\frac{5g}{8}\times0.6\) | M1 | Complete method to find \(v^2\) where \(v\) is speed at pulley; condone sign error; M0 if no new deceleration |
| \(= -5.39\) or \(-5.4488\) | A1 | Correct value for \(v^2\) |
| Since \(v^2\) must be \(\geq 0\), does not reach pulley | A1 cso |
# Question 7:
## Part 7(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1.4 = \frac{1}{2}a \times 2^2$ | M1 | Complete method to obtain equation in $a$ only; allow verification |
| $a = 0.7 \ (\text{m s}^{-2})$ **GIVEN ANSWER** | A1* | Given answer correctly obtained or verification completed correctly |
| **Total** | **(2)** | |
## Part 7(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Inextensibility of string | B1 | B0 if any extras given |
| **Total** | **(1)** | |
## Part 7(c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3g - T = 3\times0.7$ (for $B$) | M1 A1 | Equation of motion for $B$ with usual rules |
| Resultant $= 2T\cos45°$ **OR** $\sqrt{T^2+T^2}$ **OR** $\frac{T}{\cos45°}$ | M1 | Correct expression in terms of $T$ |
| $= 39$ or $38.6$ (N) | A1 | |
| **Total** | **(4)** | |
## Part 7(d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T - F = 4\times0.7$ (for $A$) **OR** $3g - F = 7\times0.7$ (whole system) | M1 A1 | Equation of motion for $A$ or whole system |
| $R = 4g$ | B1 | |
| $F = \mu \times R$ | B1 | |
| $27.3 - \mu\times4g = 4\times0.7$ **OR** $3g - \mu\times4g = 7\times0.7$ | DM1 | Solving to give equation in $\mu$ only; dependent on first M1 |
| $\mu = 0.625$ or $0.63$ | A1 | $5/8$ is A0; DM0 if $T=3g$ used |
| **Total** | **(6)** | |
## Part 7(e)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 0.7\times2$ or $v = \sqrt{2\times0.7\times1.4}$ | M1 | Finding speed or speed² of either particle when $B$ hits floor |
| $-\mu\times4g = 4a$ | M1 | Equation of motion for $A$; allow without $-$ve sign |
| $0^2 = 1.4^2 - 2\times\frac{5g}{8}s$ | M1 | Complete method to find distance moved by $A$ until stops; independent M but M0 if no new deceleration found |
| $s = 0.16$ or $0.159$ | A1 | |
| $0.16 + 1.4 < 2 \Rightarrow$ Does not reach pulley | A1 cso | Must see $'<2'$ or use 2 in working |
| **Total** | **(5)** | |
**Alternative for final 3 marks:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v^2 = 1.4^2 - 2\times\frac{5g}{8}\times0.6$ | M1 | Complete method to find $v^2$ where $v$ is speed at pulley; condone sign error; M0 if no new deceleration |
| $= -5.39$ or $-5.4488$ | A1 | Correct value for $v^2$ |
| Since $v^2$ must be $\geq 0$, does not reach pulley | A1 cso | |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{916543cb-14f7-486c-ba3c-eda9be134045-20_663_1290_260_335}
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\caption{Figure 3}
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A particle $A$ of mass 4 kg is held at rest on a rough horizontal table. Particle $A$ is attached to one end of a string that passes over a pulley $P$. The pulley is fixed the the of the table. The other end of the string is attached to a particle $B$, of mass 3 kg , which hangs freely below $P$.
The part of the string from $A$ to $P$ is perpendicular to the edge of the table and $A , P$ and $B$ all lie in the same vertical plane.
The string is modelled as being light and inextensible and the pulley is modelled as being small, smooth and light.
The system is released from rest with the string taut. At the instant of release, $A$ is 2 m from the edge of the table and $B$ is 1.4 m above a horizontal floor, as shown in Figure 3.
After descending with constant acceleration for 2 seconds, $B$ hits the floor and does not rebound.
\begin{enumerate}[label=(\alph*)]
\item Show that the acceleration of $A$ before $B$ hits the floor is $0.7 \mathrm {~ms} ^ { - 2 }$
\item State which of the modelling assumptions you have used in order to answer part (a).
\item Find the magnitude of the resultant force exerted on the pulley by the string.
The coefficient of friction between $A$ and the table is $\mu$.
\item Find the value of $\mu$.
\item Determine, by calculation, whether or not $A$ reaches the pulley.
DO NOT WRITEIN THIS AREA\\
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\includegraphics[max width=\textwidth, alt={}, center]{916543cb-14f7-486c-ba3c-eda9be134045-23_2255_50_314_34}
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\hfill \mbox{\textit{Edexcel M1 2020 Q7 [18]}}