OCR M1 2016 June — Question 6 14 marks

Exam BoardOCR
ModuleM1 (Mechanics 1)
Year2016
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeHeavier particle hits ground, lighter continues upward - inclined plane involved
DifficultyStandard +0.3 This is a standard M1 connected particles problem with friction on an inclined plane. Part (i) requires routine application of Newton's second law to find tensions before and after release. Part (ii) uses energy or kinematics after the string goes slack. Part (iii) asks for contact force (normal reaction), which is straightforward resolution perpendicular to the plane. While multi-part with several steps, all techniques are standard M1 fare with no novel insights required, making it slightly easier than average.
Spec3.03k Connected particles: pulleys and equilibrium3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle

\includegraphics{figure_6} Two particles \(P\) and \(Q\) are attached to opposite ends of a light inextensible string which passes over a small smooth pulley at the top of a rough plane inclined at \(30°\) to the horizontal. \(P\) has mass \(0.2\text{ kg}\) and is held at rest on the plane. \(Q\) has mass \(0.2\text{ kg}\) and hangs freely. The string is taut (see diagram). The coefficient of friction between \(P\) and the plane is \(0.4\). The particle \(P\) is released.
  1. State the tension in the string before \(P\) is released, and find the tension in the string after \(P\) is released. [6]
\(Q\) strikes the floor and remains at rest. \(P\) continues to move up the plane for a further distance of \(0.8\text{ m}\) before it comes to rest. \(P\) does not reach the pulley.
  1. Find the speed of the particles immediately before \(Q\) strikes the floor. [5]
  2. Calculate the magnitude of the contact force exerted on \(P\) by the plane while \(P\) is in motion. [3]

Part i
Answer/Working:
\(T(\text{before}) = 0.2g = 1.96\)
\(Fr = 0.4x0.2g\cos30 (=0.67896...)\)
\(0.2a = 0.2g - T\) Either correct
\(0.2a = T - 0.2g\sin30 - 0.4x0.2g\cos30\) Both correct
\(2T = 0.2g + 0.2g\sin30 + 0.4x0.2g\cos30\)
\(T = 1.81\) N
AnswerMarks
MarksGuidance
B1Evaluation not needed
B1Evaluation not needed, but accept 0.68
M1\(a \neq g\)
A1\(0.2g - T = T - 0.2g\sin30 - 0.4x0.2g\cos30\) is M1A1; \(0.4a=0.2g - 0.2g\sin30 - 0.4x0.2g\cos30\) is M1A1
M1Finding expression (2)T from two simultaneous equations in a and T. a = 0.7526... m s\(^{-2}\), but is not required
A1
[6]
Part ii
Answer/Working:
THIS CANNOT BE SOLVED USING a(i)
\(0.2a = +/- (0.2g\sin30 + 0.4x0.2g\cos30)\)
\(a = +/-(8.2948...)\)
\(v^2 = 2x8.29(48...)x0.8\) OR \(0 = u^2 - 2x8.29(48...)x0.8\)
\(v = 3.64\) m s\(^{-1}\) or \(u = 3.64\) m s\(^{-1}\)
AnswerMarks
MarksGuidance
M1*N2L with Fr and Weight component of P; Omitting g, M1*A0A0, D*M1A0 possible
A1
A1
D*M1Equations must lead to positive values for \(u^2\), \(v^2\)
A1
[5]
Part iii
Answer/Working:
\(R^2 = (0.2g\cos30)^2 + (0.4x0.2g\cos30)^2\)
\(R=1.83\) N
AnswerMarks
MarksGuidance
M1Applies Pythagoras to Friction and Normal Reaction
A1Omitting g, M1A0A0 possible
A1
[3]
## Part i

**Answer/Working:**
$T(\text{before}) = 0.2g = 1.96$
$Fr = 0.4x0.2g\cos30 (=0.67896...)$
$0.2a = 0.2g - T$ Either correct

$0.2a = T - 0.2g\sin30 - 0.4x0.2g\cos30$ Both correct

$2T = 0.2g + 0.2g\sin30 + 0.4x0.2g\cos30$
$T = 1.81$ N

| **Marks** | **Guidance** |
|-----------|-------------|
| B1 | Evaluation not needed |
| B1 | Evaluation not needed, but accept 0.68 |
| M1 | $a \neq g$ |
| A1 | $0.2g - T = T - 0.2g\sin30 - 0.4x0.2g\cos30$ is M1A1; $0.4a=0.2g - 0.2g\sin30 - 0.4x0.2g\cos30$ is M1A1 |
| M1 | Finding expression (2)T from two simultaneous equations in a and T. a = 0.7526... m s$^{-2}$, but is not required |
| A1 | |
| [6] | |

## Part ii

**Answer/Working:**
THIS CANNOT BE SOLVED USING a(i)
$0.2a = +/- (0.2g\sin30 + 0.4x0.2g\cos30)$
$a = +/-(8.2948...)$
$v^2 = 2x8.29(48...)x0.8$ OR $0 = u^2 - 2x8.29(48...)x0.8$
$v = 3.64$ m s$^{-1}$ or $u = 3.64$ m s$^{-1}$

| **Marks** | **Guidance** |
|-----------|-------------|
| M1* | N2L with Fr and Weight component of P; Omitting g, M1*A0A0, D*M1A0 possible |
| A1 | |
| A1 | |
| D*M1 | Equations must lead to positive values for $u^2$, $v^2$ |
| A1 | |
| [5] | |

## Part iii

**Answer/Working:**
$R^2 = (0.2g\cos30)^2 + (0.4x0.2g\cos30)^2$
$R=1.83$ N

| **Marks** | **Guidance** |
|-----------|-------------|
| M1 | Applies Pythagoras to Friction and Normal Reaction |
| A1 | Omitting g, M1A0A0 possible |
| A1 | |
| [3] | |

---
\includegraphics{figure_6}

Two particles $P$ and $Q$ are attached to opposite ends of a light inextensible string which passes over a small smooth pulley at the top of a rough plane inclined at $30°$ to the horizontal. $P$ has mass $0.2\text{ kg}$ and is held at rest on the plane. $Q$ has mass $0.2\text{ kg}$ and hangs freely. The string is taut (see diagram). The coefficient of friction between $P$ and the plane is $0.4$. The particle $P$ is released.

\begin{enumerate}[label=(\roman*)]
\item State the tension in the string before $P$ is released, and find the tension in the string after $P$ is released. [6]
\end{enumerate}

$Q$ strikes the floor and remains at rest. $P$ continues to move up the plane for a further distance of $0.8\text{ m}$ before it comes to rest. $P$ does not reach the pulley.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the speed of the particles immediately before $Q$ strikes the floor. [5]
\item Calculate the magnitude of the contact force exerted on $P$ by the plane while $P$ is in motion. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR M1 2016 Q6 [14]}}