OCR M1 2005 January — Question 6 13 marks

Exam BoardOCR
ModuleM1 (Mechanics 1)
Year2005
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeRead and interpret velocity-time graph
DifficultyModerate -0.3 This is a standard M1 two-particle meeting problem using SUVAT equations and graph interpretation. Parts (i)-(ii) involve basic kinematic calculations (v=u+at, distance from trapezium area). Parts (iii)-(iv) test graph sketching skills. Part (v) requires setting up and solving a meeting condition equation. While multi-part with several steps, all techniques are routine M1 content with no novel problem-solving required, making it slightly easier than average.
Spec3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

6 \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{5b10afa1-1c45-4370-a0e6-ad8fd626df9a-4_664_969_264_589} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} A cyclist \(P\) travels along a straight road starting from rest at \(A\) and accelerating at \(2 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) up to a speed of \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). He continues at a constant speed of \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), passing through the point \(B 20 \mathrm {~s}\) after leaving \(A\). Fig. 1 shows the ( \(t , v\) ) graph of \(P\) 's journey for \(0 \leqslant t \leqslant 20\). Find
  1. the time for which \(P\) is accelerating,
  2. the distance \(A B\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{5b10afa1-1c45-4370-a0e6-ad8fd626df9a-4_607_937_1420_605} \captionsetup{labelformat=empty} \caption{Fig. 2}
    \end{figure} Another cyclist \(Q\) travels along the same straight road in the opposite direction. She starts at rest from \(B\) at the same instant that \(P\) leaves \(A\). Cyclist \(Q\) accelerates at \(2 \mathrm {~ms} ^ { - 2 }\) up to a speed of \(8 \mathrm {~ms} ^ { - 1 }\) and continues at a constant speed of \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), passing through the point \(A 20 \mathrm {~s}\) after leaving \(B\). Fig. 2 shows the \(( t , x )\) graph of \(Q\) 's journey for \(0 \leqslant t \leqslant 20\), where \(x\) is the displacement of \(Q\) from \(A\) towards \(B\).
  3. Sketch a copy of Fig. 1 and add to your copy a sketch of the ( \(t , v\) ) graph of \(Q\) 's journey for \(0 \leqslant t \leqslant 20\).
  4. Sketch a copy of Fig. 2 and add to your copy a sketch of the \(( t , x )\) graph of \(P\) 's journey for \(0 \leqslant t \leqslant 20\).
  5. Find the value \(t\) at the instant that \(P\) and \(Q\) pass each other. \includegraphics[max width=\textwidth, alt={}, center]{5b10afa1-1c45-4370-a0e6-ad8fd626df9a-5_447_739_269_703} The upper edge of a smooth plane inclined at \(70 ^ { \circ }\) to the horizontal is joined to an edge of a rough horizontal table. Particles \(A\) and \(B\), of masses 0.3 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth pulley which is fixed at the top of the smooth inclined plane. Particle \(A\) is held in contact with the rough horizontal table and particle \(B\) is in contact with the smooth inclined plane with the string taut (see diagram). The coefficient of friction between \(A\) and the horizontal table is 0.4 . Particle \(A\) is released from rest and the system starts to move.
  6. Find the acceleration of \(A\) and the tension in the string. The string breaks when the speed of the particles is \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  7. Assuming \(A\) does not reach the pulley, find the distance travelled by \(A\) after the string breaks.
  8. Assuming \(B\) does not reach the ground before \(A\) stops, find the distance travelled by \(B\) from the time the string breaks to the time that \(A\) stops.

Question 6:
Part (i):
AnswerMarks Guidance
AnswerMark Guidance
Accelerating for \(4\) sM1, A1 [2] For using the idea that the gradient represents acceleration or for using \(v = u + at\)
Part (ii):
AnswerMarks Guidance
AnswerMark Guidance
\(AB = \frac{1}{2}(16 + 20)8\)M1, A1ft For using the idea that the distance is represented by the area of the trapezium or using suitable formulae for the two stages of the journey
Distance is \(144\) mA1 [3]
Part (iii):
AnswerMarks Guidance
AnswerMark Guidance
B1Graph is single valued and continuous and consists of two straight line segments with one segment from the origin and the other parallel to the \(t\) axis
B1 [2]Graph for \(Q\) is the reflection of the graph for \(P\) in the \(t\) axis
Part (iv):
AnswerMarks Guidance
AnswerMark Guidance
B1Graph is single valued and continuous and consists of two parts, one of which is a straight line segment, with \(x\) increasing from \(0\) for the interval \(0 < t < 20\)
B1\(x_P(20)\) appears to be equal to \(x_Q(0)\)
B1 [3]Graph for \(P\) appears to be the reflection in \(x = \text{ans(ii)} \div 2\) of graph for \(Q\)
Part (v):
AnswerMarks Guidance
AnswerMark Guidance
\(t = 20 - (\frac{1}{2} \times 144 \div 8)\) or \(16 + 8(t-4) = 128 - 8(t-4)\) or equivalentM1 For complete method of finding the required time
Value of \(t\) is \(11\)A2 [3] SR Allow B1 for \(t = 11\) without explanation
# Question 6:

## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Accelerating for $4$ s | M1, A1 [2] | For using the idea that the gradient represents acceleration or for using $v = u + at$ |

## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $AB = \frac{1}{2}(16 + 20)8$ | M1, A1ft | For using the idea that the distance is represented by the area of the trapezium or using suitable formulae for the two stages of the journey |
| Distance is $144$ m | A1 [3] | |

## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| | B1 | Graph is single valued and continuous and consists of two straight line segments with one segment from the origin and the other parallel to the $t$ axis |
| | B1 [2] | Graph for $Q$ is the reflection of the graph for $P$ in the $t$ axis |

## Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| | B1 | Graph is single valued and continuous and consists of two parts, one of which is a straight line segment, with $x$ increasing from $0$ for the interval $0 < t < 20$ |
| | B1 | $x_P(20)$ appears to be equal to $x_Q(0)$ |
| | B1 [3] | Graph for $P$ appears to be the reflection in $x = \text{ans(ii)} \div 2$ of graph for $Q$ |

## Part (v):
| Answer | Mark | Guidance |
|--------|------|----------|
| $t = 20 - (\frac{1}{2} \times 144 \div 8)$ or $16 + 8(t-4) = 128 - 8(t-4)$ or equivalent | M1 | For complete method of finding the required time |
| Value of $t$ is $11$ | A2 [3] | SR Allow B1 for $t = 11$ without explanation |

---
6

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{5b10afa1-1c45-4370-a0e6-ad8fd626df9a-4_664_969_264_589}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}

A cyclist $P$ travels along a straight road starting from rest at $A$ and accelerating at $2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ up to a speed of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. He continues at a constant speed of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, passing through the point $B 20 \mathrm {~s}$ after leaving $A$. Fig. 1 shows the ( $t , v$ ) graph of $P$ 's journey for $0 \leqslant t \leqslant 20$. Find\\
(i) the time for which $P$ is accelerating,\\
(ii) the distance $A B$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{5b10afa1-1c45-4370-a0e6-ad8fd626df9a-4_607_937_1420_605}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

Another cyclist $Q$ travels along the same straight road in the opposite direction. She starts at rest from $B$ at the same instant that $P$ leaves $A$. Cyclist $Q$ accelerates at $2 \mathrm {~ms} ^ { - 2 }$ up to a speed of $8 \mathrm {~ms} ^ { - 1 }$ and continues at a constant speed of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, passing through the point $A 20 \mathrm {~s}$ after leaving $B$. Fig. 2 shows the $( t , x )$ graph of $Q$ 's journey for $0 \leqslant t \leqslant 20$, where $x$ is the displacement of $Q$ from $A$ towards $B$.\\
(iii) Sketch a copy of Fig. 1 and add to your copy a sketch of the ( $t , v$ ) graph of $Q$ 's journey for $0 \leqslant t \leqslant 20$.\\
(iv) Sketch a copy of Fig. 2 and add to your copy a sketch of the $( t , x )$ graph of $P$ 's journey for $0 \leqslant t \leqslant 20$.\\
(v) Find the value $t$ at the instant that $P$ and $Q$ pass each other.\\
\includegraphics[max width=\textwidth, alt={}, center]{5b10afa1-1c45-4370-a0e6-ad8fd626df9a-5_447_739_269_703}

The upper edge of a smooth plane inclined at $70 ^ { \circ }$ to the horizontal is joined to an edge of a rough horizontal table. Particles $A$ and $B$, of masses 0.3 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth pulley which is fixed at the top of the smooth inclined plane. Particle $A$ is held in contact with the rough horizontal table and particle $B$ is in contact with the smooth inclined plane with the string taut (see diagram). The coefficient of friction between $A$ and the horizontal table is 0.4 . Particle $A$ is released from rest and the system starts to move.\\
(i) Find the acceleration of $A$ and the tension in the string.

The string breaks when the speed of the particles is $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) Assuming $A$ does not reach the pulley, find the distance travelled by $A$ after the string breaks.\\
(iii) Assuming $B$ does not reach the ground before $A$ stops, find the distance travelled by $B$ from the time the string breaks to the time that $A$ stops.

\hfill \mbox{\textit{OCR M1 2005 Q6 [13]}}