Edexcel M1 2024 June — Question 6 12 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2024
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFriction
TypeCoefficient from constant speed
DifficultyModerate -0.3 This is a standard M1 friction problem with two routine parts: (a) resolving forces with constant velocity to find tension, and (b) applying SUVAT with constant deceleration. Both require straightforward application of well-practiced techniques with no novel insight, making it slightly easier than average.
Spec3.02d Constant acceleration: SUVAT formulae3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces

6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7a65555e-1bb2-4947-8e70-50f267017bfd-16_272_1391_336_436} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} A box of mass \(m\) lies on a rough horizontal plane. The box is pulled along the plane in a straight line at constant speed by a light rope. The rope is inclined at an angle \(\theta\) to the plane, as shown in Figure 3.
The coefficient of friction between the box and the plane is \(\frac { 1 } { 3 }\) The box is modelled as a particle.
Given that \(\tan \theta = \frac { 3 } { 4 }\)
  1. find, in terms of \(m\) and \(g\), the tension in the rope. The rope is now removed and the box is placed at rest on the plane.
    The box is then projected horizontally along the plane with speed \(u\).
    The box is again modelled as a particle.
    When the box has moved a distance \(d\) along the plane, the speed of the box is \(\frac { 1 } { 2 } u\).
  2. Find \(d\) in terms of \(u\) and \(g\).
    VJYV SIHI NI JIIIM ION OCvauv sthin NI JLHMA LON OOV34V SIHI NI IIIIM ION OC

Question 6:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(R + T\sin\theta = mg\)M1A1 Vertical resolution, correct terms, condone sign errors and sin/cos confusion. Allow \(\sin(\frac{3}{5})\) or similar
\(T\cos\theta - F = 0\)M1A1 Horizontal resolution, correct terms, condone sign errors and sin/cos confusion. Allow \(\cos(\frac{4}{5})\) or similar
\(F = \frac{1}{3}R\)B1 Seen anywhere, including on diagram
Solve for \(T\) in terms of \(mg\)DM1 Dependent on both M marks
\(T = \frac{1}{3}mg\)A1 (7) cao. Accept \(0.33\,mg\) or better
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(F = \frac{1}{3}mg\)B1 Seen anywhere, including on diagram
\(F = \pm ma\) OR \(W.D. = \pm Fd\)B1 \(F\) is friction, allow \(+\) or \(-\)
\(\left(\frac{1}{2}u\right)^2 = u^2 - 2\left(\frac{1}{3}g\right)d\) OR \(\frac{1}{2}m\left(\frac{1}{2}u\right)^2 = \frac{1}{2}mu^2 - \frac{1}{3}mgd\)DM1A1 Complete method, dependent on previous B mark, dimensionally correct, correct no. of terms in \(d\), \(u\), \(g\), condone sign errors
\(d = \frac{9u^2}{8g}\)A1 (5) cao, must be \(d=\,\), seen or implied, but allow \(s\) in working
# Question 6:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $R + T\sin\theta = mg$ | M1A1 | Vertical resolution, correct terms, condone sign errors and sin/cos confusion. Allow $\sin(\frac{3}{5})$ or similar |
| $T\cos\theta - F = 0$ | M1A1 | Horizontal resolution, correct terms, condone sign errors and sin/cos confusion. Allow $\cos(\frac{4}{5})$ or similar |
| $F = \frac{1}{3}R$ | B1 | Seen anywhere, including on diagram |
| Solve for $T$ in terms of $mg$ | DM1 | Dependent on both M marks |
| $T = \frac{1}{3}mg$ | A1 (7) | cao. Accept $0.33\,mg$ or better |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $F = \frac{1}{3}mg$ | B1 | Seen anywhere, including on diagram |
| $F = \pm ma$ **OR** $W.D. = \pm Fd$ | B1 | $F$ is friction, allow $+$ or $-$ |
| $\left(\frac{1}{2}u\right)^2 = u^2 - 2\left(\frac{1}{3}g\right)d$ **OR** $\frac{1}{2}m\left(\frac{1}{2}u\right)^2 = \frac{1}{2}mu^2 - \frac{1}{3}mgd$ | DM1A1 | Complete method, dependent on previous B mark, dimensionally correct, correct no. of terms in $d$, $u$, $g$, condone sign errors |
| $d = \frac{9u^2}{8g}$ | A1 (5) | cao, must be $d=\,$, seen or implied, **but allow $s$ in working** |

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6.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{7a65555e-1bb2-4947-8e70-50f267017bfd-16_272_1391_336_436}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

A box of mass $m$ lies on a rough horizontal plane. The box is pulled along the plane in a straight line at constant speed by a light rope. The rope is inclined at an angle $\theta$ to the plane, as shown in Figure 3.\\
The coefficient of friction between the box and the plane is $\frac { 1 } { 3 }$\\
The box is modelled as a particle.\\
Given that $\tan \theta = \frac { 3 } { 4 }$
\begin{enumerate}[label=(\alph*)]
\item find, in terms of $m$ and $g$, the tension in the rope.

The rope is now removed and the box is placed at rest on the plane.\\
The box is then projected horizontally along the plane with speed $u$.\\
The box is again modelled as a particle.\\
When the box has moved a distance $d$ along the plane, the speed of the box is $\frac { 1 } { 2 } u$.
\item Find $d$ in terms of $u$ and $g$.

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VJYV SIHI NI JIIIM ION OC & vauv sthin NI JLHMA LON OO & V34V SIHI NI IIIIM ION OC \\
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\end{center}
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2024 Q6 [12]}}