Edexcel M1 — Question 5 12 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeMotion up then down slope
DifficultyStandard +0.3 This is a standard M1 mechanics problem involving motion on a rough inclined plane with friction. Part (a) uses kinematics (v² = u² + 2as) with given values, part (b) applies F=ma with resolved forces to find μ, part (c) uses energy or kinematics for return journey, and part (d) tests modeling awareness. All techniques are routine M1 content with straightforward application, making it slightly easier than average.
Spec3.02d Constant acceleration: SUVAT formulae3.03v Motion on rough surface: including inclined planes

5.
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A small stone is projected with speed \(7 \mathrm {~ms} ^ { - 1 }\) from \(P\), the bottom of a rough plane inclined at \(25 ^ { \circ }\) to the horizontal, and moves up a line of greatest slope of the plane until it comes to instantaneous rest at \(Q\), where \(P Q = 4 \mathrm {~m}\).
  1. Show that the deceleration of the stone as it moves up the plane has magnitude \(\frac { 49 } { 8 } \mathrm {~ms} ^ { - 2 }\).
  2. Find the coefficient of friction between the stone and the plane,
  3. Find the speed with which the stone returns to \(P\).
  4. Name one force which you have ignored in your mathematical model, and state whether the answer to part (c) would be larger or smaller if that force were taken into account.

AnswerMarks Guidance
(a) \(0 = 7^2 - 2a(4)\)M1 A1
\(a = \frac{49}{8}\) ms\(^{-2}\)
(b) Acc down plane \(= g \sin 25° + \mu g \cos 25° = 9.8(\sin 25° + \mu \cos 25°)\)M1 A1
Hence \(\sin 25° + \mu \cos 25° = 0.625\)
\(\mu = 0.223\)M1 A1
(c) Now down plane, acc. \(= g \sin 25° - \mu g \cos 25° = 0.220g\)M1 A1
\(v^2 = 0 + 2(4)(0.220g) = 17.27\)
\(v = 4.16\) ms\(^{-1}\)M1 A1
(d) Air resistance, which would make the answer smallerB1 B1 Total: 12 marks
**(a)** $0 = 7^2 - 2a(4)$ | M1 A1 |
$a = \frac{49}{8}$ ms$^{-2}$ | |

**(b)** Acc down plane $= g \sin 25° + \mu g \cos 25° = 9.8(\sin 25° + \mu \cos 25°)$ | M1 A1 |
Hence $\sin 25° + \mu \cos 25° = 0.625$ | |
$\mu = 0.223$ | M1 A1 |

**(c)** Now down plane, acc. $= g \sin 25° - \mu g \cos 25° = 0.220g$ | M1 A1 |
$v^2 = 0 + 2(4)(0.220g) = 17.27$ | |
$v = 4.16$ ms$^{-1}$ | M1 A1 |

**(d)** Air resistance, which would make the answer smaller | B1 B1 | **Total: 12 marks**
5.

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{31efa627-5114-4797-9d46-7f1311c18ff8-2_262_597_276_356}
\end{center}

A small stone is projected with speed $7 \mathrm {~ms} ^ { - 1 }$ from $P$, the bottom of a rough plane inclined at $25 ^ { \circ }$ to the horizontal, and moves up a line of greatest slope of the plane until it comes to instantaneous rest at $Q$, where $P Q = 4 \mathrm {~m}$.
\begin{enumerate}[label=(\alph*)]
\item Show that the deceleration of the stone as it moves up the plane has magnitude $\frac { 49 } { 8 } \mathrm {~ms} ^ { - 2 }$.
\item Find the coefficient of friction between the stone and the plane,
\item Find the speed with which the stone returns to $P$.
\item Name one force which you have ignored in your mathematical model, and state whether the answer to part (c) would be larger or smaller if that force were taken into account.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q5 [12]}}