Edexcel FM1 AS 2018 June — Question 2 9 marks

Exam BoardEdexcel
ModuleFM1 AS (Further Mechanics 1 AS)
Year2018
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAdvanced work-energy problems
TypeRough inclined plane work-energy
DifficultyStandard +0.3 This is a straightforward FM1 work-energy question requiring application of the work-energy principle with given values. Part (a) is a 'show that' with clear setup, part (b) is a symmetric calculation, and part (c) asks for standard modelling improvements. The calculations are routine with no conceptual surprises, making it slightly easier than average for Further Maths.
Spec6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component
2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{cfa9b998-d57d-4980-9316-1bddeac55b90-04_267_891_346_687} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a ramp inclined at an angle \(\theta\) to the horizontal, where \(\sin \theta = \frac { 2 } { 7 }\) A parcel of mass 4 kg is projected, with speed \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), from a point \(A\) on the ramp.
The parcel moves up a line of greatest slope of the ramp and first comes to instantaneous rest at the point \(B\), where \(A B = 2.5 \mathrm {~m}\).
The parcel is modelled as a particle.
The total resistance to the motion of the parcel from non-gravitational forces is modelled as a constant force of magnitude \(R\) newtons.
  1. Use the work-energy principle to show that \(R = 8.8\) After coming to instantaneous rest at \(B\), the parcel slides back down the ramp. The total resistance to the motion of the particle is modelled as a constant force of magnitude 8.8N.
  2. Find the speed of the parcel at the instant it returns to \(A\).
  3. Suggest two improvements that could be made to the model.
    VILU SIHI NI IIIUM ION OCVGHV SIHILNI IMAM ION OOVJYV SIHI NI JIIYM ION OC
Question 2:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Work-energy equation: KE lost = PE gained + Work DoneM1
\(\frac{1}{2} \times 4 \times 5^2 - 4 \times g \times 2.5 \times \sin\theta = 2.5R\)A1
\(\frac{1}{2} \times 4 \times 5^2 - 4 \times g \times 2.5 \times \frac{2}{7} = 2.5R\)A1
\(2.5R = 22 \Rightarrow R = 8.8\) *A1*
(4)
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Work-energy equation: KE after = initial KE \(- 2\)(Work Done)M1
\(\frac{1}{2} \times 4 \times v^2 = \frac{1}{2} \times 4 \times 25 - 2 \times 8.8 \times 2.5\)A1
\(\Rightarrow 2v^2 = 6,\ v = 1.7\) (m s\(^{-1}\))A1
(3)
Alternative (b) via energy from B:
AnswerMarks Guidance
Answer/WorkingMark Guidance
KE at \(B\) = PE lost \(-\) Work DoneM1
\(\frac{1}{2} \times 4 \times v^2 = 4 \times 9.8 \times \frac{2}{7} \times 2.5 - 8.8 \times 2.5\)A1
\(\Rightarrow 2v^2 = 6,\ v = 1.7\) (m s\(^{-1}\))A1
Alternative (b) via equation of motion and suvat:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(4g\sin\theta - 8.8 = 4a \quad (a = 0.6)\)M1
\(v^2 = 2 \times a \times 2.5\)A1
\(v = 1.7\) (m s\(^{-1}\))A1
Part (c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
A valid improvementB1
A second valid, distinct improvementB1
(2) 
## Question 2:

### Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Work-energy equation: KE lost = PE gained + Work Done | M1 | |
| $\frac{1}{2} \times 4 \times 5^2 - 4 \times g \times 2.5 \times \sin\theta = 2.5R$ | A1 | |
| $\frac{1}{2} \times 4 \times 5^2 - 4 \times g \times 2.5 \times \frac{2}{7} = 2.5R$ | A1 | |
| $2.5R = 22 \Rightarrow R = 8.8$ * | A1* | |
| **(4)** | | |

### Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Work-energy equation: KE after = initial KE $- 2$(Work Done) | M1 | |
| $\frac{1}{2} \times 4 \times v^2 = \frac{1}{2} \times 4 \times 25 - 2 \times 8.8 \times 2.5$ | A1 | |
| $\Rightarrow 2v^2 = 6,\ v = 1.7$ (m s$^{-1}$) | A1 | |
| **(3)** | | |

**Alternative (b) via energy from B:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| KE at $B$ = PE lost $-$ Work Done | M1 | |
| $\frac{1}{2} \times 4 \times v^2 = 4 \times 9.8 \times \frac{2}{7} \times 2.5 - 8.8 \times 2.5$ | A1 | |
| $\Rightarrow 2v^2 = 6,\ v = 1.7$ (m s$^{-1}$) | A1 | |

**Alternative (b) via equation of motion and suvat:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4g\sin\theta - 8.8 = 4a \quad (a = 0.6)$ | M1 | |
| $v^2 = 2 \times a \times 2.5$ | A1 | |
| $v = 1.7$ (m s$^{-1}$) | A1 | |

### Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| A valid improvement | B1 | |
| A second valid, distinct improvement | B1 | |
| **(2)** | | |
2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{cfa9b998-d57d-4980-9316-1bddeac55b90-04_267_891_346_687}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a ramp inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 2 } { 7 }$\\
A parcel of mass 4 kg is projected, with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, from a point $A$ on the ramp.\\
The parcel moves up a line of greatest slope of the ramp and first comes to instantaneous rest at the point $B$, where $A B = 2.5 \mathrm {~m}$.\\
The parcel is modelled as a particle.\\
The total resistance to the motion of the parcel from non-gravitational forces is modelled as a constant force of magnitude $R$ newtons.
\begin{enumerate}[label=(\alph*)]
\item Use the work-energy principle to show that $R = 8.8$

After coming to instantaneous rest at $B$, the parcel slides back down the ramp. The total resistance to the motion of the particle is modelled as a constant force of magnitude 8.8N.
\item Find the speed of the parcel at the instant it returns to $A$.
\item Suggest two improvements that could be made to the model.

\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VILU SIHI NI IIIUM ION OC & VGHV SIHILNI IMAM ION OO & VJYV SIHI NI JIIYM ION OC \\
\hline

\hline
\end{tabular}
\end{center}
\end{enumerate}

\hfill \mbox{\textit{Edexcel FM1 AS 2018 Q2 [9]}}
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