Edexcel Paper 3 2018 June — Question 7 8 marks

Exam BoardEdexcel
ModulePaper 3 (Paper 3)
Year2018
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeHorizontal force on slope
DifficultyStandard +0.3 This is a standard mechanics problem requiring resolution of forces and Newton's second law. Students must find sin α and cos α from tan α, resolve forces horizontally and vertically, calculate friction, then find acceleration. Part (b) requires conceptual understanding of how normal reaction affects friction. While multi-step, it follows a routine procedure with no novel insight required, making it slightly easier than average.
Spec3.03a Force: vector nature and diagrams3.03c Newton's second law: F=ma one dimension3.03e Resolve forces: two dimensions3.03f Weight: W=mg3.03m Equilibrium: sum of resolved forces = 03.03r Friction: concept and vector form3.03s Contact force components: normal and frictional3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes

7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{65e4b254-fb7b-45c2-9702-32f034018193-20_264_698_246_685} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} A wooden crate of mass 20 kg is pulled in a straight line along a rough horizontal floor using a handle attached to the crate.
The handle is inclined at an angle \(\alpha\) to the floor, as shown in Figure 1, where \(\tan \alpha = \frac { 3 } { 4 }\) The tension in the handle is 40 N .
The coefficient of friction between the crate and the floor is 0.14
The crate is modelled as a particle and the handle is modelled as a light rod.
Using the model,
  1. find the acceleration of the crate. The crate is now pushed along the same floor using the handle. The handle is again inclined at the same angle \(\alpha\) to the floor, and the thrust in the handle is 40 N as shown in Figure 2 below. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{65e4b254-fb7b-45c2-9702-32f034018193-20_220_923_1457_571} \captionsetup{labelformat=empty} \caption{Figure 2}
    \end{figure}
  2. Explain briefly why the acceleration of the crate would now be less than the acceleration of the crate found in part (a).

Question 7:
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
Resolve verticallyM1 3.1b
\(R + 40\sin\alpha = 20g\)A1 1.1b
Resolve horizontallyM1 3.1b
\(40\cos\alpha - F = 20a\)A1 1.1b
\(F = 0.14R\)B1 1.2
\(a = 0.396\) or \(0.40\) (m s\(^{-2}\))A1 2.2a
Notes:
- M1: Resolve vertically with usual rules applying
- A1: Correct equation. Neither \(g\) nor \(\sin\alpha\) need to be substituted
- M1: Apply \(F = ma\) horizontally, with usual rules
- A1: Neither \(F\) nor \(\cos\alpha\) need to be substituted
- B1: \(F = 0.14R\) seen (e.g. on diagram)
- A1: Either answer
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
Pushing will increase \(R\) which will increase available \(F\)B1 2.4
Increasing \(F\) will decrease \(a\) * GIVEN ANSWERB1* 2.4
Notes:
- B1: Pushing increases \(R\) which produces an increase in available (limiting) friction
- B1: \(F\) increase produces a decrease in \(a\) (need to see this)
- N.B. It is possible to score B0 B1 but for the B1, some "explanation" is needed to say why friction is increased e.g. by pushing into the ground
## Question 7:

### Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Resolve vertically | M1 | 3.1b |
| $R + 40\sin\alpha = 20g$ | A1 | 1.1b |
| Resolve horizontally | M1 | 3.1b |
| $40\cos\alpha - F = 20a$ | A1 | 1.1b |
| $F = 0.14R$ | B1 | 1.2 |
| $a = 0.396$ or $0.40$ (m s$^{-2}$) | A1 | 2.2a |

**Notes:**
- M1: Resolve vertically with usual rules applying
- A1: Correct equation. Neither $g$ nor $\sin\alpha$ need to be substituted
- M1: Apply $F = ma$ horizontally, with usual rules
- A1: Neither $F$ nor $\cos\alpha$ need to be substituted
- B1: $F = 0.14R$ seen (e.g. on diagram)
- A1: Either answer

### Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Pushing will increase $R$ which will increase available $F$ | B1 | 2.4 |
| Increasing $F$ will decrease $a$ * GIVEN ANSWER | B1* | 2.4 |

**Notes:**
- B1: Pushing increases $R$ which produces an increase in available (limiting) friction
- B1: $F$ increase produces a decrease in $a$ (need to see this)
- N.B. It is possible to score B0 B1 but for the B1, some "explanation" is needed to say why friction is increased e.g. by pushing into the ground

---
7.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{65e4b254-fb7b-45c2-9702-32f034018193-20_264_698_246_685}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

A wooden crate of mass 20 kg is pulled in a straight line along a rough horizontal floor using a handle attached to the crate.\\
The handle is inclined at an angle $\alpha$ to the floor, as shown in Figure 1, where $\tan \alpha = \frac { 3 } { 4 }$\\
The tension in the handle is 40 N .\\
The coefficient of friction between the crate and the floor is 0.14\\
The crate is modelled as a particle and the handle is modelled as a light rod.\\
Using the model,
\begin{enumerate}[label=(\alph*)]
\item find the acceleration of the crate.

The crate is now pushed along the same floor using the handle. The handle is again inclined at the same angle $\alpha$ to the floor, and the thrust in the handle is 40 N as shown in Figure 2 below.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{65e4b254-fb7b-45c2-9702-32f034018193-20_220_923_1457_571}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
\item Explain briefly why the acceleration of the crate would now be less than the acceleration of the crate found in part (a).
\end{enumerate}

\hfill \mbox{\textit{Edexcel Paper 3 2018 Q7 [8]}}