Edexcel M1 2017 January — Question 2 9 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2017
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeForces in vector form: kinematics extension
DifficultyModerate -0.8 This is a straightforward M1 mechanics question requiring basic application of F=ma in vector form, constant acceleration kinematics (v = u + at), and understanding of parallel vectors. All three parts use standard techniques with no problem-solving insight needed, making it easier than average but not trivial due to the vector notation requirement.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication3.03d Newton's second law: 2D vectors

  1. A particle \(P\) of mass 0.5 kg moves under the action of a single constant force ( \(2 \mathbf { i } + 3 \mathbf { j }\) )N.
    1. Find the acceleration of \(P\).
    At time \(t\) seconds, \(P\) has velocity \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\). When \(t = 0 , \mathbf { v } = 4 \mathbf { i }\)
  2. Find the speed of \(P\) when \(t = 3\) Given that \(P\) is moving parallel to the vector \(2 \mathbf { i } + \mathbf { j }\) at time \(t = T\)
  3. find the value of \(T\).

Question 2:
Part 2a:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Use of \(\mathbf{F} = m\mathbf{a}\): \(2\mathbf{i} + 3\mathbf{j} = 0.5\mathbf{a}\)M1 M1 for use of \(\mathbf{F} = m\mathbf{a}\)
\(\mathbf{a} = 4\mathbf{i} + 6\mathbf{j}\) (m s\(^{-2}\))A1 A1 for \(4\mathbf{i} + 6\mathbf{j}\) (m s\(^{-2}\)) isw if magnitude found
Part 2b:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Use of \(\mathbf{v} = \mathbf{u} + 3\mathbf{a}\) with their \(\mathbf{a}\)M1 First M1 for \(\mathbf{v} = 4\mathbf{i} + 3(4\mathbf{i} + 6\mathbf{j})\) with their \(\mathbf{a}\) (but M0 if they use \(2\mathbf{i} + 3\mathbf{j}\) instead of \(\mathbf{a}\))
\(= 16\mathbf{i} + 18\mathbf{j}\)A1 First A1 for \(16\mathbf{i} + 18\mathbf{j}\) seen or implied
Use of Pythagoras: speed \(= \sqrt{16^2 + 18^2}\)M1 Second M1 for finding magnitude of their \(\mathbf{v}\)
\(= \sqrt{580}\) or 24 (m s\(^{-1}\)) or betterA1 Second A1 for 24 or better (24.0831...) or \(\sqrt{580}\)
Part 2c:
AnswerMarks Guidance
Answer/WorkingMark Guidance
In component form: \(\mathbf{v} = 4\mathbf{i} + t(4\mathbf{i} + 6\mathbf{j})\)M1 First M1 for \(\mathbf{v} = 4\mathbf{i} + t(4\mathbf{i} + 6\mathbf{j})\) with their \(\mathbf{a}\) (but M0 if they use \(2\mathbf{i} + 3\mathbf{j}\) instead of \(\mathbf{a}\))
\(4 + 4T = 2 \times 6T\)M1 Second independent M1 for a correct method to give an equation in \(T\) (t) only using their \(\mathbf{v}\)
\(T = \dfrac{1}{2}\)A1 A1 for \((T) = \frac{1}{2}\)
# Question 2:

## Part 2a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{F} = m\mathbf{a}$: $2\mathbf{i} + 3\mathbf{j} = 0.5\mathbf{a}$ | M1 | M1 for use of $\mathbf{F} = m\mathbf{a}$ |
| $\mathbf{a} = 4\mathbf{i} + 6\mathbf{j}$ (m s$^{-2}$) | A1 | A1 for $4\mathbf{i} + 6\mathbf{j}$ (m s$^{-2}$) isw if magnitude found |

## Part 2b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{v} = \mathbf{u} + 3\mathbf{a}$ with their $\mathbf{a}$ | M1 | First M1 for $\mathbf{v} = 4\mathbf{i} + 3(4\mathbf{i} + 6\mathbf{j})$ with their $\mathbf{a}$ (but M0 if they use $2\mathbf{i} + 3\mathbf{j}$ instead of $\mathbf{a}$) |
| $= 16\mathbf{i} + 18\mathbf{j}$ | A1 | First A1 for $16\mathbf{i} + 18\mathbf{j}$ seen or implied |
| Use of Pythagoras: speed $= \sqrt{16^2 + 18^2}$ | M1 | Second M1 for finding magnitude of their $\mathbf{v}$ |
| $= \sqrt{580}$ or 24 (m s$^{-1}$) or better | A1 | Second A1 for 24 or better (24.0831...) or $\sqrt{580}$ |

## Part 2c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| In component form: $\mathbf{v} = 4\mathbf{i} + t(4\mathbf{i} + 6\mathbf{j})$ | M1 | First M1 for $\mathbf{v} = 4\mathbf{i} + t(4\mathbf{i} + 6\mathbf{j})$ with their $\mathbf{a}$ (but M0 if they use $2\mathbf{i} + 3\mathbf{j}$ instead of $\mathbf{a}$) |
| $4 + 4T = 2 \times 6T$ | M1 | Second independent M1 for a correct method to give an equation in $T$ (t) only using their $\mathbf{v}$ |
| $T = \dfrac{1}{2}$ | A1 | A1 for $(T) = \frac{1}{2}$ |

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\begin{enumerate}
  \item A particle $P$ of mass 0.5 kg moves under the action of a single constant force ( $2 \mathbf { i } + 3 \mathbf { j }$ )N.\\
(a) Find the acceleration of $P$.
\end{enumerate}

At time $t$ seconds, $P$ has velocity $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$. When $t = 0 , \mathbf { v } = 4 \mathbf { i }$\\
(b) Find the speed of $P$ when $t = 3$

Given that $P$ is moving parallel to the vector $2 \mathbf { i } + \mathbf { j }$ at time $t = T$\\
(c) find the value of $T$.\\

\hfill \mbox{\textit{Edexcel M1 2017 Q2 [9]}}