Question 2 - A-Level Maths

Edexcel M2 2016 October — Question 2 8 marks

Exam Board	Edexcel
Module	M2 (Mechanics 2)
Year	2016
Session	October
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Impulse and momentum (advanced)
Type	Velocity after impulse (find unknown constant)
Difficulty	Standard +0.3 This is a standard M2 impulse-momentum question requiring application of the impulse-momentum equation in vector form, followed by solving a quadratic equation using the speed condition. It's slightly above average difficulty due to the vector components and quadratic solving, but follows a well-practiced procedure with no novel insight required.
Spec	1.10a Vectors in 2D: i,j notation and column vectors 6.03e Impulse: by a force 6.03f Impulse-momentum: relation

2. A particle of mass 2 kg is moving with velocity \(3 \mathbf { i } \mathrm {~ms} ^ { - 1 }\) when it receives an impulse \(( \lambda \mathbf { i } - 2 \lambda \mathbf { j } )\) Ns, where \(\lambda\) is a constant. Immediately after the impulse is received, the speed of the particle is \(6 \mathrm {~ms} ^ { - 1 }\). Find the possible values of \(\lambda\).

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Question 2:

Answer	Marks	Guidance
Working	Mark	Notes
Form impulse-momentum equation	M1	Must be dimensionally correct
\(2\mathbf{v} - 2 \times 3\mathbf{i} = \lambda(\mathbf{i} - 2\mathbf{j})\)	A1	Correct unsimplified
\(2\mathbf{v} = (6+\lambda)\mathbf{i} - 2\lambda\mathbf{j}\) \(\left(\mathbf{v} = \frac{1}{2}(6+\lambda)\mathbf{i} - \lambda\mathbf{j}\right)\)	A1	\(\mathbf{v}\) or \(2\mathbf{v}\)
Magnitude of \(\mathbf{v}\) (or \(2\mathbf{v}\))	M1	Correct application of Pythagoras to form an equation in \(\lambda\)
\((6+\lambda)^2 + 4\lambda^2 = 144\) or equivalent	A1	Correct unsimplified equation in \(\lambda\)
\(5\lambda^2 + 12\lambda - 108 = 0\)	A1	Correct 3 term quadratic in \(\lambda\)
\((5\lambda - 18)(\lambda + 6) = 0\), solve for \(\lambda\)	M1	Solve a 3 term quadratic in \(\lambda\)
\(\lambda = -6\), \(\lambda = 3.6\)	A1	Or equivalent
(8 marks)

SC: If candidate spots \(\lambda = -6\) is a solution but cannot find second solution, allow B1 (the final M1 on epen). Max score 4/8.

# Question 2:
| Working | Mark | Notes |
|---------|------|-------|
| Form impulse-momentum equation | M1 | Must be dimensionally correct |
| $2\mathbf{v} - 2 \times 3\mathbf{i} = \lambda(\mathbf{i} - 2\mathbf{j})$ | A1 | Correct unsimplified |
| $2\mathbf{v} = (6+\lambda)\mathbf{i} - 2\lambda\mathbf{j}$ $\left(\mathbf{v} = \frac{1}{2}(6+\lambda)\mathbf{i} - \lambda\mathbf{j}\right)$ | A1 | $\mathbf{v}$ or $2\mathbf{v}$ |
| Magnitude of $\mathbf{v}$ (or $2\mathbf{v}$) | M1 | Correct application of Pythagoras to form an equation in $\lambda$ |
| $(6+\lambda)^2 + 4\lambda^2 = 144$ or equivalent | A1 | Correct unsimplified equation in $\lambda$ |
| $5\lambda^2 + 12\lambda - 108 = 0$ | A1 | Correct 3 term quadratic in $\lambda$ |
| $(5\lambda - 18)(\lambda + 6) = 0$, solve for $\lambda$ | M1 | Solve a 3 term quadratic in $\lambda$ |
| $\lambda = -6$, $\lambda = 3.6$ | A1 | Or equivalent |
| **(8 marks)** | | |

**SC:** If candidate spots $\lambda = -6$ is a solution but cannot find second solution, allow B1 (the final M1 on epen). Max score 4/8.

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2. A particle of mass 2 kg is moving with velocity $3 \mathbf { i } \mathrm {~ms} ^ { - 1 }$ when it receives an impulse $( \lambda \mathbf { i } - 2 \lambda \mathbf { j } )$ Ns, where $\lambda$ is a constant. Immediately after the impulse is received, the speed of the particle is $6 \mathrm {~ms} ^ { - 1 }$.

Find the possible values of $\lambda$.\\

\hfill \mbox{\textit{Edexcel M2 2016 Q2 [8]}}

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Q1 6 Q2 8 Q3 7 Q4 10 Q5 10 Q6 10 Q7 10 Q8 14