Edexcel M2 2021 June — Question 4 6 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2021
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImpulse and momentum (advanced)
TypeImpulse from velocity change
DifficultyModerate -0.3 This is a straightforward application of the impulse-momentum theorem in 2D, requiring resolution of vectors into components and solving simultaneous equations. While it involves multiple steps (resolving impulse, applying momentum conservation in two directions, using Pythagoras), these are standard M2 techniques with no novel insight required. The question is slightly easier than average due to its predictable structure and clear diagram.
Spec6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e6e37d85-f8de-490a-82a9-8a3c16e2fdd0-10_410_369_251_790} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} A particle \(P\) of mass 0.3 kg is moving with speed \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) along a straight line on a smooth horizontal plane. The particle receives a horizontal impulse of magnitude \(J\) Ns. The speed of \(P\) immediately after receiving the impulse is \(8 \mathrm {~ms} ^ { - 1 }\). The angle between the direction of motion of \(P\) before it receives the impulse and the direction of the impulse is \(60 ^ { \circ }\), as shown in Figure 2. Find the value of \(J\).
(6)
Question 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Use of \(J = m(v-u)\)M1 Use of \(J = m(v-u)\) parallel or perpendicular to original direction
\(J\cos 30° = 2.4\cos\theta\) or \(J\cos 60° = 2.4\sin\theta - 1.5\)A1 One correct unsimplified equation
Use of \(J = m(v-u)\)M1 Use of \(J = m(v-u)\) to form second equation
\(\begin{pmatrix}-2.4\cos\theta \\ 2.4\sin\theta\end{pmatrix} = \begin{pmatrix}-J\cos 30° \\ J\cos 60° + 1.5\end{pmatrix}\)A1 2nd correct unsimplified equation
\(2.4^2 = \dfrac{3J^2}{4} + \dfrac{J^2}{4} + 1.5J + 1.5^2\), \((J^2 + 1.5J - 3.51 = 0)\)DM1 Form an equation in \(J\) only. Dependent on previous two M1 marks
\(J = 1.3\)A1 1.3 or better (1.268…)
Question 4 (Alt 1):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Use of \(J = m(v-u)\)M1 Use of \(J = m(v-u)\) in any direction
\(J = 0.3(8\cos\alpha - 5\cos 60°)\) or \(5\sin 60° = 8\sin\alpha\)A1 Correct unsimplified equation: \(2.4\cos\alpha = J + 1.5\cos 60°\) or \(2.4\sin\alpha = 1.5\sin 60°\)
Use of \(J = m(v-u)\)M1 Use of \(J = m(v-u)\) in perpendicular direction
A1Correct unsimplified equation
\(2.4^2 = \left(J + \frac{3}{4}\right)^2 + \left(\frac{3}{2}\right)^2 \times \frac{3}{4}\) and \((J^2 + 1.5J - 3.51 = 0)\)DM1 Form an equation in \(J\) only. Dependent on previous two M1 marks
\(J = 1.3\)A1 1.3 or better (1.268…)
Could have a mixture of the first 2 alternatives. M1A1M1A1 for 2 independent equations. DM1A1 for solving (6)
Question 4 (Alt 2):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Impulse momentum triangleM1 Form dimensionally correct vector triangle (for impulse or momentum)
Use of cosine ruleM1 Use of cosine rule in momentum or velocity triangle
\(2.4^2 = J^2 + 1.5^2 - 3J\cos 120°\)A1 Unsimplified equation in \(v\) or \(mv\) with at most one error
A1Correct unsimplified equation
\(J^2 + 1.5J - 3.51 = 0\)DM1 Form a simplified equation in \(J\). Dependent on previous two M1 marks
\(J = 1.3\)A1 1.3 or better (1.268…)
(6) [6]
Question 5a:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Moments about \(A\)M1 Need all terms and dimensionally correct. Condone sign errors, incorrect angles and sin/cos confusion or complete method to form equation in \(T\) (and \(M\))
\(5a \times T\sin 55° = 4a\cos 20° \times Mg\)A1 Correct unsimplified equation in \(T\) (and \(M\))
\(T = \dfrac{4\cos 20°}{5\sin 55°}Mg\ (= 0.918Mg)\)A1 Or equivalent (Exact or \(0.92Mg\) or better)
(3)
Question 5b:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Resolve verticallyM1 Need all terms. Condone sign errors, incorrect angle and sin/cos confusion
\(\uparrow: Mg = V + T\cos 55°\) and \((V = 0.47...Mg)\)A1 Correct unsimplified equation in \(T\) or their \(T\)
Resolve horizontallyM1 Condone consistent sin/cos confusion
\(H = T\sin 55°\) and \((H = 0.75...Mg)\)A1 Correct unsimplified equation in \(T\) or their \(T\)
Resultant \(\lambda = \sqrt{(0.4736..)^2 + (0.7517..)^2}\)M1 Substitute for \(T\) and use Pythagoras
\(= 0.89\)A1 The Q asks for 2 sf
(6)
Question 5b (Alt):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Moments about \(B\): \(Mga\cos 20° + 5aH\cos 70° = 5aV\cos 20°\)M1, A1 Dimensionally correct. Need all terms. Condone sign errors and sin/cos confusion. Correct unsimplified equation
Moments about \(C\): \(5aH = 4aMg\cos 20°\)M1, A1 Dimensionally correct. Condone sign errors and sin/cos confusion. Correct unsimplified equation
Resultant \(\lambda = \sqrt{(0.4736..)^2 + (0.7517..)^2}\)M1 Use Pythagoras
\(= 0.89\)A1 The Q asks for 2 sf
M1A1M1A1 for 2 independent equations, M1A1 to solve for \(\lambda\)
Question 6a:
AnswerMarks Guidance
Answer/WorkingMark Guidance
GPE lost \(= 3g \times 2 - 2g \times 2\sin\theta\)M1 Need all terms. Condone sign errors and sin/cos confusion
\(= 6g - 4g \times \dfrac{5}{13}\)A1 Correct unsimplified. Accept \(\pm\)
\(= \dfrac{58}{13}g = 43.7\ (44)\ \text{(J)}\)A1 Must be positive. Exact multiple of \(g\) or 3 sf or 2 sf
(3)
Question 6b:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Normal reaction \(= 2g\cos\theta \left(= \dfrac{24}{13}g\right)\)B1 Condone \(\dfrac{1176}{65}\)
\(F_{\max} = \dfrac{3}{8} \times R \left(= \dfrac{9g}{13}\right)\)M1 Use \(F = \mu R\) with their \(R\): \(\left(\dfrac{441}{65}\right)\)
Work done \(= 2 \times F_{\max}\)M1 Their \(F_{\max}\)
\(= \dfrac{18g}{13} = 13.6\ \text{(J)}\ 14\text{(J)}\)A1 Exact multiple of \(g\) or 3 sf or 2 sf. Not \(\dfrac{882}{65}\)
(4)
Question 6c:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Total KE gained \(=\) GPE lost \(-\) total WD against frictionM1 Must be using work-energy. Dimensionally correct. Required terms and no extras. Condone sign errors
\(\dfrac{1}{2}(2+3)v^2 = (\text{their}(a)) - (\text{their}(b))\) and \(\left(\dfrac{5}{2}v^2 = \dfrac{58}{13}g - \dfrac{18}{13}g = \dfrac{40}{13}g\right)\)A2ft Follow their (a) and (b), \(-1\) each error
\(v = \sqrt{\dfrac{16}{13}g} = 3.47\ (\text{ms}^{-1})\) or \(3.5\ (\text{ms}^{-1})\)A1 3 sf or 2 sf (need to substitute for \(g\))
(4)
Question 6d:
AnswerMarks Guidance
Answer/WorkingMark Guidance
KE lost \(=\) GPE gained \(+\) WD against frictionM1 Must be using work-energy. Dimensionally correct. Required terms and no extras. Condone sign errors
\(\dfrac{1}{2} \times 2 \times \dfrac{16}{13}g = 2g \times d\sin\theta + \dfrac{3}{8} \times 2g \times \dfrac{12}{13}d\)A2ft Follow their (c) and their \(F_{\max}\), \(-1\) each error
\(\dfrac{16}{13}g = \left(\dfrac{10}{13}g + \dfrac{9}{13}g\right)d\)
\(d = \dfrac{16}{19}\)A1 \(g\) cancels. 0.84 or better (0.8421…)
[15]
Question 7a:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(-12 = 12 - gt\)M1 Use suvat to find time taken
\(t = \frac{24}{g} (= 2.45)\)A1
\(AB = 6t\)M1 Horizontal distance
\(= 14.7\ (15)\) (m)A1 3 sf or 2 sf. Not \(\frac{720}{49}\) (follows use of 9.8). Not \(\frac{144}{g}\) (do not accept \(g\) in denominator)
(4 marks)
Question 7b:
AnswerMarks Guidance
Working/AnswerMark Guidance
Vertical component of velocity \(= (\pm)8\)B1
\(v^2 = u^2 + 2as\)M1 Complete method using suvat to find \(h\)
\(\Rightarrow 8^2 = 12^2 - 2gh\)A1 Correct unsimplified equation
\(h = 4.08\ (4.1)\)A1 3 sf or 2 sf. Not \(\frac{200}{49}\) (follows use of 9.8). Not \(\frac{40}{g}\) (do not accept \(g\) in denominator)
(4 marks)
Alternative method:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\mathbf{v} = \begin{pmatrix}6\\12\end{pmatrix} - \begin{pmatrix}0\\g\end{pmatrix}t \Rightarrow 12 - gt = (\pm)8\)B1 Correct expression for critical value(s) of \(t\)
\(h = 12t - \frac{1}{2}gt^2\)M1 Complete method using suvat to find \(h\)
\(= \frac{48}{g} - \frac{8}{g}\) or \(= \frac{240}{g} - \frac{200}{g}\)A1 Correct unsimplified equation
\(h = 4.08\ (4.1)\)A1 3 sf or 2 sf
Alternative (energy):
AnswerMarks Guidance
Working/AnswerMark Guidance
Conservation of energyM1 Need all terms and dimensionally correct
\(mgh + \frac{1}{2}m \times 10^2 = \frac{1}{2}m(12^2 + 6^2)\)A(B)1, A1 Unsimplified equation with at most one error; Correct unsimplified equation
\(h = 4.08\ (4.1)\)A1 3 sf or 2 sf
(4 marks)
Question 7c:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\begin{pmatrix}6\\-12\end{pmatrix} \cdot \begin{pmatrix}6\\v\end{pmatrix} = 0\)M1 Complete method to find vertical component at \(C\)
\(\Rightarrow v = 3\)A1
\(\mathbf{v} = 6\mathbf{i} + 3\mathbf{j}\ (\text{ms}^{-1})\)A1 Must be a vector in terms of \(\mathbf{i}\) and \(\mathbf{j}\)
If see \(\begin{pmatrix}6\\12\end{pmatrix}\cdot\begin{pmatrix}6\\v\end{pmatrix}=0\) leading to \(\mathbf{v}=6\mathbf{i}-3\mathbf{j}\): mark as misread M1A0A0
(3 marks) [11 total]
Accept working in column vectors throughout apart from the final A1.
Question 8a:
AnswerMarks Guidance
Working/AnswerMark Guidance
Use CLM: \(4mu = 2mv + mw\)M1 Need all terms. Condone sign errors. Dimensionally correct but allow with \(m\) cancelled
\((4u = 2v + w)\)A1 Correct unsimplified. Signs correct for their \(v\), \(w\)
Use Impact lawM1 Used the right way round. Condone sign errors
\(w - v = 2ue\)A1 Correct unsimplified. Signs consistent with CLM equation
\(\Rightarrow 4u = 2(w - 2ue) + w\)DM1 Solve for \(v\) or \(w\). Dependent on previous 2 M marks
\(3w = 4u + 4ue,\ \ w = \frac{4}{3}u(1+e)\) *A1* Obtain given result from correct working
\(v = \frac{2}{3}u(2-e)\)A1 Or equivalent. Must be positive
(7 marks)
Question 8b:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(2 > e\) so \(A\) moving towards centreB1 Correct statement about direction of travel for \(A\)
\(mw - 3mu = mx + 3my\)M1 Use CLM and impact law correctly to form simultaneous equations in \(x\) and \(y\)
\(y - x = e\left(u + \frac{4u}{3} + \frac{4eu}{3}\right)\)
\(\frac{4}{3}eu - \frac{5}{3}u = x + 3y\)A1 Both equations correct unsimplified in \(u\), \(e\), \(x\) and \(y\)
\(3y - 3x = e(7u + 4ue)\)
\(4x = \frac{4}{3}eu - \frac{5}{3}u - 7ue - 4ue^2\)DM1 Solve for \(x\)
\(x = -\frac{5}{12}u - \frac{17}{12}ue - ue^2\)A1 Allow for a correct constant multiple of \(x\)
\(e > 0,\ u > 0\) so \(B\) moving towards centre from opposite direction, hence they collide.*A1* Obtain given answer from correct working
(6 marks)
Alternative for last 3 marks:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(C\) moving towards centre implies \(B\) moving towards centre, so collision. \(C\) moving away from centre, so \(y > 0\): \(x = w - 3u - 3y = -\frac{8u}{3} + \frac{4eu}{3} - 3y\)DM1 Consider direction of \(C\)
\(= -\frac{u}{3}(8 - 4e) - 3y\)A1
\(< 0\) because \(e \leq 1\) and \(y > 0\) hence \(B\) moving towards centre from opposite direction, and they will collide.*A1* Obtain given answer from correct working
(13 marks total for Q8)
## Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $J = m(v-u)$ | M1 | Use of $J = m(v-u)$ parallel or perpendicular to original direction |
| $J\cos 30° = 2.4\cos\theta$ or $J\cos 60° = 2.4\sin\theta - 1.5$ | A1 | One correct unsimplified equation |
| Use of $J = m(v-u)$ | M1 | Use of $J = m(v-u)$ to form second equation |
| $\begin{pmatrix}-2.4\cos\theta \\ 2.4\sin\theta\end{pmatrix} = \begin{pmatrix}-J\cos 30° \\ J\cos 60° + 1.5\end{pmatrix}$ | A1 | 2nd correct unsimplified equation |
| $2.4^2 = \dfrac{3J^2}{4} + \dfrac{J^2}{4} + 1.5J + 1.5^2$, $(J^2 + 1.5J - 3.51 = 0)$ | DM1 | Form an equation in $J$ only. Dependent on previous two M1 marks |
| $J = 1.3$ | A1 | 1.3 or better (1.268…) |

## Question 4 (Alt 1):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $J = m(v-u)$ | M1 | Use of $J = m(v-u)$ in any direction |
| $J = 0.3(8\cos\alpha - 5\cos 60°)$ or $5\sin 60° = 8\sin\alpha$ | A1 | Correct unsimplified equation: $2.4\cos\alpha = J + 1.5\cos 60°$ or $2.4\sin\alpha = 1.5\sin 60°$ |
| Use of $J = m(v-u)$ | M1 | Use of $J = m(v-u)$ in perpendicular direction |
| | A1 | Correct unsimplified equation |
| $2.4^2 = \left(J + \frac{3}{4}\right)^2 + \left(\frac{3}{2}\right)^2 \times \frac{3}{4}$ and $(J^2 + 1.5J - 3.51 = 0)$ | DM1 | Form an equation in $J$ only. Dependent on previous two M1 marks |
| $J = 1.3$ | A1 | 1.3 or better (1.268…) |
| Could have a mixture of the first 2 alternatives. M1A1M1A1 for 2 independent equations. DM1A1 for solving | | **(6)** |

---

## Question 4 (Alt 2):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Impulse momentum triangle | M1 | Form dimensionally correct vector triangle (for impulse or momentum) |
| Use of cosine rule | M1 | Use of cosine rule in momentum or velocity triangle |
| $2.4^2 = J^2 + 1.5^2 - 3J\cos 120°$ | A1 | Unsimplified equation in $v$ or $mv$ with at most one error |
| | A1 | Correct unsimplified equation |
| $J^2 + 1.5J - 3.51 = 0$ | DM1 | Form a simplified equation in $J$. Dependent on previous two M1 marks |
| $J = 1.3$ | A1 | 1.3 or better (1.268…) |
| | | **(6) [6]** |

---

## Question 5a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $A$ | M1 | Need all terms and dimensionally correct. Condone sign errors, incorrect angles and sin/cos confusion or complete method to form equation in $T$ (and $M$) |
| $5a \times T\sin 55° = 4a\cos 20° \times Mg$ | A1 | Correct unsimplified equation in $T$ (and $M$) |
| $T = \dfrac{4\cos 20°}{5\sin 55°}Mg\ (= 0.918Mg)$ | A1 | Or equivalent (Exact or $0.92Mg$ or better) |
| | | **(3)** |

---

## Question 5b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically | M1 | Need all terms. Condone sign errors, incorrect angle and sin/cos confusion |
| $\uparrow: Mg = V + T\cos 55°$ and $(V = 0.47...Mg)$ | A1 | Correct unsimplified equation in $T$ or their $T$ |
| Resolve horizontally | M1 | Condone consistent sin/cos confusion |
| $H = T\sin 55°$ and $(H = 0.75...Mg)$ | A1 | Correct unsimplified equation in $T$ or their $T$ |
| Resultant $\lambda = \sqrt{(0.4736..)^2 + (0.7517..)^2}$ | M1 | Substitute for $T$ and use Pythagoras |
| $= 0.89$ | A1 | The Q asks for 2 sf |
| | | **(6)** |

---

## Question 5b (Alt):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $B$: $Mga\cos 20° + 5aH\cos 70° = 5aV\cos 20°$ | M1, A1 | Dimensionally correct. Need all terms. Condone sign errors and sin/cos confusion. Correct unsimplified equation |
| Moments about $C$: $5aH = 4aMg\cos 20°$ | M1, A1 | Dimensionally correct. Condone sign errors and sin/cos confusion. Correct unsimplified equation |
| Resultant $\lambda = \sqrt{(0.4736..)^2 + (0.7517..)^2}$ | M1 | Use Pythagoras |
| $= 0.89$ | A1 | The Q asks for 2 sf |
| M1A1M1A1 for 2 independent equations, M1A1 to solve for $\lambda$ | | |

---

## Question 6a:

| Answer/Working | Mark | Guidance |
|---|---|---|
| GPE lost $= 3g \times 2 - 2g \times 2\sin\theta$ | M1 | Need all terms. Condone sign errors and sin/cos confusion |
| $= 6g - 4g \times \dfrac{5}{13}$ | A1 | Correct unsimplified. Accept $\pm$ |
| $= \dfrac{58}{13}g = 43.7\ (44)\ \text{(J)}$ | A1 | Must be positive. Exact multiple of $g$ or 3 sf or 2 sf |
| | | **(3)** |

---

## Question 6b:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Normal reaction $= 2g\cos\theta \left(= \dfrac{24}{13}g\right)$ | B1 | Condone $\dfrac{1176}{65}$ |
| $F_{\max} = \dfrac{3}{8} \times R \left(= \dfrac{9g}{13}\right)$ | M1 | Use $F = \mu R$ with their $R$: $\left(\dfrac{441}{65}\right)$ |
| Work done $= 2 \times F_{\max}$ | M1 | Their $F_{\max}$ |
| $= \dfrac{18g}{13} = 13.6\ \text{(J)}\ 14\text{(J)}$ | A1 | Exact multiple of $g$ or 3 sf or 2 sf. Not $\dfrac{882}{65}$ |
| | | **(4)** |

---

## Question 6c:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Total KE gained $=$ GPE lost $-$ total WD against friction | M1 | Must be using work-energy. Dimensionally correct. Required terms and no extras. Condone sign errors |
| $\dfrac{1}{2}(2+3)v^2 = (\text{their}(a)) - (\text{their}(b))$ and $\left(\dfrac{5}{2}v^2 = \dfrac{58}{13}g - \dfrac{18}{13}g = \dfrac{40}{13}g\right)$ | A2ft | Follow their (a) and (b), $-1$ each error |
| $v = \sqrt{\dfrac{16}{13}g} = 3.47\ (\text{ms}^{-1})$ or $3.5\ (\text{ms}^{-1})$ | A1 | 3 sf or 2 sf (need to substitute for $g$) |
| | | **(4)** |

---

## Question 6d:

| Answer/Working | Mark | Guidance |
|---|---|---|
| KE lost $=$ GPE gained $+$ WD against friction | M1 | Must be using work-energy. Dimensionally correct. Required terms and no extras. Condone sign errors |
| $\dfrac{1}{2} \times 2 \times \dfrac{16}{13}g = 2g \times d\sin\theta + \dfrac{3}{8} \times 2g \times \dfrac{12}{13}d$ | A2ft | Follow their (c) and their $F_{\max}$, $-1$ each error |
| $\dfrac{16}{13}g = \left(\dfrac{10}{13}g + \dfrac{9}{13}g\right)d$ | | |
| $d = \dfrac{16}{19}$ | A1 | $g$ cancels. 0.84 or better (0.8421…) |
| | | **[15]** |

## Question 7a:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $-12 = 12 - gt$ | M1 | Use suvat to find time taken |
| $t = \frac{24}{g} (= 2.45)$ | A1 | |
| $AB = 6t$ | M1 | Horizontal distance |
| $= 14.7\ (15)$ (m) | A1 | 3 sf or 2 sf. Not $\frac{720}{49}$ (follows use of 9.8). Not $\frac{144}{g}$ (do not accept $g$ in denominator) |

**(4 marks)**

---

## Question 7b:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Vertical component of velocity $= (\pm)8$ | B1 | |
| $v^2 = u^2 + 2as$ | M1 | Complete method using suvat to find $h$ |
| $\Rightarrow 8^2 = 12^2 - 2gh$ | A1 | Correct unsimplified equation |
| $h = 4.08\ (4.1)$ | A1 | 3 sf or 2 sf. Not $\frac{200}{49}$ (follows use of 9.8). Not $\frac{40}{g}$ (do not accept $g$ in denominator) |

**(4 marks)**

**Alternative method:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mathbf{v} = \begin{pmatrix}6\\12\end{pmatrix} - \begin{pmatrix}0\\g\end{pmatrix}t \Rightarrow 12 - gt = (\pm)8$ | B1 | Correct expression for critical value(s) of $t$ |
| $h = 12t - \frac{1}{2}gt^2$ | M1 | Complete method using suvat to find $h$ |
| $= \frac{48}{g} - \frac{8}{g}$ or $= \frac{240}{g} - \frac{200}{g}$ | A1 | Correct unsimplified equation |
| $h = 4.08\ (4.1)$ | A1 | 3 sf or 2 sf |

**Alternative (energy):**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Conservation of energy | M1 | Need all terms and dimensionally correct |
| $mgh + \frac{1}{2}m \times 10^2 = \frac{1}{2}m(12^2 + 6^2)$ | A(B)1, A1 | Unsimplified equation with at most one error; Correct unsimplified equation |
| $h = 4.08\ (4.1)$ | A1 | 3 sf or 2 sf |

**(4 marks)**

---

## Question 7c:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}6\\-12\end{pmatrix} \cdot \begin{pmatrix}6\\v\end{pmatrix} = 0$ | M1 | Complete method to find vertical component at $C$ |
| $\Rightarrow v = 3$ | A1 | |
| $\mathbf{v} = 6\mathbf{i} + 3\mathbf{j}\ (\text{ms}^{-1})$ | A1 | Must be a vector in terms of $\mathbf{i}$ and $\mathbf{j}$ |
| If see $\begin{pmatrix}6\\12\end{pmatrix}\cdot\begin{pmatrix}6\\v\end{pmatrix}=0$ leading to $\mathbf{v}=6\mathbf{i}-3\mathbf{j}$: mark as misread M1A0A0 | | |

**(3 marks) [11 total]**

Accept working in column vectors throughout apart from the final A1.

---

## Question 8a:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Use CLM: $4mu = 2mv + mw$ | M1 | Need all terms. Condone sign errors. Dimensionally correct but allow with $m$ cancelled |
| $(4u = 2v + w)$ | A1 | Correct unsimplified. Signs correct for their $v$, $w$ |
| Use Impact law | M1 | Used the right way round. Condone sign errors |
| $w - v = 2ue$ | A1 | Correct unsimplified. Signs consistent with CLM equation |
| $\Rightarrow 4u = 2(w - 2ue) + w$ | DM1 | Solve for $v$ or $w$. Dependent on previous 2 M marks |
| $3w = 4u + 4ue,\ \ w = \frac{4}{3}u(1+e)$ * | A1* | Obtain **given result** from correct working |
| $v = \frac{2}{3}u(2-e)$ | A1 | Or equivalent. Must be positive |

**(7 marks)**

---

## Question 8b:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $2 > e$ so $A$ moving towards centre | B1 | Correct statement about direction of travel for $A$ |
| $mw - 3mu = mx + 3my$ | M1 | Use CLM and impact law correctly to form simultaneous equations in $x$ and $y$ |
| $y - x = e\left(u + \frac{4u}{3} + \frac{4eu}{3}\right)$ | | |
| $\frac{4}{3}eu - \frac{5}{3}u = x + 3y$ | A1 | Both equations correct unsimplified in $u$, $e$, $x$ and $y$ |
| $3y - 3x = e(7u + 4ue)$ | | |
| $4x = \frac{4}{3}eu - \frac{5}{3}u - 7ue - 4ue^2$ | DM1 | Solve for $x$ |
| $x = -\frac{5}{12}u - \frac{17}{12}ue - ue^2$ | A1 | Allow for a correct constant multiple of $x$ |
| $e > 0,\ u > 0$ so $B$ moving towards centre from opposite direction, hence they collide.* | A1* | Obtain **given answer** from correct working |

**(6 marks)**

**Alternative for last 3 marks:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $C$ moving towards centre implies $B$ moving towards centre, so collision. $C$ moving away from centre, so $y > 0$: $x = w - 3u - 3y = -\frac{8u}{3} + \frac{4eu}{3} - 3y$ | DM1 | Consider direction of $C$ |
| $= -\frac{u}{3}(8 - 4e) - 3y$ | A1 | |
| $< 0$ because $e \leq 1$ and $y > 0$ hence $B$ moving towards centre from opposite direction, and they will collide.* | A1* | Obtain **given answer** from correct working |

**(13 marks total for Q8)**
4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{e6e37d85-f8de-490a-82a9-8a3c16e2fdd0-10_410_369_251_790}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

A particle $P$ of mass 0.3 kg is moving with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along a straight line on a smooth horizontal plane. The particle receives a horizontal impulse of magnitude $J$ Ns. The speed of $P$ immediately after receiving the impulse is $8 \mathrm {~ms} ^ { - 1 }$. The angle between the direction of motion of $P$ before it receives the impulse and the direction of the impulse is $60 ^ { \circ }$, as shown in Figure 2.

Find the value of $J$.\\
(6)\\

\hfill \mbox{\textit{Edexcel M2 2021 Q4 [6]}}
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