Notes

B1: Correct only, \(3\mathbf{i}\). Accept a magnitude of 3, if seen with correct direction indicated. Accept trig components with 3 and direction indicated
E.g. \(v\cos\theta = \sqrt{13}\cos\alpha = \sqrt{13} \times \dfrac{3}{\sqrt{13}} = 3\)

M1: Correct use of the impact law perpendicular to \(RS\) with \(\dfrac{1}{3}\) and 2. Allow \(\pm\).
M0 if separation and approach are on the wrong side.
May also use scalar product eg \(\begin{pmatrix} 3 \\ 2 \end{pmatrix} \bullet \begin{pmatrix} 0 \\ 1 \end{pmatrix} = -\dfrac{1}{3}\begin{pmatrix} x \\ y \end{pmatrix} \bullet \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) or magnitude and direction with trig components eg \(v\sin\theta = \dfrac{\sqrt{13}\sin\alpha}{\frac{1}{3}} = 3 \times \sqrt{13} \times \dfrac{2}{\sqrt{13}} = 6\)

A1: Correct perpendicular component of \(\mathbf{v}\), \(-6\mathbf{j}\).
Accept a magnitude of 6, if seen with correct direction indicated.
A0 for \(+6\mathbf{j}\)

M1: Correct method for change in KE. Allow \(\pm 6\)

A1: Correct unsimplified equation for change in KE (allow \(\pm 6\))

A1*: Obtain given answer from complete and correct working including consideration of parallel components.
NB

• If the given answer is obtained by only considering the KE of the perpendicular components ie \(\dfrac{1}{2} \times 0.25(6^2) - \dfrac{1}{2} \times 0.25(2^2)\), the method is incomplete.
  Max score B1M1A1M1A1A0*
• If the given answer is obtained by considering the KE of the perpendicular components, \(\dfrac{1}{2} \times 0.25(6^2) - \dfrac{1}{2} \times 0.25(2^2)\) and states that the KE parallel to \(RS\) is unchanged, all marks are available.
• If the solution only consists of the energy equation, this is insufficient working for a given answer \(\dfrac{1}{2} \times 0.25(3^2 + 6^2) - \dfrac{1}{2} \times 0.25(3^2 + 2^2)\). Max score B1M0A0M1A1A0*

Notes

ALT 1: Method [1] Accept column vectors throughout working but must be in i-j form for final A mark.

M1: Method [1] using scalar product and components of \(\mathbf{w} = a\mathbf{i} + b\mathbf{j}\) and \((3\mathbf{i} + 2\mathbf{j})\) parallel to \(ST\) \((3\mathbf{i} + 4\mathbf{j})\) i.e. any multiple of \((3\mathbf{i} + 4\mathbf{j})\) to form an equation in \(a\) and \(b\).
Allow without \(\dfrac{1}{5}\).

A1: Correct equation in \(a\) and \(b\).

M1: Method [1] using scalar product and components of \(\mathbf{w} = a\mathbf{i} + b\mathbf{j}\) and \((3\mathbf{i} + 2\mathbf{j})\) perpendicular to \(ST\) \((-4\mathbf{i} + 3\mathbf{j})\) i.e. any multiple of \((-4\mathbf{i} + 3\mathbf{j})\) to form an equation in \(a\) and \(b\). Must have \(\pm\dfrac{1}{3}\). Allow without \(\dfrac{1}{5}\).
N.B. It is possible to form 2 simultaneous equations using 2 perpendicular equations instead of 1 parallel and 1 perpendicular
NEL:   \(\dfrac{1}{5}(-4\mathbf{i} + 3\mathbf{j}).(a\mathbf{i} + b\mathbf{j}) = -\dfrac{1}{3}\left[\dfrac{1}{5}(-4\mathbf{i} + 3\mathbf{j}).(3\mathbf{i} + 2\mathbf{j})\right] \;\Rightarrow\; -4a + 3b = 2\)
Perp.:   \(\mathbf{I} = 0.25\begin{pmatrix} a - 3 \\ b - 2 \end{pmatrix}\) and \(\mathbf{I} \bullet \begin{pmatrix} 3 \\ 4 \end{pmatrix} = 0 \;\Rightarrow\; 3a + 4b = 17\)

A1: Correct equation in \(a\) and \(b\).

dM1: Dependent on previous 2 M’s. Complete method [1] to solve simultaneous equations and obtain \(\mathbf{w}\).

A1: Correct i-j form. Accept \(\dfrac{43}{25}\mathbf{i} + \dfrac{74}{25}\mathbf{j}\) or \((1.72\mathbf{i} + 2.96\mathbf{j})\)

ALT 2

ALT 2: Method [2] Accept column vectors throughout working but must be in i-j form for final A mark.

M1: Method [2] using scalar product and component of \((3\mathbf{i} + 2\mathbf{j})\) and unit vector parallel to \(ST\)

A1: Correct expression

M1: Method [2] using scalar product and component of \((3\mathbf{i} + 2\mathbf{j})\) and unit vector perpendicular to \(ST\)

A1: Correct expression.

dM1: Dependent on previous 2 M’s. Complete method [2] to combine and form \(\mathbf{w}\). Must see \(\pm\dfrac{1}{3}\) with \((-4\mathbf{i} + 3\mathbf{j})\) component.

A1: Correct i-j form. Accept \(\dfrac{43}{25}\mathbf{i} + \dfrac{74}{25}\mathbf{j}\) or \((1.72\mathbf{i} + 2.96\mathbf{j})\)

ALT 3

ALT 3: Method [3] Accept column vectors throughout working but must be in i-j form for final A mark.

M1: Method [3] finding coefficients to express \((3\mathbf{i} + 2\mathbf{j})\) as multiples of \((3\mathbf{i} + 4\mathbf{j})\) and \((-4\mathbf{i} + 3\mathbf{j})\)

A1: Correct expression

M1: Solve to obtain the coefficients.

A1: Correct only.

dM1: Dependent on previous 2 M’s. Complete method [3] to \(\mathbf{w}\). Must see \(\pm\dfrac{1}{3}\) with \((-4\mathbf{i} + 3\mathbf{j})\).

A1: Correct i-j form. Accept \(\dfrac{43}{25}\mathbf{i} + \dfrac{74}{25}\mathbf{j}\) or \((1.72\mathbf{i} + 2.96\mathbf{j})\)

ALT 4

ALT 4: Method [4] Accept column vectors throughout working but must be in i-j form for final A mark.

M1: Method [4] using rotation matrix on \(3\mathbf{i} + 2\mathbf{j}\)

A1: Correct matrix expression

M1: Method [4] using \(\pm\dfrac{1}{3}\) with the correct component to find velocity relative to \(ST\).

A1: Correct only.

dM1: Dependent on previous 2 M’s. Complete method [4] by reversing rotation to form \(\mathbf{w}\)

A1: Correct i-j form. Accept \(\dfrac{43}{25}\mathbf{i} + \dfrac{74}{25}\mathbf{j}\) or \((1.72\mathbf{i} + 2.96\mathbf{j})\)

ALT 5

Scheme	Marks	AO
Method [5] using speed and direction parallel to \(ST\)	M1	3.1b
\(w\cos\delta = \sqrt{13}\cos\gamma \quad (= 3.4)\)	A1	1.1b
Method [5] using speed and direction perpendicular to \(ST\) with \(\pm\dfrac{1}{3}\)	M1	3.1b
\(w\sin\delta = \dfrac{1}{3}\sqrt{13}\sin\gamma \quad (= 0.4)\)	A1	1.1b
Complete method [5] to combine modulus with angle \((\beta + \delta)\) and form \(\mathbf{w}\) \(\mathbf{w} = \dfrac{\sqrt{293}}{5}\cos(\beta + \delta)\mathbf{i} + \dfrac{\sqrt{293}}{5}\sin(\beta + \delta)\mathbf{j}\)	dM1	2.1
\(= \dfrac{43}{25}\mathbf{i} + \dfrac{74}{25}\mathbf{j}\)    (i-j notation required)	A1	2.2a
	(6)

ALT 5: Method [5] Answer must be in i-j form for final A mark.

M1: Method [5] using speed and direction parallel to \(ST\). Use \(\sqrt{3^2 + 2^2}\) with \(\gamma\) where \(\gamma = \tan^{-1}\left(\dfrac{4}{3}\right) - \tan^{-1}\left(\dfrac{2}{3}\right)\)

A1: \(w\cos\delta = \sqrt{3^2 + 2^2}\cos\gamma \quad (= 3.4)\)

M1: Method [5] using speed and direction perpendicular to \(ST\). Use \(\pm\dfrac{1}{3}\) and \(\sqrt{3^2 + 2^2}\) with \(\gamma\) where \(\gamma = \tan^{-1}\left(\dfrac{4}{3}\right) - \tan^{-1}\left(\dfrac{2}{3}\right)\)

A1: \(w\sin\delta = \dfrac{1}{3}\sqrt{3^2 + 2^2}\sin\gamma \quad (= 0.4)\)

dM1: Dependent on previous 2 M’s. Complete method [5] to combine the modulus with the relevant angle and form \(\mathbf{w}\)
\(\mathbf{w} = \dfrac{\sqrt{293}}{5}\cos(\beta + \delta)\mathbf{i} + \dfrac{\sqrt{293}}{5}\sin(\beta + \delta)\mathbf{j}\)

A1: Correct i-j form. Accept \(\dfrac{43}{25}\mathbf{i} + \dfrac{74}{25}\mathbf{j}\) or \((1.72\mathbf{i} + 2.96\mathbf{j})\)

Notes

Allow column vectors throughout the question

M1: Complete method to find the direction of the normal (perpendicular to the wall).
Eg

• Main scheme: use of impulse equation with difference in momenta, \(m\) is not required so may see only the difference in velocities.
• Find the direction of the wall first by equating parallel components then find a perpendicular direction using the scalar product or otherwise.
  Eg \((4\mathbf{i} - \mathbf{j}) \cdot (x\mathbf{i} + y\mathbf{j}) = (\mathbf{i} + 3\mathbf{j}) \cdot (x\mathbf{i} + y\mathbf{j}) \;\Rightarrow\; 3x = 4y\)
  Wall direction // \((4\mathbf{i} + 3\mathbf{j}) \Rightarrow\) Impulse direction // \((-3\mathbf{i} + 4\mathbf{j})\)

A1: Any correct vector parallel to impulse \((-3\mathbf{i} + 4\mathbf{j})\)

M1: Complete method to resolve both velocities parallel to impulse

• Using the scalar product (main scheme)
• Using trig: velocity components from shown on diagram

  

  where \(\tan\theta = \tfrac{3}{4}\)    \(\tan\alpha = \tfrac{1}{4}\)    \(\tan\beta = \tfrac{3}{1}\)
• Using rotation matrix: velocity components shown on RHS of matrix equation

  

  \[\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} 4 \\ -1 \end{pmatrix} = \begin{pmatrix} \mathbf{3.2} \\ 2.6 \end{pmatrix}\]\[\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} \mathbf{-1.8} \\ 2.6 \end{pmatrix}\]

A1: Both correct velocity components parallel to impulse, allow positive or negative values in both cases.

DM1: Dependent on previous M. Use of Impact Law parallel to impulse. Ratio of speeds correct way up (separation \(\div\) approach)

A1: Correct answer (0.56 or better)

Notes

Allow column vectors throughout the question

M1: Method to find velocity after impact with second wall: \(\mathbf{i}\) component unchanged, \(\mathbf{j}\) component: \(\tfrac{1}{3}(3)\). Condone sign errors.

A1: Correct velocity components after impact, may be given if seen on a diagram.

DM1: Complete method to find the angle of deflection for their velocity after second impact. Dependent on previous M.

A1: Correct answer, must be rounded to nearest degree.

Notes

B1: \(e_1U\sin\alpha\) seen from a relevant equation or on a diagram.

B1: \(U\cos\alpha\) seen in a relevant equation or on a diagram.

M1: A clear method using two equations to eliminate \(U\) and \(V\).

A1*: GIVEN answer correctly obtained. Must include two equations showing how to reach both \(\tan\beta\) and \(e_1\tan\alpha\). It is not sufficient to use the side lengths of the snooker eg using \(\tan\beta = \dfrac{CQ}{PC}\) oe is not sufficient.
Accept \(\tan\beta = e_1\tan\alpha\) or \(e_1\tan\alpha = \tan\beta\)

Notes

This part states ‘hence’ so \(\beta\) must be used.

B1: Form a correct expression for \(\tan\gamma\) or \(\cot\gamma\) in terms of \(e_2\) and \(\beta\) or \((90 - \beta)\). May quote result from (a) or obtain again.

M1: Use result from (a) to eliminate \(\tan\beta\) and form an equation in \(\alpha\), \(\gamma\), \(e_1\), \(e_2\)

A1*: Given answer correctly obtained. The solution must include the replacement of \(\tan\beta\) and rearrangement to the correct form.
Accept \(e_1\tan\alpha = e_2\cot\gamma\) or \(e_2\cot\gamma = e_1\tan\alpha\)

Notes

M1: Clear attempt to find angle sum (condone slips) or another relevant starting point eg an expression for angle \(PAQ\)

M1: Clear statement to form an inequality eg

• the correct angle sum < 180 is acceptable
• angle \(PAQ > 0\)

A1: Correct simplified inequality in correct form

M1: Correct method to form an inequality in tan or cot

M1: Using part (b) to eliminate the angles

A1*: Given answer correctly obtained

7(c) alt: Use of trig identity

M1: (angle \(APQ\) + angle \(AQP\)) = \((180^\circ - \alpha - \beta) + \{180^\circ - (90^\circ - \beta) - \gamma\}\) = \(270 - \alpha - \gamma\)

M1: To return to \(A\), (angle \(APQ\) + angle \(AQP\)) < \(180^\circ\), since \(APQ\) is a triangle

A1: \(\tan(\alpha + \gamma) = \dfrac{\tan\alpha + \tan\gamma}{1 - \tan\alpha\tan\gamma}\)   and   \(\tan\alpha = \dfrac{e_2\cot\gamma}{e_1}\)   or   \(\tan\alpha = \dfrac{e_2}{e_1\tan\gamma}\)
Leads to
\(\tan(\alpha + \gamma) = \dfrac{e_2 + e_1\tan^2\gamma}{e_1\tan\gamma - e_2\tan\gamma}\)   oe

M1: \(180 > (\alpha + \gamma) > 90 \;\Rightarrow\; \tan(\alpha + \gamma) < 0 \;\Rightarrow\; \dfrac{e_2 + e_1\tan^2\gamma}{e_1\tan\gamma - e_2\tan\gamma} < 0\)
Condone if ‘180 >’ is not stated again.

M1: Since numerator > 0
\(e_1\tan\gamma - e_2\tan\gamma < 0\)

A1: \(e_2 > e_1\)   *

Notes

B1: Use the given information in (b) to make any equivalent statement with a correct reason and no incorrect statements.

• \(\alpha = 90^\circ - \gamma\), so it moves parallel to \(AP\)
• \(\alpha = 90^\circ - \gamma\), so it moves parallel to the initial velocity

Notes

M1: Correct no. of terms, dimensionally correct, mass × velocity, condone sin/cos confusion.

A1: Correct equation

M1: Dimensionally correct. Must be subtracting, but condone subtracting in the wrong order and sin/cos confusion

A1: Correct unsimplified equation

A1*: Given answer correctly obtained. Must be EXACT factorisation.

6(a) alt1

Scheme	Marks	AO
Impulse-momentum vertically.	M1 M1	3.1a 3.1a
\(I\cos\alpha = m(0 - -u)\)	A1 A1	1.1b 1.1b
\(I = mu\sec\alpha\)   *	A1*	2.2a
	(5)

6(a) alt 2

Scheme	Marks	AO
Introduce and use an expression for \(e\)
CLM along the plane:	M1	3.1a
\(u\sin\alpha\) unchanged	A1	1.1b
Finds an expression for \(e\) together with Impulse-momentum perpendicular to the plane \(\quad \tan\alpha = \dfrac{eu\cos\alpha}{u\sin\alpha} \Rightarrow e = \tan^2\alpha\)   and \(I = m(eu\cos\alpha - (-u\cos\alpha))\)	M1	3.1a
\(I = m(u\cos\alpha\tan^2\alpha - (-u\cos\alpha))\)	A1	1.1b
\(I = m\left(\dfrac{u\sin^2\alpha}{\cos\alpha} + u\cos\alpha\right) = \dfrac{mu}{\cos\alpha}(\sin^2\alpha + \cos^2\alpha) = mu\sec\alpha\) *	A1*	2.2a
	(5)

6(a) alt 3

Scheme	Marks	AO
Use a vector approach and magnitude of impulse
CLM along the plane:	M1	3.1a
\((m)u\sin\alpha = (m)v\cos\alpha\)    (this leads to \(v = u\tan\alpha\))	A1	1.1b
Impulse-momentum as a vector equation followed by Pythagoras to find the magnitude. \(I = m\begin{pmatrix} -v \\ u \end{pmatrix}\)   and   \(|I| = m\sqrt{v^2 + u^2}\)	M1	3.1a
\(|I| = m\sqrt{u^2\tan^2\alpha + u^2}\)	A1	1.1b
\(I = m\sqrt{u^2(1 + \tan^2\alpha)} = m\sqrt{u^2\sec^2\alpha} = mu\sec\alpha\) *	A1*	2.2a
	(5)

Notes

M1: Attempt at NEL

M1: Squaring and adding their expressions for \(v\sin\alpha\) and \(v\cos\alpha\) to obtain \(v^2\).

A1*: Given answer correctly obtained. Must be EXACT.

Notes

M1: Expression for difference of KE in terms of \(m\), \(u\), \(\alpha\) and \(e\)

A1*: Given answer correctly obtained. Factorisation must be EXACT.

Notes

M1: Complete method to eliminate \(\alpha\) e.g. using \(\tan^2\alpha = e\) to eliminate \(\alpha\)
Any trig identity used must be correct eg \(\sec^2\alpha = 1 + e\) or \(\cos^2\alpha = \dfrac{1}{1 + e}\)

A1: Correct answer.

Notes

B1: Component parallel to the wall unchanged. Could be on a diagram or implied if they use \(e\tan\alpha = \tan\beta\)

B1: Use of impact law perpendicular to the wall. Could be on a diagram or implied if they use \(e\tan\alpha = \tan\beta\)

B1: Use the direction to determine the range for \(e\). (could come via \(e\tan\alpha = \tan\beta \lt 1/2\))

Notes

B1: Correct vector perpendicular to \(ST\) seen or implied
\(\mu\) can have any scalar value

M1: Use scalar product to find component of \(\mathbf{v}\) parallel to \(ST\). \(\lambda\) can have any scalar value

A1: Correct unsimplified expression for the magnitude

M1: Use scalar product and impact law perpendicular to \(ST\) to find magnitude of component perpendicular to the wall. For their perpendicular vector. Must clearly be using \(e = \tfrac{1}{2}\). \(\gamma\) can have any scalar value.

A1: Correct unsimplified expression for the perpendicular component. Allow \(\pm\)

M1: Combine the magnitudes and directions to obtain the velocity. The perpendicular should now be in the correct direction.

A1: Correct simplified velocity.

Alternative 1 (8b alt 1)

B1: Correct vector perpendicular to \(ST\) seen or implied.
\(\mu\) can have any scalar value

M1: Correct method for component parallel to \(ST\)

A1: Correct equation in \(a\) and \(b\)

M1: Correct method for component perpendicular to \(ST\)
Allow \(\pm\) For their perpendicular vector

A1: Correct equation in \(a\) and \(b\)

M1: Solve for \(a\) and \(b\) to obtain velocity. Using the correct direction for the perpendicular component

A1: Correct simplified answer.

Alternative 2 (8b alt 2)

B1: Correct vector perpendicular to \(ST\) seen or implied.
\(\mu\) can have any scalar value

M1: Split \(\mathbf{v}\) into components parallel and perpendicular to \(ST\)

A1: Two equations in \(p\) and \(q\)

M1: Use the impact law perpendicular to \(ST\) For their perpendicular vector

A1: Correct unsimpified perpendicular component.
With \(q\) or their \(q\)

M1: Solve for \(p\) and \(q\) to obtain velocity Using the correct direction for the perpendicular component

A1: Correct simplified total.

Alternative 3 (8b alt 3)

Scheme	Marks	AO
\(\alpha - \beta = 8.1\ldots^\circ\)	B1
Component of \(\mathbf{w}\) parallel to \(ST\) is \(|\mathbf{v}|\cos(\alpha - \beta)\)	M1
\(= \sqrt{40}\cos(\alpha - \beta)\left(= \sqrt{40} \times \dfrac{7}{\sqrt{50}} = 6.26..\right)\)	A1
Component of \(\mathbf{w}\) perpendicular to \(ST\) is \(\dfrac{1}{2}|\mathbf{v}|\sin(\alpha - \beta)\)	M1
\(\pm\dfrac{1}{2} \times \sqrt{40}\sin(\alpha - \beta)\left(= \dfrac{\sqrt{40}}{2} \times \dfrac{1}{\sqrt{50}} = 0.447\ldots\right)\)	A1
\(\mathbf{w} = |\mathbf{w}|\cos(\alpha + \theta)\mathbf{i} + |\mathbf{w}|\sin(\alpha + \theta)\mathbf{j}\)	M1
\(\mathbf{w} = \left(\dfrac{27}{5}\mathbf{i} + \dfrac{16}{5}\mathbf{j}\right)\) or \((5.4\mathbf{i} + 3.2\mathbf{j})\)   \((\text{m s}^{-1})\)	A1
	(7)

B1: Seen or implied. \(\sin(\alpha - \beta) = \dfrac{1}{\sqrt{50}},\ \cos(\alpha - \beta) = \dfrac{7}{\sqrt{50}},\ \tan(\alpha - \beta) = \dfrac{1}{7}\)

M1: Correct use of their \(|\mathbf{v}|\) and their \(\alpha - \beta\)

A1: Correct unsimplified

M1: Correct use of \(\dfrac{1}{2}\), their \(|\mathbf{v}|\) and their \(\alpha - \beta\)

A1: Correct unsimplified

M1: Use of Pythagoras and correct method for \(\theta + \alpha\).
\(\cos(\alpha + \theta) = \dfrac{27}{\sqrt{5}\sqrt{197}},\ \sin(\alpha + \theta) = \dfrac{16}{\sqrt{5}\sqrt{197}}\)
\(\alpha + \theta = 30.65^\circ\)

A1: Correct simplified total.

Notes

M1: Use of scalar product or equivalent. Allow M1 if not using unit vector

A1: Correct unsimplified expression for component parallel to wall

M1: Correct use of impact law perpendicular to the wall. Condone sign error

A1: Correct unsimplified expression for component perpendicular to wall

M1: Complete method to solve for \(\mathbf{v}\)

A1*: Obtain given result from correct working

Alternative (a) alt

Alternative (a) alt 2

Notes

M1: Use of \(\mathbf{I} = m\mathbf{v} - m\mathbf{u}\) with velocities or perpendicular components of velocities. Must be subtracting but allow subtraction in either order.

M1: Correct use of Pythagoras to find modulus

A1: Accept 1.4 Ns or better

Notes

B1 B1: Parallel component correct
Perpendicular component correct Check the diagram

M1: Equation for KE in \(v, \theta\). Dimensionally correct. Includes all components. Condone \(\tfrac{1}{2}\) used on wrong side

M1: Form and solve equation in \(\cos\theta\)

A1: Or exact equivalent

Alternative (a)

Notes

B1 B1: Parallel component correct
Perpendicular component correct

M1: Equation for KE in \(x, y\). Dimensionally correct. Includes all components. Condone \(\tfrac{1}{2}\) used on wrong side

M1: Use their \(\cos\theta\) to form and solve equation in \(e\)

A1: Or exact equivalent

Alternative (b)

Notes

B1 B1 B1 B1: One mark for each component correct.

M1: Form expression for \(\tan\dfrac{\alpha}{2}\) in terms of \(\tan\alpha\)

M1: Form and solve equation in \(\tan\dfrac{\alpha}{2}\)

A1*: Obtain given answer from correct working

NB: This is a “Show that ..” question. A candidate who assumes, without proof, that \(\tan\dfrac{\alpha}{2} = e^2\tan\alpha\) can only score the last two marks.

Notes

B1: Correct use of \(t = \dfrac{1}{3}\)   Must be seen / used in part (b)

M1: Dimensionally correct expression for change in KE
NB note that they may not show component parallel to the wall

M1: Dimensionally correct expression for the percentage of KE lost.

A1: Accept 29 (%) or better   Accept \(\dfrac{260}{9}\)

Notes

Correction: In Alternative (a) 2 the ratio is \(b = \dfrac{3}{2}a\) (corrected from the printed mark scheme: \(b = \dfrac{2}{3}a\)).

M1: Must be finding the difference between two momenta or two velocities

A1: Correct unsimplified equation for the impulse or for change in velocity

M1: Use of scalar product or equivalent. In the alt method allow full marks if \(\sqrt{13}\) not used.

A1*: Reach given conclusion from correct working

If working with angles, score
M1 for correct method to find components parallel to the wall
A1 for \(\sqrt{53}\cos 40.36..^\circ\) and \(\sqrt{37}\cos 24.23..^\circ\)
M1 for comparing the two values
A0 because the work has involved decimal approximations (since working towards an exact given answer).
Alternative: Could use \(e\tan 24.2..^\circ = \tan 40.36..^\circ\)

Alternative (a)

Alternative (a) 2

Notes

M1: Condone if not using a unit vector

A1 A1: One correct value
Second correct values

If working with angles, score M1A1A1 for \(\sqrt{53}\sin 40.36..^\circ\) and \(\sqrt{37}\sin 24.23...^\circ\)

M1: Use their components the right way round in the impact law. Condone sign error.

A1: 0.53 or better (0.52941……)

Notes

B1: Use model to find component parallel to the wall

M1: Use model and impact law perpendicular to the wall

A1: Correct perpendicular component

M1: Use \(\tfrac{1}{3}\) and their components to form equation in \(e\)   \(\left(v = \dfrac{6\sqrt{10}}{5} = 3.79\right)\)

A1*: Correct answer from correct exact working

If only see \(e\tan\alpha = \tan\beta\) with no explanation of where it comes from then score 0/5

Notes

B1: First component correct

B1: Second component correct

Alternative: B1 for speed of impact with \(BC = 3.79\ldots\)
B1 for path on leaving \(BC\) at \(56.3\ldots^\circ\) to \(BC\)

M1: Use Pythagoras’ theorem or trigonometry to find the speed

A1: Any equivalent form. 2.2 or better \((2.1633\ldots)\)

Notes

B1 B1: Two independent refinements relating to the modelling e.g.

• Include friction between the floor and the ball
• Include friction between the ball and the walls
• Give the ball dimensions.
• Consider air resistance
• Spin / rotation

2 Alt

Notes

M1 Use of scalar product. Do not need to see \(\dfrac{1}{5}\)

A1 *Given Answer*

4a alt

\(\theta + \alpha = 46.3\ldots^\circ\)

M1 Equate components and form an equation in \(a\) and \(\theta\)

Notes

M1 Scalar product of \(-7\mathbf{j}\) and unit vector parallel to plane

M1 Use Pythagoras to find components perpendicular to the plane

A1 Both correct

DM1 Use if impact law. Dependent on preceding M mark

Alt 4b

Alt 4b

\(\theta + \alpha = 46.3\ldots^\circ,\ v = \sqrt{37}\)

Alt 4b

Alt 4b

M1A1 Pair of equations

M1 Square and substitute to eliminate \(\theta + \alpha\)

M1A1 Substitute for \(\theta\) to obtain \(e\).

Notes

M1 Correct use of impact law. Condone trig. confusion

B1 In terms of \(U\) and \(\alpha\)

B1 In terms of \(U\) and \(\alpha\)

B1 Seen or implied

A1 Equation in \(k\) and \(\alpha\)

A1 From correct work only

NB: A candidate who makes a false assumption about an angle \(\alpha\) in triangle CXY can score a maximum B1B1B1 B0B0 B1 M1 A0 (6/8)

Notes

M1 Eliminate \(\theta\) and solve for \(v\).

A1 Obtain given answer correctly with no errors seen

Notes

M1 Components parallel to \(BC\)

M1 Components perpendicular to \(BC\)

A1 Equations correct for their 19.1

M1 Square and add or divide to find \(\tan\phi\)

A1 \((\phi = 9.83^\circ)\) Follow their 19,1?

7balt

Notes

M1 Use the speeds to form an equation in \(x\) & \(y\) (or equivalent)

A1 Correct unsimplified

A1 Correct ratio for \(x\) & \(y\) (any equivalent form)

M1 To find the correct angle

A1 Correct in \(x\) & \(y\)

(Corrected from the printed mark scheme: the M1 A1 line is printed as \(\frac{1}{2}\left(x^2 + y^2\right) = x^2 + \frac{1}{9}y^2\) and the next line as \(9\left(x^2 + y^2\right) = 2\left(9x^2 + y^2\right),\ 9x^2 = 7y^2,\ x = \frac{\sqrt{7}}{3}y\); the speed after is half the speed before, so the factor is \(\frac{1}{4}\), as in alt2 and in the printed final line.)

alt2

M1 Use the speeds to form an equation in \(u\) & \(\theta\) (or equivalent)

A1 Correct unsimplified

A1 Correct trig ratio for \(\theta\) (or equivalent)

M1 To find the correct angle

A1 Correct in \(\theta\) (or equivalent)

Notes

M1 A1 Must be in correct direction but condone trig confusion

M1 A1 Condone consistent trig confusion

Notes

B1 Requires scalar product or gradient diagram.

Notes

M1 Requires all terms present and of the correct structure

A2 -1 each error

7(b) alt

M1 A2 Parallel to the wall. Condone trig confusion? -1 each error. Both angles in same variable?

B1 When seen in (b). Implied by 26.6 or 18.4

M1 Perpendicular to the wall. Condone consistent trig confusion?

A1 \(e = \sqrt{\dfrac{10a^2}{b^2} - 4}\)

A1 0.67 or better

Notes

M1 Resolve parallel to the wall

Alt: reasonable attempt at equation connecting two variables

A1 Correct as above or equivalent

equation correct

A1 \(u\) in terms of \(v\) or v.v. – not necessarily simplified.

or ration of the two variables correct

M1 expression for KE lost

DM1 expression in one variable for fraction of KE lost – could be \(u/v\) as above

A1 cao

The first three marks can be awarded in (b) if not seen in (a)

Notes

M1 Use NIL perpendicular to the wall and form equation in \(e\)

A1 Correct unsimplified expression as above or \(eu\sin 60^o = v\sin 30^o\) or equivalent

DM1 Substitute values for trig functions or use relationship from (a) and rearrange to e = …..

A1 cao accept decimals to at least 3sf

The first two marks can be awarded in (a)

Notes

The published mark scheme for this paper is handwritten.

Notes

The published mark scheme for this paper is handwritten.

Notes

The published mark scheme for this paper is a set of worked answers: no mark allocation is printed.

Notes

(Corrected from the printed mark scheme: \(\mathbf{v}_2\) is printed as \(\tfrac{39}{25}\mathbf{i} - \tfrac{64}{25}\mathbf{j}\) and the final velocity as \(4\mathbf{i} - \mathbf{i}\).)