Notes

M1: Correct use of Hooke’s law with \(\dfrac{8mg}{3}\) and \(2a\) substituted.

M1: Resolve parallel to the slope with all required terms and no extras. Dimensionally correct. Weight must be resolved but condone sin/cos confusion. \(T\) does not need to be replaced.

A1: Correct unsimplified equilibrium equation, \(T\) does not need to be replaced.

A1*: A complete and correct method using Hooke’s law with the parallel equilibrium equation to obtain the given answer. There must be at least one line of working between the initial equations and the given answer. Must see ‘\(AB = \ldots\)’ Accept fractions \(\dfrac{9a}{4}\) or \(\dfrac{9}{4}a\).

Notes

M1: Correct method for difference in EPE at \(B\) and \(C\), allow either way round. Dimensionally correct and of the correct structure. No need to substitute \(\lambda\). For M mark, condone denominator of \(2a\). Must use extensions \((3a - 2a)\) and \(\left(\dfrac{9a}{4} - 2a\right).\)

A1: Correct unsimplified expression for change in EPE. Allow \(\pm\)

A1: Correct answer, o.e. ISW. Accept \(0.625mga\) and \(0.63mga\). Must be positive but allow a negative expression to change to a positive expression without justification.

Notes

M1: Use of conservation of energy principle from \(C\) to \(B\) to form an equation with all terms of the correct structure and dimensionally correct. Condone sign errors. Condone sin/cos confusion on vertical height.
For GPE, \(mg\left(3a - \dfrac{9a}{4}\right)\sin\theta\) o.e for example \(mg\dfrac{3a}{4}\sin\theta,\ \ mg\dfrac{3a}{4}\left(\dfrac{1}{3}\right),\ \ mg\dfrac{a}{4}\)
For EPE change, may use their answer from (b) if dimensionally correct (of the form \(kma\) where \(k\) is a constant) or start again with 2 EPE terms:
\(\dfrac{\frac{8mg}{3}(3a - 2a)^2}{2(2a)}\) and \(\dfrac{\frac{8mg}{3}\left(\frac{9}{4}a - 2a\right)^2}{2(2a)}\).
For M mark, condone denominator of \(2a\).
(Corrected from the printed mark scheme: the second EPE term is printed with \(\left(3a - \frac{9}{4}a\right)^2\); the extension at \(B\) is \(\frac{9}{4}a - 2a\).)

A1ft: Correct unsimplified equation. Terms must be correct when the equation is formed but follow their answer to (b) for EPE if used.

A1: Correct answer in terms of \(\sqrt{ag}\) ISW. Accept eg \(\dfrac{1}{2}\sqrt{3ag}\), \(0.87\sqrt{ag}\) or better.
N.B. If the final mark in (b) is A0 due to substituting \(g = 9.8\ \text{m s}^{-2}\), do not penalise again for the same reason here.

Notes

M1: Use of GPE for unknown distance from \(E\) to instantaneous rest. May be implied by a difference of 2 GPE terms.

M1: Use of conservation of energy principle to form an equation with one KE term, one GPE term, two EPE terms. For the method mark only we will, condone only one EPE term. However, all terms are required for A marks. Dimensionally correct energy equation (energy terms must have the correct structure). Note there are common rearrangements.
Eg Initial Energy = Final Energy, Energy Loss = Energy Gain oe
Condone \(\pm\) sign errors. M0 for use of suvat. M0 if ‘E’ or similar is used to represent unknown EPE unless recovered.

A1: All 4 terms present in an energy equation with one unknown length.
At most one error. A0 if an energy term is missing.

A1: Fully correct equation in one unknown length.

A1: cao

M1: Use of EPE formula at least once. May be seen here or in earlier working. EPE must have the form \(\dfrac{\lambda x^2}{ka}\) where \(\lambda\) is modulus of elasticity, \(k\) is a constant and \(x\) is extension.

A1: Correct answer

Notes

B1: Hooke’s Law seen with \(4mg\) and \(2a\) substituted.

M1: Resolving vertically. Correct number of terms.

A1: cao for extension.

A1: cao for \(OE\). Note that if the extension is \((OE - 2a)\) in their equation, \(OE\) can be found directly and both A’s can be earned together.

Notes

B1: GPE term seen, ignore sign.

B1: Work term seen \(\dfrac{mga}{4}\), ignore sign.
Allow B1 for the case where WD = \(\dfrac{5mga}{4}\). This is a special case where the work done against resistance is included within the term. \(\dfrac{5mga}{4}\) = WD against resistance + WD against weight.

M1: Use of EPE formula. Accept EPE in the form \(\dfrac{\lambda x^2}{ka}\)

A1: Difference between two correct EPE terms seen, unsimplified.

M1: Work-energy equation is formed with all relevant terms and no extras.: KE, GPE, 2EPE, WD. Condone sign errors.
M0: For work-energy equation with WD = \(\dfrac{5mga}{4}\) and a GPE term. This is because weight is considered twice and so the equation contains an extra term.

A1: Correct unsimplified equation

A1: Correct answer in terms of \(a\) and \(g\), do not allow 9.8 for \(g\) \(\quad v = \sqrt{\dfrac{7ag}{2}}\), \(\ v = \dfrac{1}{2}\sqrt{14ag}\)

Notes

M1: Vertical equilibrium equation or equation of motion with \(a = 0\). Condone sign errors. Correct no. of terms - all 3 forces must be included although \(\left(mg \pm \dfrac{mg}{4}\right)\) may already be simplified.
Hooke’s Law does not need to be substituted but M0 if the equilibrium position from (a) is used.

A1: Correct equation in one unknown.

A1: cao

A1: cao
Note that if the extension is \((OB - 2a)\) in their equation, \(OB\) can be found directly and both A’s can be earned together.

4(c) Alt 1: Using differentiation with a Work - energy equation from the point of release

M1: Forming work-energy equation with the usual rules: all relevant terms to be included and of the correct form and no extra terms.
\(\dfrac{1}{2}mv^2 = mgh - \dfrac{4mg(h - 2a)^2}{2(2a)} - \dfrac{mgh}{4}\)

A1: Correct equation for \(v^2\) and \(h\) (may use a different letter)

A1: Correct equation after differentiating \(v^2\) or \(v\) with respect to \(h\) and setting it equal to zero.
\(\dfrac{\mathrm{d}}{\mathrm{dh}}\left(v^2\right) = 0 \quad \rightarrow \quad \dfrac{3g}{2} = \dfrac{4g(h - 2a)}{a}\)   oe

A1: Correct answer \(OB = \dfrac{19a}{8}\)

Notes

M1: Correct method for EPE seen or implied
Need something of the form \(\dfrac{1}{2}kx^2\) where \(k = \dfrac{\lambda}{a}\)
Must be using the formula for EPE correctly at least once

M1: Require all terms. Dimensionally correct. Condone their EPE. Condone sign errors

A1 A1: Unsimplified equation with at most one error. A repeated error in EPE formula is one error
Correct unsimplified equation.

A1*: Obtain given answer from correct working

Notes

M1: Use correct method for tension to find the extension at equilibrium. Need to see the formula for tension used.
Allow verification with an appropriate conclusion
If they use SHM they must use \(F = ma\) to prove that \(P\) is moving with SHM, otherwise 0/2.

A1*: Correct answer from correct work
Allow verification with an appropriate conclusion

M1: Use given \(x\) to form work-energy equation. Need all terms, and dimensionally correct. Condone sign errors.
Accept with values of \(\lambda\) and \(x\) not substituted

A1 A1: Unsimplified equation with at most one error. Need given \(\lambda\) and given \(x\) substituted at some point. A repeated error in the formula for EPE is one error.
Correct unsimplified equation with given \(\lambda\) and given \(x\) substituted at some point

A1: Use correct method for tension to find the extension at equilibrium. Any equivalent form. \(2.1\sqrt{ag}\) or better

Alternative for the first M1A1

Alt: M1: Or an equivalent method for finding the turning point of a quadratic

Alt: A1*: Correct answer from correct work

Notes

B1: Any valid response.
B0 if answer includes an additional incorrect factor. Must be specific e.g. not just “the GPE changes”, but the GPE of the system changes is OK.
Must relate to an effect on the energy equation
E.g. for B0
The extension changes
\(AB\) will increase
The tension/energy/GPE/work done etc would increase
The KE/GPE/EPE/acceleration/extension/velocity changes
The mass of the spring would drag down and the EPE would change
The EPE/KE/GPE etc would be variable
There would be tension in the spring as well
It has weight
The velocity would decrease as energy is converted

Notes

B1: Correct unsimplified expression for the thrust

M1: Equation of motion. All required terms and no extras. Dimensionally correct. Condone sign errors and sin/cos confusion

A1ft A1ft: Unsimplified equation with at most one error (in \(T\) or their \(T\))
Correct unsimplified equation (in \(T\) or their \(T\))

A1*: Obtain given result from correct working

Notes

B1 B1 B1: Use model to obtain one correct term
Use model to obtain two correct terms
Use model to obtain three correct terms

M1: All required terms and no extras. Dimensionally correct. Condone sign errors and sin/cos confusion.

A1: Correct unsimplified equation

A1: Accept \(0.73\sqrt{gl}\)

Notes

B1 B1 B1: Use model to obtain one correct term
Use model to obtain two correct terms
Use model to obtain three correct terms

M1: Work-energy equation. Need all terms and no extras. Dimensionally correct. Condone sign errors and sin/cos confusion.

A1: Correct unsimplified equation

A1*: Obtain given result from correct working

NB: The use of suvat equations is not a valid alternative method because the acceleration is not constant

Notes

B1: Correct unsimplified expression for the tension in the string

M1: Equation of motion. Need all terms and no extras. Condone sign errors and sin/cos confusion Allow with \(T\) or their \(T\)

A1 A1: Unsimplified equation with at most one error
Correct unsimplified equation

A1: Exact answer or accept 12.8 or 13 \((\text{m s}^{-2})\)

Notes

M1: Use conservation of energy with EPE \(= \dfrac{\lambda x^2}{2a}\). (Condone EPE \(= \dfrac{\lambda x^2}{a}\) here). All three terms required. Must be dimensionally correct. Condone sign errors.

A1: Correct unsimplified equation in \(k\)

A1*: Derive given result from correct working.

Notes

M1: Use \(T = \dfrac{\lambda x}{a}\) and N2L to form equation of motion. All terms required. Dimensionally correct. Condone sign errors

A1: Correct unsimplified equation

A1: Correct only ISW. Condone \(1.7g\) or better. Accept + / -

Notes

M1: Maximum speed at equilibrium seen or implied, and correct method to find \(e\)

A1: Correct \(e\)

Alternative: form energy equation for movement through a height of \(h\) and differentiate \(v^2\) wrt \(h\) to find \(h\) for max \(v\)   M1
\(h = \dfrac{5a}{4}\)   A1

M1: Form energy equation for movement from \(A\) to equilibrium position. Need all 4 terms. Correct form for EPE. Dimensionally correct. Condone sign errors. Allow in \(a\), \(g\) and \(e\) (with \(e\) defined)

A1ft A1ft: Unsimplified equation in their \(e\) with at most one error
Correct unsimplified equation (using their \(e\)) for \(v\)

A1: Any equivalent form. Accept \(1.44\sqrt{ag}\) or \(1.4\sqrt{ag}\)

SHM is not on this specification, but you might see some candidates using it. See below for SHM alternative for parts (b) and (c)

SHM alternative for parts (b) and (c)

Notes

B1 Correct elastic energy

B1 Correct work done by friction

M1 Attempt a work-energy equation. Must have 3 terms:  work done by friction, elastic energy, GPE.
EPE term must be of the form \(= k\lambda\dfrac{x^2}{l}\)   \(k = 1, 2\) or \(\dfrac{1}{2}\)
Work done term must be of the form   distance \(\times\ \mu mg\cos\) or \(\sin\alpha\)

A1ft Correct equation, ft their EPE and work terms

dM1 Solve their 3 term quadratic to obtain a value for the extension as a multiple of \(l\). Award if correct answer follows a correct quadratic. If the quadratic is incorrect award only if working shown (ie general formula shown explicitly and used or by implication through substitution correct for their equation, pos root only needed)

A1 Correct extension  decimal or exact

A1 Complete by adding 1 to the numerical multiple of \(l\) Must be 3 significant figures.

(Corrected from the printed mark scheme: the fifth line is printed as \(l^2 + 4lx + 5x^2 = 3l^2 + 3lx\).)

ALT

Using dist moved:  \(kl\) or \(x\)

EPE \(= \dfrac{1}{2} \times 2mg\dfrac{(kl - l)^2}{l}\),    WD \(= kl\mu mg\cos\alpha\)    B1, B1

\(kl\mu mg\cos\alpha + \dfrac{1}{2} \times 2mg\dfrac{(kl - l)^2}{l} = klmg\sin\alpha\)    M1A1ft

\(5k^2 - 12k + 5 = 0\)

\(k = 1.86\)    M1A2 (Give A1 for correct ans with more than 3 sf or exact)

Notes

M1 Total PE

A1 *given answer*

Notes

M1 Differentiate

M1 \(\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = 0\) and solve for \(\theta\)

Notes

M1 Differentiate to obtain second derivative

DM1 Find value of second derivative when \(\theta = \cos^{-1}\dfrac{6}{7}\). Dependent on preceding M1

A1 \((14.8\ldots mga)\)

Notes

M1 Attempt Hooke's Law using the whole string or a half string.

A1 Correct equation.

A1cso Correct given value of \(\lambda\) obtained with no errors seen.

Notes

M1 Use Hooke's Law with the new longer length for the string or half string.  \(\lambda\) must be 16, but length need not be correct but use of 0.2 for extension of full string or 0.1 for extension of half string scores M0.

A1 Obtain \(T = 24\)

M1 Resolve parallel to \(F\) or in another direction which gives an equation connecting \(T\) and \(F\).

A1 Obtain the correct value of \(F\)

Notes

B1 Correct initial or final EPE with one string (\(l = 0.4\)) or two half strings (\(l = 0.2\))

M1 Attempt an energy equation with the difference of 2 EPE terms and a KE term. The EPE terms must be of the form \(k\dfrac{\lambda x^2}{l}\).

A1A1 Deduct one mark per error.   (A1A1, A1A0 or A0A0)

dM1 Solve for \(v\). Depends on the previous M mark.

A1cso Correct value of \(v\), min 2 sf or exact value.
Energy terms wrong way round in the equation will lose this mark even if modulus sign inc here.

NB If the energy terms are subtracted the wrong way round, max score is B1M1A1A0M1A0

Notes

M1 GPE of rod + particle (relative to a fixed point)

A1 Correct total

M1 Use of EPE \(= \dfrac{\lambda x^2}{2a}\)

M1 Total PE expressed in \(\sin\theta\)

A1 Given answer as printed

Notes

M1 Differentiate \(V\) – condone errors but do not accept integration

DM1 Set \(V' = 0\) and solve for \(\text{trig}(\theta) = \mathrm{f}(k)\). Dependent on the first M1

A1 \(\cos\theta = 0 \Rightarrow \theta = \dfrac{\pi}{2}\) need not be seen

B1ft ft on their \(\sin\theta\)

M1 Solve right hand inequality for \(k\)

A1 Given answer

Notes

M1 Substitute \(k = 10\) and find second derivative of \(V\)

A1 Any equivalent form

DM1 Dependent on the previous M1. Substitute their trig. values. Need to be considering the whole of \(V''\)

Accept 14.3 or better

A1 With no errors seen

Notes

B1 Correct EPE when extension is \(x\)

M1 Equating EPE to GPE lost  EPE to be of the form \(k\dfrac{\lambda x^2}{l}\), where \(k\) is a rational no.

A1ft Correct equation ft their EPE. Use of unknown must be consistent.

DM1 Solve their equation (3TQ) (quadratic formula must be correct)

A1 \(x = 2.069\ldots\) neg value not needed

A1 Add 1.5 to 2.069... and give final answer to 2 or 3 sf.   (No "exact" answers allowed here due to use of \(g\).)

ALT 1: Using extension \((x - 1.5)\):

B1 Correct EPE when extension is \((x - 1.5)\) where \(x\) is total length

M1 Equating EPE to GPE lost   EPE of form shown above

A1ft Correct equation ft their EPE. Use of unknown must be consistent. No simplification needed. Equation is \(\dfrac{14.7(x - 1.5)^2}{3} = 0.6gx\)

A1 Simplify to \(x^2 - 4.2x + 2.25 = 0\) or equivalent 3TQ

DM1 Solve their equation (3TQ) (quadratic formula must be correct)

A1 \(x = 3.57\) or 3.6  Must be 2 or 3 sf

ALT 2:

Use \(v^2 = u^2 + 2as\) or energy to obtain speed at natural length, then energy to \(A\).

B1 for EPE at \(A\)
No more marks until an energy equation with EPE, GPE and KE terms seen.

M1A1ft Energy equation ft their EPE and initial speed   EPE of form shown above

DM1A1 A1 As main scheme

NB: Solution of quadratic by calculator: Method mark only available if solution is correct ( 2.069  or  3.569)

Notes

M1 Use NL2 at \(A\) inc use of Hooke's law.  Formula for HL to be correct. Ext can be \((3.569 - 1.5)\)

A1ft Correct numbers in the equation,  ft their extension

A1 24 or 24.0 only.  (No negatives allowed.)

Notes

M1 Use Hooke's law to obtain the tension and equate to the weight

A1 Correct equation

A1 Solve to get \(\lambda = 19.6\)   Accept 20 or \(2g\)

Notes

M1 Attempt an energy equation with the difference of 2 EPE terms and a loss of GPE
EPE formula must be of the form \(k\dfrac{\lambda x^2}{l}\)

A1ft EPE terms correct follow through their \(\lambda\)

A1 GPE term correct, including all signs in the equation correct   If \(x\) used for EPE and GPE A0 here

A1 Correct length \(AC\)   If \(\lambda = 20\) is used, this is p.a. and so scores A0

ALT

Methods depending on SHM must prove SHM first, but if correct answer only is given award B1 (M1 on e-PEN)

By integration: Integrating and substituting yields an equation equivalent to the one shown - mark from here    M1A1A1ft  -1 each error  ft on \(\lambda\)

Notes

B1 correct equation perpendicular to the plane

B1 correct expression for work done against friction

M1 work-energy equation

A2 fully correct; A1 one error;

A1 correct expression for \(x\)   no errors in the working

Notes

B1 use Hooke's law to obtain a correct expression for \(T\)

M1 using NL2 parallel to the plane to set up an inequality for situation where no motion

A1 correct inequality

DM1 solving to get an inequality for \(\mu\)

A1 correct inequality and no errors in the working

If only error is use of \(<\) instead of \(\leqslant\), deduct final A mark only

Notes

B1 Or \(-2mgr\cos^2\theta\), or \(-mgr(1 + \cos 2\theta)\) or equivalent

M1 Correct unsimplified total

A1 In terms of \(r\) & \(\theta\)

A1 Reach given answer correctly

Notes

M1 Differentiate \(V\)

M1 Derivative = 0 and solve for \(\theta\)

A3 -1 for each missing solution

M1 Second derivative of \(V\)

A1 Need to see \(-2mgr\) or equivalent

A1 Do not need to consider the symmetrical position as well

Notes

M1 for an energy equation with an EPE term of the form \(\dfrac{kmgx^2}{l}\) and a GPE term. If a KE term is included it must become 0 later.

A1 for a correct EPE term

B1 for a correct GPE term. This can be in terms of the distance moved down the plane or the vertical distance fallen

M1 dep for solving their equation to obtain the distance moved or using the vertical distance and obtaining the distance moved along the plane.

A1 for \(x = \dfrac{4}{5}l\) oe eg \(x = \dfrac{12}{15}l\)

If \(m\) used instead of \(2m\), assuming correct otherwise: (a) M1A1B0M1A0   (so 2 penalties for mis-read)

Alternative for Question 4 (a): Using NL2

Notes

B1 for resolving perpendicular to the plane to obtain \(R = 2mg\cos\alpha\). May only be seen in an equation.

M1 for an work-energy equation with an EPE term of the form \(\dfrac{kmgx^2}{l}\), a GPE term and the work done against friction. The work term must include a distance along the plane.

A1 for EPE and GPE terms correct and work subtracted from the GPE

B1 ft for the work term ft their \(R\)

M1 dep for solving to obtain a value for \(\mu\)

A1 cso for \(\mu = \dfrac{3}{8}\) oe inc 0.375 but not 0.38

If \(m\) used instead of \(2m\), assuming correct otherwise:

(b) B1 \(R = mg\cos\alpha\)

M1, A1 Equation, with EPE correct and \(mg \times \dfrac{2}{5}l \times \dfrac{3}{5}\)

B1 ft \(\mu\dfrac{4mg}{5} \times \dfrac{2}{5}l\)

DM1, A1 \(\mu = 0\)

Alternative for Question 4 (b): Using NL2

If SHM methods are used, SHM must be proved first.

Notes

M1 EPE term needs to be dimensionally correct. Need all three terms.

A2 Correct unsimplified

M1 All in \(\sin\theta\)

A1 Given Answer

Notes

M1 Differentiate

M1 Set derivative = 0

M1 Use of \(\sin\theta \leqslant \dfrac{1}{2}\)

Notes

M1 Second derivative (\(8mgl\) or \(24mgl\) not needed) [or differentiate \(8mgl(3\sin 2\theta - 3\cos\theta)\)]

A1 Numerical unsimplified

M1 by numerical evaluation or justification from trig terms \((36mgl)\)

A1 CSO

Notes

M1 for attempting Hooke's Law, formula must be correct, either explicitly or by correct substitution.

A1 for \(20 = \dfrac{\lambda \times 0.3}{1.2}\)

A1 for obtaining \(\lambda = 80\)

B1 for the initial EPE \(\dfrac{"\lambda" \times 0.3^2}{2.4}\ \ (= 3\text{ J})\)  their value for \(\lambda\) allowed. May only be seen in the eqaution.

M1 for a work-energy equation with one EPE term, one KE term and work done against friction (Award if second EPE/KE terms included provided these become 0). The EPE must be dimensionally correct, but need not be fully correct (eg denominator 1.2 instead of 2.4)

A1ft for a completely correct equation   follow through their EPE

A1 cao for \(v = 0.80\) or 0.805  must be 2 or 3 sf

NB: This is damped harmonic motion (due to friction) so all SHM attempts lose the last 4 marks.

Notes

M1 for any complete method leading to a value for either \(BC\). If the distance travelled after the string becomes slack is found the work must be completed by adding 0.3  Their EPE found in (a) used in energy methods.

MS method is energy from \(B\) to \(C\) ie work done against friction = loss of EPE.

OR Energy from point where the string becomes slack to \(C\)  ie work done against friction = loss of KE  and completed for the required distance

OR NL2 to obtain the acceleration \(\left(-\dfrac{2g}{5}\right)\) while the string is slack and \(v^2 = u^2 + 2as\) to find the distance and completed for the required distance

A1cso for \(BC = 0.38\) or 0.383 (m) must be 2 or 3 sf

Alternatives:

(Corrected from the printed mark scheme: \(x\) is printed as \(0.8265\ldots\).)

Notes

Alternative

Notes

Alternative

Alternative to (b)

Alternative approach using SHM for (b) and (c)

If SHM is used mark (b) and (c) together placing the marks in the gird as shown.

Alternative approach using SHM for (b) and (c)

If SHM is used mark (b) and (c) together placing the marks in the gird as shown.

Notes

M1 A1 attempt at both equations

Alternative approach to (b) using calculus with energy.

Alternative approach to (b) using calculus with Newton’s second law.

Notes

In part (a)

DM1 requires EE, PE and KE to have been included in the energy equation.

If sign errors lead to \(V^2 = -\dfrac{ga}{2}\), the last two marks are M0 A0

In parts (a) and (b) A marks need to have the correct signs

Notes

In parts (a) and (b) A marks need to have the correct signs

In part (b) for M1 need one KE term in energy equation of at least 3 terms with distance \(\dfrac{3a}{4}\) to indicate first method, and two KE terms in energy equation of at least 4 terms with distance \(\dfrac{a}{4}\) to indicate second method.

Alternative (using point of projection and point where string becomes slack):

SHM approach in part (b). (Condone this method only if SHM is proved)

Notes

Special case \(T\sin\theta = mg\) giving \(\theta = 30\) is M1 A0 A0 unless there is evidence that they think \(\theta\) is with horizontal – then M1 A1 A0

Notes

↓ marks a mark that depends on the M mark above it (an arrow in the scheme).

1st M1: allow for attempt to find work done by frictional force (i.e. not just finding friction).
2nd M1: “relevant” terms, i.e. energy or work terms!
A1 f.t. on their work done by friction

Notes

The published mark scheme for this paper is handwritten.

Notes

The scheme prints a direction symbol before the alternative equation; it resolves perpendicular to the string.

Notes

(Corrected from the printed mark scheme: the unit is printed as (cm).)

Notes

The published mark scheme for this paper is handwritten.

Notes

The scheme brackets the two M1 marks for the energy equation and solving for \(V^2\): the second depends on the first.

Alternative using Newton’s second law

HL   \(T_1 = \dfrac{4mg(l+x)}{l}, \quad T_2 = \dfrac{4mg(l-x)}{l}\)
N2L   \(m\ddot{x} = T_2 - T_1 = -\dfrac{8mg}{l}x\)	M1 A1
This is SHM, centre \(M\)
\(a = \dfrac{l}{3}, \quad \omega^2 = \dfrac{8g}{l}\)	A1, A1ft
\(v^2 = \omega^2(a^2 - x^2) \Rightarrow v^2 = \dfrac{8g}{l}\left(\dfrac{l^2}{9} - x^2\right)\)   Depends on showing SHM	M1
At \(M\), \(x = 0\),   \(V^2 = \dfrac{8gl}{9}, \ V = \left(\dfrac{8gl}{9}\right)^{\frac{1}{2}}\)   or exact equivalents	M1, A1

The scheme brackets the last two M1 marks: the second depends on the first.

Alternative using Newton’s second law

The scheme brackets these two B1 marks.

Notes

The scheme brackets the last two M1 marks: the M1 for \(0.8g \times 0.6 + 4 = 0.8a\) depends on the previous M1.

Notes

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

(Corrected from the printed mark scheme: the extension in the elastic potential energy is printed as \(2a\cos 2\theta - \tfrac{5}{4}a\); the string length is \(FB = 2a\cos\theta\), as used in the next line.)