June 2025 Paper 3 Q11

Notes

B1: Correct expression for the horizontal displacement

oe e.g. \(t = 0.5x\)

M1*: Applying \(v = u + at\) correctly with \(v = \pm 8\), \(t = 2\) and \(a = \pm g, \pm 10, \pm 9.8\) - for reference if correct then \(u = 12\). If \(u = 12\) stated from either incorrect or no working, then no further marks can be awarded (so B1 max). Some candidates are using e.g. \(u\sin\theta\) or \(u\cos\alpha\) and correctly obtaining \(u\sin\theta = 12\) which is M1

or other complete method to find an equation/expression for the initial vertical component of the velocity. If using a vector approach ignore horizontal component (unless this is merged with the vertical component in some way)

M1dep*: Applying \(s = ut + \frac{1}{2}at^2\) correctly with their value of \(u\) and \(a = \pm g, \pm 10, \pm 9.8\)

M1: Eliminate \(t\) using \(x = 2t\) to form an equation in \(y\) and \(x\) only (so must be using their value of \(u\)) – dependent on both previous M marks

Allow with \(a = \pm g, \pm 10, \pm 9.8\)

A1: AG www - correct simplification to given answer. Note that the correct parametric equations (so scoring the first three marks) followed by \(y = 6x - \frac{5}{4}x^2 \Rightarrow y = 6x - 1.25x^2\) then scores M1 A0 only

Allow verification of correct equation from \(x = 2t\) and \(y = 12t - 5t^2\) for full marks

Notes

B1: Correct derivative of given expression for \(y\)

B1 for \(6x - 1.25x^2 = 0 \Rightarrow x_{\max} = 4.8 \therefore x = \frac{4.8}{2}\)

M1: Set their derivative of the form \(6 + kx\) (where \(k \lt 0\)) equal to zero, solve for \(x\) and substitute to find corresponding \(y\)

Alternative 1 for first two marks in part (b)

M1: Must use \(v = 0, a = \pm 10\) or \(\pm 9.8\) or \(\pm g\) with any non-zero value of \(u\) (and possibly \(t\)). If \(u = 12\) stated from either incorrect or no working in part (a), then M1 max

Or other complete suvat/energy method to find max. height – if using a vector method then condone poor notation and see guidance in part (a)

A1: Correct equation for max. height but allow \(a = -10\) or \(-9.8\) or \(-g\)

Must be a correct scalar equation for this mark

Alternative 2 for first two marks in part (b)

M1: For attempting to complete the square – must be of the form \(-1.25\left[(x \pm 2.4)^2 \pm \ldots\right]\)

Or M1 for \(\left(-\dfrac{b}{2a} =\right) -\dfrac{6}{2 \times (-1.25)}\)

A1: cao

A1 for \(6(2.4) - 1.25(2.4)^2\)

A1: www cao e.g. \(\frac{36}{5}\) but do not ISW if then finding the distance \(OP\)

Must be 7.2

Notes

M1: Substitute \(x = 2.5\) into \(y^{\prime} = 6 + kx\) (where \(k \lt 0\)) and equate numerical value to \(\tan\theta\)

If correct, then \(y^{\prime} = 6 - 2.5x\)

A1: Or equivalent e.g. 76 to the downward vertical, angle must be positive, and description of direction must be in words so not just on a diagram (bearing only is A0)

or in radians (0.24497… horizontal or 1.3258…vertical) to at least 2 sf

M1*: Correct method to find vertical component velocity at \(x = 2.5\) – must be of the form \(u + a \times 1.25\) for any non-zero \(u\) and \(a = \pm g, \pm 10, \pm 9.8\) or for component squared from \(v^2 = u^2 + 2 \times a \times \frac{115}{16}\) for any non-zero \(u\) and \(a = \pm g, \pm 10, \pm 9.8\)

If correct, then vertical component is \(\pm 0.5\)
If using vertical distance, then must be correct e.g. \(\frac{115}{16}\) or 7.1875 or 7.19 or 7.2

M1dep*: Correct method to find the speed at \(x = 2.5\) - must be of the form \(\sqrt{2^2 + (u + a \times 1.25)^2}\) for any non-zero \(u\) and \(a = \pm g, \pm 10, \pm 9.8\)

A1: awrt 2.06 or exact e.g. \(\frac{\sqrt{17}}{2}, \sqrt{4.25}\) - allow from a positive vertical component of 0.5 from \(v^2 = u^2 + 2as\)

2.061552… If \(u = 12\) stated from either incorrect or no working in part (a), then A0

Alternative for part (c)

M1*: Correct method to find the vertical component of velocity at \(x = 2.5\) – must be of the form \(u + a \times 1.25\) for any non-zero \(u\) and \(a = \pm g, \pm 10, \pm 9.8\) or for component squared from \(v^2 = u^2 + 2 \times a \times \frac{115}{16}\) for any non-zero \(u\) and \(a = \pm g, \pm 10, \pm 9.8\)

If correct, then vertical component is \(\pm 0.5\)
If using vertical distance, then must be correct e.g. \(\frac{115}{16}\) or 7.1875 or 7.19 or 7.2

M1dep*: Correct method to find the speed at \(x = 2.5\) - must be of the form \(\sqrt{2^2 + (u + a \times 1.25)^2}\) for any non-zero \(u\) and \(a = \pm g, \pm 10, \pm 9.8\)

A1: awrt 2.06 or exact e.g. \(\frac{\sqrt{17}}{2}, \sqrt{4.25}\) - allow from a positive vertical component of 0.5 from \(v^2 = u^2 + 2as\)

If \(u = 12\) stated from either incorrect or no working in part (a), then neither A mark can be awarded

M1dep*: For \(\tan\theta = \pm\dfrac{u + a \times 1.25}{2}\) (or reciprocal) for any non-zero \(u\) and \(a = \pm g, \pm 10, \pm 9.8\)

oe e.g. using positive/negative vertical component from \(v^2 = u^2 + 2as\)

A1: Or equivalent e.g. 76 to the downward vertical, angle must be positive, description of direction must be in words so not just on a diagram (bearing only is A0)

or in radians (0.24497… horizontal or 1.3258…vertical) to at least 2 sf

Notes

B1: Must mention that the height or the answer to part (b) would be less

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