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June 2025 Paper 3 Mechanics Q5

EdexcelCurrent spec8 marksProjectiles

Figure 3: stone projected at 14 m/s at angle theta above the horizontal from O at the top of a cliff of height H m above N; it lands at A on the horizontal ground with NA = 40 m — Figure 3

A small stone is projected with speed \(14\ \text{m s}^{-1}\) from a point \(O\) on the top of a cliff.
The point \(O\) is \(H\) metres vertically above the point \(N\).

Point \(N\) is on horizontal ground.

The stone is projected at an angle \(\theta\) above the horizontal, where \(\tan\theta = \dfrac{1}{2}\)

The stone strikes the horizontal ground at the point \(A\), where \(NA = 40\) m, as shown in Figure 3.

The stone is modelled as a particle moving freely under gravity.

Using this model, find

(a) the value of \(H\) (5)

(b) the maximum height of the stone above the horizontal ground. (3)

Mark scheme (a)

Scheme	Marks	AO
Using horizontal motion	M1	3.3
\(14\cos\theta \times t = 40\)	A1	1.1b
Using vertical motion	M1	3.4
\((s =)\, 14\sin\theta \times t - \dfrac{1}{2}gt^2\) oe with down as positive.	A1	1.1b
OR
Using the equation of the path
\((s =)\, 40\tan\theta - \dfrac{g \times 40^2}{2 \times 14^2\cos^2\theta}\) oe	M2 A2
\((H =)\ 30\) or 30.0	A1	1.1b
	(5)

Notes

N.B. Penalise the use of \(g = 9.81\) once (the first time it is used) for the whole question.
All marks available if they use a different letter for \(\theta\)

M1: Complete method, using the horizontal motion, to give an equation in \(t\) and \(\theta\) only, condone sin/cos confusion and sign errors
(\(\theta\) may or may not be substituted for)

A1: Correct equation (\(t\) and/or \(\theta\) may or may not be substituted for)

M1: Complete method, using the vertical motion, to give an equation in \(t\) and \(\theta\) only, condone sin/cos confusion and sign errors
(\(t\) and/or \(\theta\) may or may not be substituted for)

A1: Correct equation (\(t\) and/or \(\theta\) may or may not be substituted for)
N.B. Just mark the RHS of the equation for this A mark.

M2 A2: Quoting and using correctly the equation of the path.
N.B. No marks if they misquote the equation and no marks if they put in 14 and/or 40 incorrectly.
These 4 marks are indivisible i.e. all 4 or 0.

A1: \(H = 30\) or 30.0 to 2 or 3 sf, after use of \(g = 9.8\), obtained correctly.

Mark scheme (b)

Scheme	Marks	AO
Using vertical motion OR Using cons. of energy	M1	2.1
\(0^2 = (14\sin\theta)^2 - 2gh\) OR \(\dfrac{1}{2}m\left(14^2 - (14\cos\theta)^2\right) = mgh\)	A1	1.1b
32 (m)	A1ft	1.1b
	(3)
	(8 marks)

Notes

N.B. Penalise the use of \(g = 9.81\) once (the first time it is used) for the whole question.
All marks available if they use a different letter for \(\theta\)

M1: Complete method to give equation in \(h\) and \(\theta\) only, condone sin/cos confusion and sign errors
They may find \(t\) first: \(0 = 14\sin\theta - gt \Rightarrow t = \dfrac{14\sin\theta}{g}\)
then \(h = (14\sin\theta)\dfrac{14\sin\theta}{g} - \dfrac{1}{2}g\left(\dfrac{14\sin\theta}{g}\right)^2\)
or \(h = -\dfrac{1}{2}(-g)\left(\dfrac{14\sin\theta}{g}\right)^2\)
or \(h = \dfrac{(14\sin\theta + 0)}{2}\left(\dfrac{14\sin\theta}{g}\right)\)

A1: Correct equation

A1ft: \(\lvert\text{their } H\rvert + 2\), to 2sf or 3sf

June 2024 Paper 3 Mechanics Q5

EdexcelCurrent spec12 marksProjectiles

Figure 4: stone projected from O at 35 m/s at angle alpha above the horizontal ground, following a parabolic path that lands at A — Figure 4

At time \(t = 0\), a small stone is projected with velocity \(35\ \text{m s}^{-1}\) from a point \(O\) on horizontal ground.

The stone is projected at an angle \(\alpha\) to the horizontal, where \(\tan\alpha = \dfrac{3}{4}\)

In an initial model

the stone is modelled as a particle \(P\) moving freely under gravity
the stone hits the ground at the point \(A\)

Figure 4 shows the path of \(P\) from \(O\) to \(A\).

For the motion of \(P\) from \(O\) to \(A\)

at time \(t\) seconds, the horizontal distance of \(P\) from \(O\) is \(x\) metres
at time \(t\) seconds, the vertical distance of \(P\) above the ground is \(y\) metres

(a) Using the model, show that\[y = \frac{3}{4}x - \frac{1}{160}x^2\] (6)

(b) Use the answer to (a), or otherwise, to find the length \(OA\). (2)

Using the model, the greatest height of the stone above the ground is found to be \(H\) metres.

The model is refined to include air resistance.

Using this new model, the greatest height of the stone above the ground is found to be \(K\) metres.

(d) State which is greater, \(H\) or \(K\), justifying your answer. (1)

(e) State one limitation of this refined model. (1)

Mark scheme (a)

Scheme	Marks	AO
Using horizontal motion, \(s = ut\), with 35 resolved	M1	3.3
\(x = 35\cos\alpha \times t\)	A1	1.1b
Using vertical motion, \(s = ut + \dfrac{1}{2}at^2\), with 35 resolved	M1	3.4
\(y = 35\sin\alpha \times t - \dfrac{1}{2}gt^2\)	A1	1.1b
Eliminate \(t\): \(y = 35\sin\alpha \times \dfrac{x}{35\cos\alpha} - \dfrac{1}{2}g\left(\dfrac{x}{35\cos\alpha}\right)^2\)	DM1	3.1b
\(y = \dfrac{3}{4}x - \dfrac{1}{160}x^2\) *	A1*	1.1b
N.B. No marks available if they just quote the equation of the path.
ALTERNATIVE: they do (b) and/or (c) first using a suvat method Assume \(y = ax^2 + bx + c\) Use any three of (0,0), (120,0) from part (b), (60,22.5) from part (c) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{4}\) at \(x = 0\) to find \(a\), \(b\), and \(c\). M1A1, M1A1, DM1A1* for finding each of \(a\), \(b\) and \(c\) and stating final answer in correct form. N.B. If they realise that \(c = 0\), and just use \(y = ax^2 + bx\), that could score M1A1. Enter marks on ePEN in the order in which \(a\), \(b\) and \(c\) are found: e.g. \(x = 0,\ y = 0 \Rightarrow c = 0\) M1A1 \(x = 0,\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2ax + b = \dfrac{3}{4} \Rightarrow b = \dfrac{3}{4}\) M1 A1 \(x = 120,\ y = 0 \Rightarrow 0 = 120^2a + 120 \times \dfrac{3}{4} \Rightarrow a = \dfrac{-1}{160}\) DM1 so, \(y = \dfrac{3}{4}x - \dfrac{1}{160}x^2\) A1*
	(6)

Notes

M1: Correct terms but condone sin/cos confusion and sign errors
Available if they use \(s\) instead of \(x\)

A1: Correct equation in \(x\) and \(t\)
N.B. they may have the wrong value for \(\cos\alpha\)

M1: Correct terms but condone sin/cos confusion and sign errors
Available if they use a different letter for \(y\) provided it’s not the same as they’ve used for \(x\).
N.B. M0 if they subsequently use a value for \(y\) e.g. 0

A1: Correct equation in \(y\) and \(t\)
N.B. they may have the wrong value for \(\sin\alpha\)
They may have \(t\) in terms of \(x\), from their first equation.

DM1: Dependent on the two previous M marks for eliminating \(t\) to give an equation in \(y\) and \(x\) only.

A1*: Given answer correctly obtained, with at least one further line of working with trig ratios and \(g = 9.8\) explicitly seen or 4.9 oe used for \(0.5g\).
Allow \(y = \dfrac{3x}{4} - \dfrac{x^2}{160}\) and with \(y\) on the RHS

Mark scheme (b)

Scheme	Marks	AO
ALT 1 \(0 = \dfrac{3}{4}x - \dfrac{1}{160}x^2\) and solve for \(x\) ALT 2 \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{4} - \dfrac{x}{80} = 0 \Rightarrow x = 60\) and \(OA = 2 \times 60\) ALT 3 A complete suvat method to find \(OA\): e.g. \(0 = 35\sin\alpha \times t - \dfrac{1}{2}gt^2\) or \(0 = 35\sin\alpha - g\dfrac{t}{2}\) or \(-35\sin\alpha = 35\sin\alpha - gt\) to find \(t\) \(\left(= \dfrac{70\sin\alpha}{g} = \dfrac{30}{7}\right)\) AND \((OA =)\, 35\cos\alpha \times t = 35\cos\alpha \times \dfrac{70\sin\alpha}{g}\) N.B. OR use the calculator to input the equation of the path which then gives \(y_{\max} = 45/2\) when \(x = 60\) with no working, so \(OA = 2 \times 60\)	M1	3.1b
\((OA =)\ 120\) (m)	A1	1.1b
	(2)

Notes

M1: ALT 1: Use of \(y = 0\) in equation of path and solve for \(x\).
ALT 2: Use calculus to find the \(x\)-coordinate of the max point and double it
ALT 3: Any other complete suvat method to find \(OA\), condone sin/cos confusion and sign errors

A1: cao

Mark scheme (c)

Scheme	Marks	AO
ALT 1 \(H = \dfrac{3}{4} \times 60 - \dfrac{1}{160} \times 60^2\) ALT 2 \(y = \dfrac{-1}{160}(x^2 - 120x) = 22.5 - \dfrac{1}{160}(x - 60)^2\) so max \(y = 45/2\) or 22.5 ALT 3 \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3}{4} - \dfrac{2x}{160} = 0 \Rightarrow x = 60\) then find \(y\) when \(x = 60\) ALT 4 A complete suvat method: e.g. \(H = \dfrac{(35\sin\alpha)^2}{2g}\) or \(0 = 35\sin\alpha - gt\) to find the time to top, \(t = \dfrac{35\sin\alpha}{g}\), or use half their time they found in (b) AND \(H = 35\sin\alpha \times \frac{35\sin\alpha}{g} - \dfrac{1}{2}g\left(\frac{35\sin\alpha}{g}\right)^2\) or \(\left(\dfrac{35\sin\alpha + 0}{2}\right) \times \dfrac{35\sin\alpha}{g}\) N.B. OR use the calculator to input the equation of the path which then gives \(y_{\max} = 45/2\) (when \(x = 60\)) with no working.	M1	3.1b
\((H =)\ 22.5\) Accept 23	A1	1.1b
	(2)

Notes

M1: ALT 1: Use of \(x = \left(\dfrac{1}{2} \times \text{their } OA\right)\) in equation of path
ALT 2: Complete the square for the equation of the path and deduce maximum value of \(y\): Need to see \(p - q(x - r)^2\) with \(p\) as the answer
ALT 3: Use \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) to find the \(x\)-coordinate of the max point and then use the path equation to find the \(y\)-coordinate
ALT 4: Any complete suvat method to find \(H\), condone sin/cos confusion and sign errors

A1: A0 for \(\dfrac{45}{2}\)

Mark scheme (d)

Scheme	Marks	AO
\(H\) is greater (or \(K\) is smaller), as air resistance would slow the particle down oe.	B1	3.5a
	(1)

Notes

B1: Possible justifications for \(H\) greater (or \(K\) smaller):
air resistance will provide an extra force acting against the stone or opposing the motion
air resistance would take away energy from the stone
work would be done against air resistance
air resistance will mean the acceleration will be less (or deceleration greater)
air resistance would reduce the velocity/speed
Allow ‘it’ for air resistance
B0 if no justification given.
B0 for air resistance would reduce the initial velocity of the stone
B0 for air resistance will limit the vertical height (not a reason) or oppose the vertical force.
Ignore extras.

Mark scheme (e)

Scheme	Marks	AO
e.g. the inaccuracy of using \(9.8\ \text{m s}^{-2}\) for \(g\)	B1	3.5b
	(1)
	(12 marks)

Notes

B1: Any single correct answer.

Acceptable answer:
the inaccuracy of using \(9.8\ \text{m s}^{-2}\) for \(g\)
or wind or weather effects
or the spin of the stone
or the size (or shape or surface area) of the stone oe
or the stone is still modelled as a particle
N.B. Allow if the stone is referred to as e.g. a ball

B0 if any incorrect extras

Unacceptable answers:
The model (it) does not take account of:
the mass or weight of the stone
air resistance.
The ground may not be horizontal.

B0 for consequences of air resistance being included e.g.
the path won’t be a parabola
the path won’t be symmetrical

June 2023 Paper 3 Mechanics Q5

EdexcelCurrent spec11 marksProjectiles

Figure 2: ball projected from O at 28 m/s at angle alpha to horizontal ground, passing through A, which is 40 m horizontally and 20 m vertically from O — Figure 2

A small ball is projected with speed \(28\ \text{m s}^{-1}\) from a point \(O\) on horizontal ground.

After moving for \(T\) seconds, the ball passes through the point \(A\).

The point \(A\) is 40 m horizontally and 20 m vertically from the point \(O\), as shown in Figure 2.

The motion of the ball from \(O\) to \(A\) is modelled as that of a particle moving freely under gravity.

Given that the ball is projected at an angle \(\alpha\) to the ground, use the model to

(a) show that \(T = \dfrac{10}{7\cos\alpha}\) (2)

(b) show that \(\tan^2\alpha - 4\tan\alpha + 3 = 0\) (5)

(c) find the greatest possible height, in metres, of the ball above the ground as the ball moves from \(O\) to \(A\). (3)

The model does not include air resistance.

(d) State one other limitation of the model. (1)

Mark scheme (a)

Scheme	Marks	AO
Use horizontal motion to give an equation in \(T\) and \(\alpha\) only: \(28\cos\alpha \times T = 40\)	M1	3.4
\(T = \dfrac{10}{7\cos\alpha}\) *	A1*	1.1b
	(2)

Notes

N.B. In this question, allow misread of \(\alpha\) for \(a\).

M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors

A1*: Correct printed answer correctly obtained.
Allow \(\dfrac{10}{7\cos\alpha} = T\) OR \(T = \dfrac{40}{28\cos\alpha} = \dfrac{10}{7\cos\alpha}\) OR \(\dfrac{40}{28\cos\alpha} = \dfrac{10}{7\cos\alpha} = T\)
OR \(t\) instead of \(T\)

Mark scheme (b)

Scheme	Marks	AO
Use vertical motion to give an equation in \(T\) and \(\alpha\) only	M1	3.3
\(20 = (28\sin\alpha)T - \dfrac{1}{2}gT^2\)	A1	1.1b
Eliminate \(T\) to give an unsimplified equation in \(\alpha\) only: \(20 = (28\sin\alpha) \times \dfrac{10}{7\cos\alpha} - \dfrac{1}{2}g\left(\dfrac{10}{7\cos\alpha}\right)^2\)	M1	1.1b
Use \(\sec^2\alpha = 1 + \tan^2\alpha\) oe to give an unsimplified equation in tan \(\alpha\) only: \(20 = 40\tan\alpha - \dfrac{1}{2}g \times \dfrac{100}{49}(1 + \tan^2\alpha)\)	M1	3.1b
\(\tan^2\alpha - 4\tan\alpha + 3 = 0\) * (allow \(0 = \tan^2\alpha - 4\tan\alpha + 3\))	A1*	2.2a
	(5)

Notes

N.B. In this question, allow misread of \(\alpha\) for \(a\).

M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors

A1: Correct equation

M1: Eliminate \(T\), using either the given answer in (a) or their own \(T\) expression, from their equation to give an unsimplified equation in \(\alpha\) only

M1: Use \(\sec^2\alpha = 1 + \tan^2\alpha\) to produce an equation in \(\tan\alpha\) only

A1*: Given answer correctly obtained.
N.B. Must be \(\alpha\) (or \(a\)) in the final answer but allow a different angle in the working.

Mark scheme (c)

Scheme	Marks	AO
Solve and use of \(\tan\alpha = 3\) or \(\sin\alpha = \dfrac{3}{\sqrt{10}}\) or \(\alpha = 71.565..^\circ\) to find an equation in \(H\) only.	M1	3.1b
\(0 = (28\sin\alpha)^2 - 2gH\) where \(\tan\alpha = 3\) \((\alpha = 71.565..^\circ)\)	M1	3.4
\(H = 36\) or 36.0 (m)	A1	1.1b
	(3)

Notes

N.B. In this question, allow misread of \(\alpha\) for \(a\).

M1: Solve given equation and select larger value of \(\tan\alpha\) and use it to try to obtain an equation in \(H\) only.

M1: Complete method to give an equation in H only, using larger value of \(\alpha\), correct no. of terms, dim correct, condone sin/cos confusion and sign errors.

A1: cao. Must be positive, (allow a negative value, changed to a positive answer).
N.B. This answer comes from use of \(g = 9.8\), so must be rounded to 2 or 3 sf.

Mark scheme (d)

Scheme	Marks	AO
e.g. spin of the ball, the wind, the dimensions or shape of the ball, ball is modelled as a particle, uses an inaccurate value of \(g\), motion takes place in 3D not in 2D, \(g\) could be variable. B0 if mass or weight are mentioned. B0 for ground may not be horizontal.	B1	3.5b
	(1)
	(11 marks)

Notes

B1: B0 if any incorrect extras

June 2022 Paper 3 Mechanics Q5

EdexcelCurrent spec12 marksProjectiles

Figure 3: golf ball projected from A at angle alpha to horizontal ground, landing at B, where AB = 120 m — Figure 3

A golf ball is at rest at the point \(A\) on horizontal ground.

The ball is hit and initially moves at an angle \(\alpha\) to the ground.

The ball first hits the ground at the point \(B\), where \(AB = 120\) m, as shown in Figure 3.

The motion of the ball is modelled as that of a particle, moving freely under gravity, whose initial speed is \(U\ \text{m s}^{-1}\)

Using this model,

(a) show that \(U^2\sin\alpha\cos\alpha = 588\) (6)

The ball reaches a maximum height of 10 m above the ground.

(b) Show that \(U^2 = 1960\) (4)

In a refinement to the model, the effect of air resistance is included.

The motion of the ball, from \(A\) to \(B\), is now modelled as that of a particle whose initial speed is \(V\ \text{m s}^{-1}\)

This refined model is used to calculate a value for \(V\)

(d) State one further refinement to the model that would make the model more realistic. (1)

Mark scheme (a)

Scheme	Marks	AO
Using horizontal motion	M1	3.3
Whole Motion: \(U\cos\alpha \times t = 120\) OR Half way: \(U\cos\alpha \times t = 60\)	A1	1.1b
Using vertical motion	M1	3.4
Whole Motion: \(U\sin\alpha \times t - \dfrac{1}{2}gt^2 = 0\) OR Half way: \(0 = U\sin\alpha - gt\)	A1	1.1b
Attempt to solve problem by eliminating \(t\)	DM1	3.1b
\(U^2\sin\alpha\cos\alpha = 588\) *	A1*	2.2a
	(6)

Notes

N.B. No credit given if they use the given answer from (b).

N.B. Could score 2/6 for any one of the 4 given equations if there is no corresponding second equation or there is an attempt but it’s incorrect.

M1: Complete method to give equation in \(U\), \(\alpha\) and \(t\) only, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved

A1: Correct equation

M1: Complete method to give equation in \(U\), \(\alpha\) and \(t\) only, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved

A1: Correct equation

DM1: Eliminate \(t\), dependent on first and second M1’s

A1*: Given answer correctly obtained, with no wrong working seen.
Allow \(588 = U^2\sin\alpha\cos\alpha\) but nothing else

Mark scheme (b)

Scheme	Marks	AO
Using vertical motion OR conservation of energy	M1	3.4
\(0^2 = (U\sin\alpha)^2 - 2g \times 10\) OR \(\dfrac{1}{2}mU^2 - \dfrac{1}{2}m(U\cos\alpha)^2 = mg \times 10\)	A1	1.1b
ALTERNATIVE 1: If \(t\) is time to top: use of \(10 = \dfrac{1}{2}gt^2\) oe \(\left(t = \dfrac{10}{7}\right)\) to obtain an equation in \(U\) and \(\alpha\) only M1 \(U\sin\alpha = 14\) or \(U\cos\alpha = 42\) A1 ALTERNATIVE 2: If \(t\) is time to top: use of : \(10 = U\sin\alpha\, t - \dfrac{1}{2}gt^2\) with \(t = \dfrac{60}{U\cos\alpha}\) substituted to obtain an equation in \(U\) and \(\alpha\) only : M1 \(10 = U\sin\alpha \times \dfrac{60}{U\cos\alpha} - \dfrac{1}{2}g\left(\dfrac{60}{U\cos\alpha}\right)^2\) A1
Attempt to solve problem by eliminating \(\alpha\) : e.g. \(U\sin\alpha = 14 \Rightarrow U\cos\alpha = 42\), from part (a) or from using \(t = \dfrac{10}{7}\), then square and add to give result OR: \(U^2\sin^2\alpha = 20g = 196\) and \(U^2\sin\alpha\cos\alpha = 588\), divide to give \(\tan\alpha = \dfrac{1}{3}\) then \(\sin^2\alpha = \dfrac{1}{10}\), hence result OR in ALTERNATIVE 2: sub for \(U^2\) using part (a), to give \(\tan\alpha = \dfrac{1}{3}\) then \(\sin^2\alpha = \dfrac{1}{10}\), hence result N.B. Just stating that \(\sin^2\alpha = \dfrac{1}{10}\), with no working is DM0A0.	DM1	3.1b
\(U^2 = 1960\) *	A1*	2.2a
N.B. Verification (i.e. starting with \(U^2 = 1960\) and trying to work backwards) is not an acceptable method for this question.
	(4)

Notes

M1: Complete method to give equation in \(U\) and \(\alpha\) only with correct no. of terms, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved

A1: Correct equation

DM1: Eliminate \(\alpha\) and rearrange, dependent on first M1

A1*: Given answer correctly obtained with no wrong working seen
(N.B. If they use a value for \(\alpha\) \((18.43.^\circ)\) they lose the final A1*)

Scheme	Marks	AO
\(V\), since air resistance has to be overcome, or just ‘because of air resistance’ isw	B1	3.5a
	(1)

Mark scheme (d)

Scheme	Marks	AO
e.g. wind effects, more accurate value of \(g\), spin of ball, size of ball, shape of ball, dimensions of ball, not a particle, variable acceleration, surface area of ball, humidity. Allow wind resistance and rotational resistance (Ignore any mention of air resistance or drag)	B1	3.5c
	(1)
	(12 marks)

Notes

B1: B0 if there is an incorrect extra e.g. mass or weight

October 2021 Paper 3 Mechanics Q4

EdexcelCurrent spec10 marksProjectiles

Figure 3: stone projected from O at the top of a vertical cliff 70 m above N, with speed 65 m/s at angle alpha above the horizontal, landing at A on the horizontal ground — Figure 3

A small stone is projected with speed \(65\ \text{m s}^{-1}\) from a point \(O\) at the top of a vertical cliff.

Point \(O\) is 70 m vertically above the point \(N\).

Point \(N\) is on horizontal ground.

The stone is projected at an angle \(\alpha\) above the horizontal, where \(\tan\alpha = \dfrac{5}{12}\)

The stone hits the ground at the point \(A\), as shown in Figure 3.

The stone is modelled as a particle moving freely under gravity.

The acceleration due to gravity is modelled as having magnitude \(\textbf{10 m s}^{\mathbf{-2}}\)

Using the model,

(a) find the time taken for the stone to travel from \(O\) to \(A\), (4)

(b) find the speed of the stone at the instant just before it hits the ground at \(A\). (5)

One limitation of the model is that it ignores air resistance.

Mark scheme (a)

Scheme	Marks	AO
Use \(s = ut + \dfrac{1}{2}at^2\) vertically or any complete method to give an equation in \(t\) only	M1	3.4
\(-70 = 65\sin\alpha \times t - \dfrac{1}{2} \times g \times t^2\)	A1 M(A)1	1.1b 1.1b
\(t = 7\) (s)	A1	1.1b
	(4)

Notes

Note that \(g = 10\); penalise once for whole question if \(g = 9.8\)

M1: Complete method, correct no. of terms, condone sign errors and sin/cos confusion

A1: Correct equation in \(t\) only with at most one error

M(A)1: Correct equation in \(t\) only

N.B. For ‘up and down’ methods etc, the two A marks are for all the equations that they use, lose a mark for each error.

A1: Cao (\(g = 9.8\), 7.1 or 7.11) (\(g = 9.81\), 7.1 or 7.12)

Mark scheme (b)

Scheme	Marks	AO
Horizontal velocity component at \(A = 65\cos\alpha\) (60)	B1	3.4
Complete method to find vertical velocity component at \(A\)	M1	3.4
\(65\sin\alpha - g \times 7\) OR \(\sqrt{(-25)^2 + 2g \times 70}\) (45)	A1ft	1.1b
Sub for trig and square, add and square root: \(\sqrt{60^2 + (-45)^2}\)	M1	3.1b
75 Accept 80 \((\text{m s}^{-1})\)	A1	1.1b
	(5)

Notes

Note that \(g = 10\); penalise once for whole question if \(g = 9.8\)

B1: Seen, including on a diagram.

M1: Condone sign errors and sin/cos confusion

A1ft: Correct expression; accept negative of this, follow their \(t\)

M1: Sub for trig and use Pythagoras

A1: Cao (\(g = 9.8\) or 9.81, 75 or 74.8)

Mark scheme (c)

Scheme	Marks	AO
e.g. an approximate value of \(g\) has been used, the dimensions of the stone could affect its motion, spin of the stone, \(g = 10\) instead of 9.8 has been used, \(g\) has been assumed to be constant, wind effect, shape of the stone	B1	3.5b
	(1)
	(10 marks)

Notes

B1: B0 if incorrect extras

October 2020 Paper 3 Mechanics Q5

EdexcelCurrent spec11 marksProjectiles

Figure 2: ball projected from O at the top of a vertical cliff 25 m above N, with speed U m/s at 45 degrees above the horizontal, landing at A on horizontal ground 100 m from N — Figure 2

A small ball is projected with speed \(U\ \text{m s}^{-1}\) from a point \(O\) at the top of a vertical cliff.

The point \(O\) is 25 m vertically above the point \(N\) which is on horizontal ground.

The ball is projected at an angle of \(45^\circ\) above the horizontal.

The ball hits the ground at a point \(A\), where \(AN = 100\) m, as shown in Figure 2.

The motion of the ball is modelled as that of a particle moving freely under gravity.

Using this initial model,

(a) show that \(U = 28\) (6)

(b) find the greatest height of the ball above the horizontal ground \(NA\). (3)

In a refinement to the model of the motion of the ball from \(O\) to \(A\), the effect of air resistance is included.

This refined model is used to find a new value of \(U\).

(d) State one further refinement to the model that would make the model more realistic. (1)

Mark scheme (a)

Scheme	Marks	AO
Using horizontal motion	M1	3.3
\(U\cos 45^\circ t = 100\)	A1	1.1b
Using vertical motion	M1	3.4
\(U\sin 45^\circ t - \dfrac{1}{2}gt^2 = -25\)	A1	1.1b
Solve problem by eliminating \(t\) and solving for \(U\)	M1	3.1b
\(U = 28\)*	A1*	1.1b
	(6)

Notes

M1: Complete method to give equation in \(U\) and \(t\) only, condone sin/cos confusion and sign errors

A1: Correct equation

M1: Complete method to give equation in \(U\) and \(t\) only, condone sin/cos confusion and sign errors

A1: Correct equation (\(g\) does not need to be substituted)

M1: Must have earned the previous two M marks.
Eliminate \(t\) and solve for \(U\).
N.B. They may solve for \(t\) first \(\left(100 - \dfrac{1}{2}gt^2 = -25\right)\) and then use it to find \(U\).

A1*: Exact given answer correctly obtained with no wrong working (e.g. \(g = 9.81\) used) or approximation seen.

Mark scheme (b)

Scheme	Marks	AO
Using vertical motion	M1	3.4
\(0^2 = (28\sin 45^\circ)^2 - 2gh\)	A1	1.1b
Greatest height = 45 m	A1	1.1b
	(3)

Notes

M1: Complete method to give equation in \(h\) only (allow if \(U\) not substituted), condone sin/cos confusion and sign errors

A1: Correct equation (\(g\) does not need to be substituted) (A0 if \(U\) is used instead of 28)

A1: cao

Scheme	Marks	AO
New value \(\gt 28\)	B1	3.5a
	(1)

Mark scheme (d)

Scheme	Marks	AO
e.g. wind effects, more accurate value of \(g\), spin of ball, include size of the ball, not model as a particle, shape of ball	B1	3.5c
	(1)
	(11 marks)

Notes

B1: Penalise incorrect extras i.e. B0 if there are incorrect extras.
The ground being horizontal, the cliff being vertical, .. are not part of the model so B0
Include weight/mass of the ball B0

June 2019 Paper 3 Mechanics Q5

EdexcelCurrent spec13 marksProjectiles

Figure 3: points A and B 50 m apart on horizontal ground; ball P projected from A at 20 m/s at 30 degrees to AB, ball Q projected from B at u m/s at angle theta to BA — Figure 3

The points \(A\) and \(B\) lie 50 m apart on horizontal ground.

At time \(t = 0\) two small balls, \(P\) and \(Q\), are projected in the vertical plane containing \(AB\).

Ball \(P\) is projected from \(A\) with speed \(20\ \text{m s}^{-1}\) at \(30^\circ\) to \(AB\).

Ball \(Q\) is projected from \(B\) with speed \(u\ \text{m s}^{-1}\) at angle \(\theta\) to \(BA\), as shown in Figure 3.

At time \(t = 2\) seconds, \(P\) and \(Q\) collide.

Until they collide, the balls are modelled as particles moving freely under gravity.

(a) Find the velocity of \(P\) at the instant before it collides with \(Q\). (6)

(b) Find

(i) the size of angle \(\theta\),

(ii) the value of \(u\). (6)

Mark scheme (a)

Scheme	Marks	AO
Horizontal speed \(= 20\cos 30^\circ\)	B1	3.4
Vertical velocity at \(t = 2\)	M1	3.4
\(= 20\sin 30^\circ - 2g\)	A1	1.1b
\(\theta = \tan^{-1}\left(\pm\dfrac{9.6}{10\sqrt{3}}\right)\)	M1	1.1b
Speed \(= \sqrt{100 \times 3 + 9.6^2}\) or e.g. speed \(= \dfrac{9.6}{\sin\theta}\)	M1	1.1b
19.8 or 20 \((\text{m s}^{-1})\) at \(29.0^\circ\) or \(29^\circ\) to the horizontal oe	A1	2.2a
	(6)

Notes

In this question mark parts (a) and (b) together.

B1: Seen or implied, possibly on a diagram

M1: Use of \(v = u + at\) or any other complete method using \(t = 2\)
Condone sign errors and sin/cos confusion.

A1: Correct unsimplified equation in \(v\) or \(v^2\)

M1: Correct use of trig to find a relevant angle for the direction.
Must have found a horizontal and a vertical velocity component

M1: Use Pythagoras or trig to find the magnitude
Must have found a horizontal and a vertical velocity component

A1: Or equivalent. Need magnitude and direction stated or implied in a diagram.
(0.506 or 0.51 rads)

Mark scheme (b)

Scheme	Marks	AO
Using sum of horizontal distances \(= 50\) at \(t = 2\)	M1	3.3
\((u\cos\theta) \times 2 + (20\cos 30^\circ) \times 2 = 50\) \((u\cos\theta = 25 - 20\cos 30^\circ)\)	A1	1.1b
Vertical distances equal	M1	3.4
\(\Rightarrow (20\sin 30^\circ) \times 2 - \dfrac{g}{2} \times 4 = (u\sin\theta) \times 2 - \dfrac{g}{2} \times 4\) \((20\sin 30^\circ = u\sin\theta)\)	A1	1.1b
Solving for both \(\theta\) and \(u\)	M1	3.1b
\(\theta = 52^\circ\) or better \((52.47756849....^\circ)\) \(u = 13\) or better (12.6085128…)	A1	2.2a
	(6)

Notes

In this question mark parts (a) and (b) together.

M1: First equation, in terms of \(u\) and \(\theta\) (could be implied by subsequent working), using the horizontal motion with \(t = 2\) used
Condone sign errors and sin/cos confusion

A1: Correct unsimplified equation – any equivalent form

M1: Second equation, in terms of \(u\) and \(\theta\) (could be implied by subsequent working), using the vertical motion – equating distances or just vertical components of velocities.
Condone sign errors and sin/cos confusion

A1: Correct unsimplified equation – any equivalent form

M1: Complete strategy: all necessary equations formed and solve for \(u\) and \(\theta\)
N.B. This is an independent method mark but can only be earned if 50 m has been used in their solution.

A1: Both values correct. (Here we accept 2SF or better, since the \(g\)’s cancel)
Allow radians for \(\theta\): 0.92 or better (0.915906..) rads.

Mark scheme (c)

Scheme	Marks	AO
It does not take account of the fact that they are not particles (moving freely under gravity) It does not take account of the size(s) of the balls It does not take account of the spin of the balls It does not take account of the wind \(g\) is not exactly \(9.8\ \text{m s}^{-2}\) N.B. If they refer to the mass or weight of the balls give B0	B1	3.5b
	(1)
	(13 marks)

Notes

B1: Any factor related to the model as stated in the question.
Penalise incorrect extras but ignore consequences
e.g. ‘\(AB\) (or the ground) is not horizontal’ should be penalised
or ‘they do not move in a vertical plane’ should be penalised

June 2025 Paper 2 Q14

AQACurrent spec4 marksProjectiles

14 In this question use \(g = 9.8\ \text{m s}^{-2}\)

An arrow is projected from a point \(P\) which is at a height of 2.5 metres above the horizontal ground.

The arrow has an initial velocity of \(\begin{bmatrix}40\\25\end{bmatrix}\ \text{m s}^{-1}\)

The arrow lands on the horizontal ground at a point \(Q\)

The path of the arrow is shown in the diagram.

The arrow is projected from P, 2.5 m above horizontal ground, with velocity (40, 25) m/s, follows a parabolic path and lands at Q on the ground

Find the speed of the arrow at point \(Q\) [4 marks]

Mark scheme

Scheme	Marks	AO
Substitutes \(u = 25,\ a = \pm 9.8\) and \(s = \pm 2.5\) into \(v^2 = u^2 + 2as\) PI \(\begin{bmatrix}k\\674\end{bmatrix}\) or \(\begin{bmatrix}\sqrt{k}\\\sqrt{674}\end{bmatrix}\) Or Substitutes \(u = 25,\ a = -9.8\) and \(s = -2.5\) into \(s = ut + \frac{1}{2}at^2\) to find \(t\) and substitutes their \(t\) into \(v = u + at\)	M1	3.3
Obtains correct \(v^2 = 674\) or \(v = \sqrt{674}\) If inconsistent signs are used for \(u\), \(a\) and \(s\) score A0 PI AWRT \(\pm\) 25.96 If vector form is used, both components must be correct	A1	1.1b
Finds the magnitude of the resultant vector of their vertical component of velocity and 40 Do not allow 25 for their vertical component of velocity	M1	1.1a
Obtains the correct speed of the arrow. Accept AWRT 48 If inconsistent signs are used for \(u\), \(a\) and \(s\) score A0	A1	1.1b
	(4 marks)

Typical solution

\[v^2 = 25^2 + 2 - 9.8 \times -2.5\]\[v^2 = 674\]\[\text{Speed} = \sqrt{40^2 + 674}\]\[\text{Speed} = 47.7\ \text{m s}^{-1}\]\[\text{Speed} = 48\ \text{m s}^{-1}\]

June 2024 Paper 2 Q19

AQACurrent spec8 marksProjectiles SUVAT

19 In this question use \(g = 9.8\) m s⁻²

A toy shoots balls upwards with an initial velocity of 7 m s⁻¹

The advertisement for this toy claims the balls can reach a maximum height of 2.5 metres from the ground.

(a) Suppose that the toy shoots the balls vertically upwards.

(i) Verify the claim in the advertisement. [2 marks]

(ii) State two modelling assumptions you have made in verifying this claim. [2 marks]

(b) In fact the toy shoots the balls anywhere between 0 and 11 degrees from the vertical.

The range of maximum heights, \(h\) metres, above the ground which can be reached by the balls may be expressed as

\[k \lt h \leqslant 2.5\]

Find the value of \(k\) [4 marks]

Mark scheme (a)

Scheme	Marks	AO
(i) Substitutes three of the four given values into \(v^2 = u^2 + 2as\) Condone inconsistent signs for the substituted values	M1	3.1b
Completes reasoned argument to obtain the correct fourth value and concludes the claim is correct Must have clearly stated three out of the following four \(v = 0\) \(u = 7\) \(a = -9.8\) or \(-g\) \(s = 2.5\) Allow a consistent swapping of signs AG	R1	2.1
	(2)
(ii) States one valid assumption from: No air resistance The ball is a particle The ball is shot from the ground	E1	3.5b
States a second valid assumption from No air resistance The ball is a particle The ball is shot from the ground	E1	3.5b
	(2)

Typical solution

(i)

\[v = 0,\ u = 7,\ a = -9.8\]

Using \(v^2 = u^2 + 2as\) then:

\[0 = 49 - 2(9.8)h\]\[h = \frac{49}{2(9.8)} = 2.5\text{ m}\]

The claim is correct

(ii)

The ball experiences no air resistance
Ball projected from ground level

Mark scheme (b)

Scheme	Marks	AO
States or uses 7cos11 for the vertical component of velocity OE	B1	2.2a
Uses \(v^2 = u^2 + 2as\) with \(u\) = their vertical component of velocity, \(v\) = 0 and \(a = -9.8\)	M1	3.1b
Obtains \(0 = 49\cos^2 11 + 2(-9.8)k\) Accept any variable for \(k\)	A1	1.1b
Obtains AWRT 2.4	A1	1.1b
	(4)
	(8 marks)

Typical solution

\[s = k\]\[u = 7\cos 11\]\[v = 0\]\[a = -9.8\]\[v^2 = u^2 + 2as\]\[0 = 49\cos^2 11 + 2(-9.8)k\]\[k = 2.4\]

June 2023 Paper 2 Q20

AQACurrent spec7 marksProjectiles SUVAT

20 In this question use \(g = 9.8\) m s⁻²

Nell and her pet dog Maia are visiting the beach.

The beach surface can be assumed to be level and horizontal.

Nell and Maia are initially standing next to each other.

Nell throws a ball forward, from a height of 1.8 metres above the surface of the beach, at an angle of 60° above the horizontal with a speed of 14 m s⁻¹

Exactly 0.2 seconds after the ball is thrown, Maia sets off from Nell and runs across the surface of the beach, in a straight line with a constant acceleration \(a\) m s⁻²

Maia catches the ball when it is 0.3 metres above ground level as shown in the diagram below.

Nell throws a ball from 1.8 m above the beach at 14 m s^−1 at 60° above the horizontal; the ball follows a curved path and is caught by the dog Maia 0.3 m above the ground

Find \(a\) [7 marks]

Mark scheme

Scheme	Marks	AO
States or uses \(14\cos 60\) for the horizontal component.	B1	3.1b
States or uses \(14\sin 60\) for the vertical component.	B1	3.1b
Uses \(s = ut + \dfrac{1}{2}at^2\) with \(u\) = their vertical component of velocity, \(a = -g\) and \(s = \pm 1.5\) OE PI by \(t\) = AWFW [2.54, 2.60]	M1	3.3
Obtains \(t\) = 2.592 AWFW [2.54, 2.60] Exact value is \(\dfrac{3\sqrt{10} + 5\sqrt{3}}{7}\)	A1	1.1b
Multiplies their \(t\) value by their horizontal component provided their \(t\) > 0.2	M1	1.1b
Substitutes \(u\) = 0, their \(t\) – 0.2 into \(ut + \dfrac{1}{2}at^2\) to obtain an expression for the horizontal distance travelled by the dog.	M1	3.3
Obtains \(6.3\) CAO	A1	3.2a
	(7 marks)

Typical solution

\[u_H = 14\cos 60 = 7\]\[u_V = 14\sin 60 = 7\sqrt{3}\]\[s = 7\sqrt{3}\,t - \frac{g}{2}t^2\]\[-1.5 = 7\sqrt{3}\,t - 4.9t^2\]\[4.9t^2 - 12.124t - 1.5 = 0\]

\(t = 2.592\) seconds

\[7t = 18.144\]\[ut + \frac{1}{2}at^2 = 0.5a(2.392)^2\]\[2.860832a = 18.144\]\[a = 6.3\]

June 2022 Paper 2 Q13

AQACurrent spec6 marksProjectiles

13 In this question use \(g\) = 9.8 m s⁻²

A ball is projected from a point on horizontal ground with an initial velocity of 7 m s⁻¹ at an angle \(\theta\) above the horizontal.

The ball reaches a maximum vertical height of \(h\) metres above the ground.

(a) Show that\[h = 2.5\sin^2\theta\] [3 marks]

(b) Hence, given that \(0^\circ \leqslant \theta \leqslant 60^\circ\), find the maximum value of \(h\). [2 marks]

State whether Nisha is correct, giving a reason for your answer. [1 mark]

Mark scheme (a)

Scheme	Marks	AO
Obtains \(7\sin\theta\) for vertical component of initial velocity	B1	1.1b
Uses \(v^2 = u^2 + 2as\) with \(v = 0\) Or uses appropriate constant acceleration equations that form a complete method to obtain \(h\) eg finds \(t\) then subs into \(s = ut + \dfrac{1}{2}at^2\)	M1	3.3
Completes argument substituting \(s = h\), \(v = 0\), \(u = 7\sin\theta\) and \(a = -9.8\) to show the given result. Accept negative values used consistently AG	R1	2.1
	(3)

Typical solution

\[u = 7\sin\theta\]\[v^2 = u^2 + 2as\]\[0 = 49\sin^2\theta - 19.6h\]

\[h = 2.5\sin^2\theta\]

Mark scheme (b)

Scheme	Marks	AO
Substitutes any value of \(\theta\) in the range \(0 \lt \theta \leqslant 60\) to obtain a height greater than zero	M1	1.1a
Deduces \(h\) = 1.9 AWRT 1.9 or \(\dfrac{15}{8}\)	A1	2.2a
	(2)

Typical solution

\[\theta = 60^\circ\]\[h = 2.5\sin^2 60\]\[h = 1.9\]

Mark scheme (c)

Scheme	Marks	AO
States that Nisha is incorrect and refers to air resistance or ball modelled as a particle	E1	3.5a
	(1)
	(6 marks)

Typical solution

Nisha is incorrect

The model ignores air resistance which gets greater as ball gets larger

June 2025 Paper 3 Q11

OCR ACurrent spec14 marksProjectiles

11 In this question you should take the acceleration due to gravity to be 10 m s⁻².

The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in a vertical plane with \(\mathbf{i}\) horizontal and \(\mathbf{j}\) vertically upwards.

A small ball \(P\) is projected from a point \(O\) on horizontal ground into the air. When \(P\) is in the air, it is to be modelled as a particle moving under the influence of gravity only.

At time \(t = 2\) seconds, \(P\) has velocity \((2\mathbf{i} - 8\mathbf{j})\) m s⁻¹.
The position vector of a point on the trajectory of \(P\) is \((x\mathbf{i} + y\mathbf{j})\) m relative to \(O\).

(a) Show that \(y = 6x - 1.25x^2\). [5]

(b) Determine the maximum height of \(P\) above the ground during its motion. [3]

the speed of \(P\)
the direction of motion of \(P\) [5]

In reality \(P\) does not move only under the influence of gravity but is also subject to air resistance.

(d) Explain how this would affect your answer to part (b). [1]

Mark scheme (a)

Scheme	Marks	AO
\(x = 2t\)	B1	1.1
\(-8 = u + (-10)(2)\) or \(\begin{pmatrix}2\\-8\end{pmatrix} = \begin{pmatrix}u_1\\u_2\end{pmatrix} + 2\begin{pmatrix}0\\-10\end{pmatrix}\)	M1*	3.4
\(y = \text{‘}12\text{’}t + \frac{1}{2}(-10)t^2\)	M1dep*	3.3
\(y = \text{‘}12\text{’}\left(\frac{x}{2}\right) + \frac{1}{2}(-10)\left(\frac{x}{2}\right)^2\)	M1	1.1
\(y = 6x - 5\left(\frac{x}{2}\right)^2 \Rightarrow y = 6x - 1.25x^2\)	A1	2.1
	[5]

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

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6 - 2.5x\)	B1	1.1
\(6 - 2.5x = 0 \Rightarrow x = 2.4\) and substitute in \(y = 6x - 1.25x^2\) to find \(y\)	M1	1.1
Maximum height is 7.2 (m)	A1	1.1
	[3]

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)

Scheme	Marks	AO
Apply \(v^2 = u^2 + 2as\) vertically or \(\mathbf{s} = 1.2\begin{pmatrix}2\\12\end{pmatrix} + \frac{1}{2}\begin{pmatrix}0\\-10\end{pmatrix} \times 1.2^2\)	M1
\(0 = 12^2 + 2 \times (-10) \times h\) or \(h = 12 \times 1.2 + \frac{1}{2} \times (-10) \times 1.2^2\)	A1

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)

Scheme	Marks	AO
\(-1.25x^2 + 6x = -1.25\left[(x - 2.4)^2 - 5.76\right]\)	M1
\(= -1.25(x - 2.4)^2 + 7.2\)	A1

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

Mark scheme (c)

Scheme	Marks	AO
When \(x = 2.5\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -0.25\) so \(\tan\theta = -0.25\)	M1	3.1b
\(14(.0^\circ)\) below the horizontal	A1	3.2a
Vertical component of velocity at \(x = 2.5\) is given by \(\text{‘}12\text{’} + (-10)(1.25)\)	M1*	1.1
Speed is given by \(\sqrt{2^2 + (\text{‘}12\text{’} + (-10)(1.25))^2}\)	M1dep*	1.1
\(2.06\) (m s⁻¹)	A1	1.1
	[5]

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)

Scheme	Marks	AO
Vertical component of velocity at \(x = 2.5\) is given by \(\text{‘}12\text{’} + (-10)(1.25)\)	M1*
Speed is given by \(\sqrt{2^2 + (\text{‘}12\text{’} + (-10)(1.25))^2}\)	M1dep*
\(2.06\) (m s⁻¹)	A1
\(\tan\theta = (\pm)\dfrac{0.5}{2}\)	M1dep*
\(14(.0^\circ)\) below the horizontal	A1

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

Mark scheme (d)

Scheme	Marks	AO
The maximum height will be less than the value found in part (b)	B1	3.5a
	[1]

Notes

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

Ignore all other comments

June 2024 Paper 3 Q13

OCR ACurrent spec12 marksProjectiles Resolving Forces

A particle at A at the bottom of a slope inclined at angle theta to horizontal ground, projected up the slope at 6 m s to the minus 1; the slope AB has length 1.375 m; a dashed curved path leaves B and lands at C on the ground

The points \(A\) and \(B\) are the lower and upper ends, respectively, of a line of greatest slope on a plane inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = 0.6\) and \(AB = 1.375\) m (see diagram).

A particle \(P\) is projected up the plane with speed \(6\,\mathrm{m\,s^{-1}}\) from \(A\) towards \(B\).

The plane at \(A\) is fixed to the ground which is horizontal.

The surface of the plane is rough and the coefficient of friction between \(P\) and the plane is 0.25.

(a) Show that the speed of \(P\) at \(B\) is \(3.8\,\mathrm{m\,s^{-1}}\). [6]

The particle leaves the slope at \(B\) and moves freely under gravity.

The particle first lands at a point \(C\) on the horizontal ground. The time taken for \(P\) to travel from \(A\) to \(C\) is \(T\) seconds.

(b) Determine the value of \(T\). [6]

Mark scheme (a)

Scheme	Marks	AO
\(R(\text{perp. plane}): R = mg\cos 36.86\ldots\) or \(mg \times 0.8\)	B1	3.3
\(F = 0.25 \times mg\cos 36.86\ldots\) or \(0.25 \times mg \times 0.8\)	M1*	3.4
N2L (parallel to plane): \(-0.25 \times mg \times 0.8 - mg \times 0.6 = ma\) or \(-0.25 \times mg \times \cos 36.86\ldots - mg \times \sin 36.86\ldots = ma\)	M1dep*	3.1b
\(a = -7.84\) or \(7.84\)	A1	1.1
\(v^2 = 6^2 + 2 \times -7.84 \times 1.375\)	M1dep*	3.4
\(v^2 = 14.44 \Rightarrow v = 3.8\ (\mathrm{m\,s^{-1}})\)	A1	2.2a
	[6]

Notes

B1: Allow using an angle of 37 or better
soi (possibly in N2L)

M1*: \(F = 0.25R\) with their \(R\) (with \(R \ne mg\)) – using an angle of 37 or better – allow sin/cos mix
\(R\) must be a component of weight soi (possibly in N2L)

M1dep*: Applying N2L parallel to the plane with correct number of terms and weight term resolved, dimensionally consistent – if down the plane is the positive direction expect \(0.25 \times mg \times 0.8 + mg \times 0.6 = ma\)
Allow sign errors and sin/cos mix, using an angle of 37 or better

A1: \(\pm 0.8g\) - allow awrt \(\pm 7.84\) or awrt \(\pm 7.85\) (using an angle of 37) www
Condone \(\pm 7.8\) or \(\pm 7.9\) www

M1dep*: Use of \(v^2 = u^2 + 2as\) with correct \(u\) and \(s\) and their negative \(a\) (\(\ne -9.8\)) - dependent on both previous M marks – must see the values substituted into the formula (as value for \(v\) is AG) – therefore just seeing \(v^2 = \frac{361}{25}\) or 14.44 is M0
Condone finding a positive value of \(a\) and then using the negative version

A1: AG – do not award this mark if using any non-exact values for \(\sin\theta / \cos\theta\) (so seeing e.g. \(\cos 36.86\ldots\) in their working loses this final A mark)
Must see either \(v^2 = 14.44\) or \(v^2 = 36 - 21.56\) or \(v^2 = 6^2 - 21.56\) or a correct expression for \(v\) before given answer

SC – if \(m\) absent throughout (so implying \(m = 1\)) or \(m\) given a numerical value then at most B0 M1 M1 A1 M1 A0 can be awarded.

Alternative for final two marks

Scheme	Marks
\(1.375 = 6t + \frac{1}{2} \times -7.84 \times t^2\) \(t = \frac{55}{196}\) (or \(t = 1.25\)) \(v = 6 + (-7.84) \times \frac{55}{196}\)	M1dep*
\(v = 3.8\ (\mathrm{m\,s^{-1}})\)	A1

M1dep*: Use of \(s = ut + \frac{1}{2}at^2\) with correct \(u\) and \(s\) and their negative \(a\) (\(\ne -9.8\)), and selects the smaller of the two positive \(t\) values and uses this value correctly in \(v = u + at\) with correct \(u\) and their negative \(a\) - dependent on both previous M marks
So, using the second positive \(t\) (if correct this is 1.25) is M0

A1: AG – do not award this mark if using any non-exact values for \(\sin\theta / \cos\theta\)
Must see a correct expression for \(v\) before given answer

Alternative energy approach to 13(a) (Appendix)

Scheme	Marks
\(R(\text{perp. plane}): R = mg\cos 36.86\ldots\) or \(mg \times 0.8\)	B1
\(F = 0.25 \times mg\cos 36.86\ldots\) or \(0.25 \times mg \times 0.8\)	M1
Work done by gravity/PE term \(= \pm mg \times (1.375 \times 0.6)\) or \(\pm mg \times (1.375 \times \sin 36.86\ldots)\) OR Work done by friction \(\pm 1.375 \times (0.25 \times mg \times 0.8)\) or \(\pm 1.375 \times (0.25 \times mg \times \cos 36.86\ldots)\)	B1
\(\frac{1}{2}mv^2 - \frac{1}{2}m \times 6^2 = -mg(1.375 \times 0.6) - 1.375 \times (0.25 \times mg \times 0.8)\)	M1
Correct W-E equation	A1
\(v^2 = 14.44 \Rightarrow v = 3.8\ (\mathrm{m\,s^{-1}})\)	A1

B1: Allow using an angle of 37 or better
soi (possibly in Work-Energy equation)

M1: Use of \(F = 0.25R\) with their \(R\) (with \(R \ne mg\)) – using an angle of 37 or better – allow sin/cos mix
\(R\) must be a component of weight not mass
soi (possibly in W-E equation)

B1: Correct expression for either WD by gravity or friction – using an angle of 37 or better
soi (possibly in W-E equation)

M1: Attempt at W-E principle – equation with the correct number of relevant terms, dimensionally correct, allow sign errors and sin/cos mix,
M0 if PE term is \(\pm 1.375mg\)

A1: Correct W-E equation

A1: AG – do not award this mark if using any non-exact values for \(\sin\theta / \cos\theta\)
Must see either \(v^2 = 14.44\) or \(v^2 = 36 - 21.56\) or \(v^2 = 6^2 - 21.56\) or a correct expression for \(v\) before given answer

SC – if \(m\) absent throughout (so implying \(m = 1\)) or \(m\) given a numerical value then at most B0 M1 B1 M1 A1 A0 can be awarded.

Mark scheme (b)

Scheme	Marks	AO
Time from \(A\) to \(B\) is \(t_1\) where \(3.8 = 6 + (-7.84)t_1\)	M1	3.4
\([t_1 =]\ 0.281\) (s)	A1	1.1
[Height of \(P\) at \(B\) is] 0.825 (m)	B1	3.1b
Equation for time \(t_2\) (from \(B\) to \(C\))	M1	3.1b
\(-0.825 = (3.8 \times 0.6)t_2 + 0.5(-g)t_2^{\,2}\) or \(-0.825 = (3.8 \times \sin 36.86\ldots)t_2 + 0.5(-g)t_2^{\,2}\) \(\left(4.9t_2^{\,2} - 2.28t_2 - 0.825 = 0\right)\)	A1	1.1
\(T\,[= 0.280\ldots + 0.704\ldots] = 0.985\)	A1	2.2a
	[6]

Notes

M1: Use of \(v = u + at\) with correct \(u\) and \(v\), and their negative \(a\) from (a) (or any other complete method e.g. \(1.375 = \left(\dfrac{6 + 3.8}{2}\right)t_1\))
Give BOD if \(6 = 3.8 + 7.84t_1\) seen but if \(v = u + at\) stated before then M0

A1: awrt 0.281 or accept 0.28 www. If exact then expect \(\frac{55}{196}\)
0.2806122...

B1: Must be exact e.g. \(\frac{33}{40}\) – allow un-simplified e.g. \(1.375 \times 0.6\)
Allow negative

M1: M1 for an equation for time \(t_2\) (from \(B\) to \(C\)) with awrt \(\pm 0.83\) for \(s\), \(a = \pm g\) and either \(u = 3.8 \times \sin\theta\) or \(u = 3.8 \times \cos\theta\) with \(\theta = 37\) (or better) substituted (or corresponding exact value) – condone sign errors

A1: A1 for a correct (un-simplified) equation for \(t_2\) - must be using 0.825 but allow \(\theta = 37\) or better
For reference: \(t_2 = 0.704346599\ldots\)

A1: awrt 0.985 only – dependent on all previous marks in this part
0.984958…

June 2023 Paper 3 Q12

OCR ACurrent spec13 marksProjectiles Variable Acceleration (Calculus)

12 In this question you should take the acceleration due to gravity to be \(10\,\mathrm{m\,s^{-2}}\).

A ball is projected from point A, 20 m vertically above point B on horizontal ground, with speed 39 m s^-1 at angle theta above the horizontal; its path curves up and then down to land at point C on the ground

A small ball \(P\) is projected from a point \(A\) with speed \(39\,\mathrm{m\,s^{-1}}\) at an angle of elevation \(\theta\), where \(\sin\theta = \frac{5}{13}\) and \(\cos\theta = \frac{12}{13}\). Point \(A\) is \(20\,\mathrm{m}\) vertically above a point \(B\) on horizontal ground. The ball first lands at a point \(C\) on the horizontal ground (see diagram).

The ball \(P\) is modelled as a particle moving freely under gravity.

(a) Find the maximum height of \(P\) above the ground during its motion. [3]

The time taken for \(P\) to travel from \(A\) to \(C\) is \(T\) seconds.

(b) Determine the value of \(T\). [3]

(c) State one limitation of the model, other than air resistance or the wind, that could affect the answer to part (b). [1]

At the instant that \(P\) is projected, a second small ball \(Q\) is released from rest at \(B\) and moves towards \(C\) along the horizontal ground.

At time \(t\) seconds, where \(t \geqslant 0\), the velocity \(v\,\mathrm{m\,s^{-1}}\) of \(Q\) is given by

\(v = kt^3 + 6t^2 + \frac{3}{2}t,\)

where \(k\) is a positive constant.

(d) Given that \(P\) and \(Q\) collide at \(C\), determine the acceleration of \(Q\) immediately before this collision. [6]

Mark scheme (a)

Scheme	Marks	AO
\(0 = (39\sin\theta)^2 + 2(-10)h\)	M1	3.3
\(0 = \left(39 \times \frac{5}{13}\right)^2 + 2(-10)h\)	A1	1.1
Max. height \(= 20 + h = 31.25\) (m)	A1	2.2a
	[3]

Notes

M1: Using \(v^2 = u^2 + 2as\) with \(v = 0\), \(a = \pm 10\) or \(\pm 9.8\) or \(\pm g\), and \(u = 39\sin\theta\) or \(u = 39\cos\theta\).
Accept any other complete method (using correct suvat equations) to find the maximum height e.g. \(0 = 39\sin\theta - 10t\) and with \(t\) then substituted into \(s = (39\sin\theta)t + 0.5(-10)t^2\)
Condone \(g = 9.8\) for full marks in (a) and (b)

A1: A correct equation with the correct value of sin substituted or for \(0 = (39\sin(22.6\ldots))^2 + 2(-10)h\)
Allow using \(g\) (and not replaced with 10 or 9.8) for this A mark

A1: www

accept awrt 31.2 (using \(22.6\ldots\) as the angle and \(g = 10\))
accept awrt 31.5 (coming from exact value of sin or \(22.6\ldots\) and \(g = 9.8\))
condone 31.3 (3 sf)

Mark scheme (b)

Scheme	Marks	AO
\(-20 = (39\sin\theta)T + \frac{1}{2}(-10)T^2\)	M1	3.3
\(-20 = \left(39 \times \frac{5}{13}\right)T + \frac{1}{2}(-10)T^2\)	A1	1.1
\(T = 4\) only	A1	1.1
	[3]

Notes

M1: Applying \(s = ut + \frac{1}{2}at^2\) with \(s = \pm 20\), \(a = \pm 10\) or \(\pm 9.8\) or \(\pm g\) and \(u = 39\sin\theta\) or \(u = 39\cos\theta\)
Condone \(g = 9.8\) for full marks in (a) and (b)
Accept any other complete method to find \(T\)

A1: With correct value of sine or \(\sin(22.6)\)
Allow \(-20 = \left(39 \times \frac{5}{13}\right)T + \frac{1}{2}(-9.8)T^2\)
Allow using \(g\) (and not replaced with 10 or 9.8) for this A mark

A1: BC

accept awrt 4.07 (using \(g = 9.8\) and exact value of sine)
accept awrt 4.06 (using \(g = 9.8\) and \(22.6\ldots\))
a value of \(4(.00\ldots)\) coming from \(3.998\ldots\) (using \(g = 10\) and \(22.6\ldots\) for sine)

Condone \(t = 4\)

Mark scheme (c)

Scheme	Marks	AO
Examples of possible limitations The ball will have dimensions/volume The spin/rotational forces of the ball A 1sf approx. to \(g\) was used Other weather conditions (ignore wind and air resistance) The ball is not a particle \(g\) is modelled as a (universal) constant	B1	3.5b
	[1]

Notes

B1: Allow any correct limitation
B0 if referring to

Air resistance and/or wind (only)
The ground is unlikely to be horizontal (only)
The mass or weight or shape of \(P\) (only)
The angle/heights/speeds may not be as quoted (only)
Modelling the problem as 3D rather than in 2D (only)

If multiple limitations given, and any are incorrect, then B0

Mark scheme (d)

Scheme	Marks	AO
\(a = 3kt^2 + 12t + \frac{3}{2}\)	M1	3.4
\(BC = \left(39 \times \frac{12}{13}\right)\text{‘}T\text{’}\ (= 144)\)	M1*	3.1b
\(s = \frac{1}{4}kt^4 + 2t^3 + \frac{3}{4}t^2\ (+c)\)	M1*	2.1
\(\frac{1}{4}k(4)^4 + 2(4)^3 + \frac{3}{4}(4)^2 = 144\)	M1dep*	3.4
\(k = \frac{1}{16}\)	A1	1.1
\(a = 52.5\ (\mathrm{m\,s^{-2}})\)	A1	2.2a
	[6]

Notes

M1: Differentiate given \(v\) (at least two terms correct)

M1*: Applying \(s = ut\) horizontally to find distance \(BC\) with correct value of cos (or \(\cos(22.6\ldots)\)) and their value of \(T\) from (b)
Allow \(g = 9.8\) which if correct leads to \(146.34\ldots\)

M1*: Integrate given \(v\) (at least two terms correct)

M1dep*: Puts their integrated expression for \(s\), with \(t =\) their \(T\) from (b), equal to their distance for \(BC\) to form an equation in \(k\) only – dependent on the two previous M marks only
Must not include a \(+c\) unless dealt with as part of a definite integral. However, if a \(+c\) is included then subsequently ignored/set equal to zero without justification then give bod for this and any subsequent A marks (if earned)

A1: Correct exact value for \(k\) (oe e.g. 0.0625)
Final two marks can only be awarded if \(g = 10\) used

A1: oe

June 2022 Paper 3 Q13

OCR ACurrent spec14 marksProjectiles

13 A small ball \(B\) moves in the plane of a fixed horizontal axis \(Ox\), which lies on horizontal ground, and a fixed vertically upwards axis \(Oy\). \(B\) is projected from \(O\) with a velocity whose components along \(Ox\) and \(Oy\) are \(U\,\mathrm{m\,s^{-1}}\) and \(V\,\mathrm{m\,s^{-1}}\), respectively. The units of \(x\) and \(y\) are metres.

\(B\) is modelled as a particle moving freely under gravity.

(a) Show that the path of \(B\) has equation \(2U^2y = 2UVx - gx^2\). [3]

During its motion, \(B\) just clears a vertical wall of height \(\frac{1}{2}a\) m at a horizontal distance \(a\) m from \(O\). \(B\) strikes the ground at a horizontal distance \(3a\) m beyond the wall.

(b) Determine the angle of projection of \(B\). Give your answer in degrees correct to 3 significant figures. [5]

(d) Hence find the maximum height of \(B\) above the ground during its motion. [3]

(e) State one refinement of the model, other than including air resistance, that would make it more realistic. [1]

Mark scheme (a)

Scheme	Marks	AO
\(x = Ut\) \(y = Vt - \dfrac{1}{2}gt^2\)	M1*	3.3
\(y = V\left(\dfrac{x}{U}\right) - \dfrac{1}{2}g\left(\dfrac{x}{U}\right)^2\)	M1dep*	3.4
\(y = \dfrac{Vx}{U} - \dfrac{gx^2}{2U^2} \Rightarrow 2U^2y = 2UVx - gx^2\)	A1	2.2a
	[3]

Notes

M1*: Setting up expressions for \(x\) and \(y\) using \(s = ut + \frac{1}{2}at^2\) with \(a = 0\) in \(x\) and \(\pm g\) in \(y\) oe. M0 if using \(U\) instead of \(V\) vertically
Allow sign errors. May use \(t = x/U\) in \(y = Vt + \frac{1}{2}at^2\)

M1dep*: Eliminating both \(t\) terms in \(y\) to get an equation in \(y\), \(x\), \(U\), \(V\) (and possibly \(g\))

A1: AG so sufficient working must be shown
www

Mark scheme (b)

Scheme	Marks	AO
\(B\) passes through \(\left(a, \frac{1}{2}a\right) \Rightarrow 2U^2\left(\dfrac{a}{2}\right) = 2UVa - ga^2\) \((\Rightarrow U^2 = 2UV - ga)\)	B1	3.4
\(B\) passes through \((4a, 0) \Rightarrow 2UV(4a) - g(4a)^2 = 0\) \((\Rightarrow UV - 2ga = 0)\)	B1	3.1b
\(U = \sqrt{3ga}\), \(V = \dfrac{2\sqrt{ga}}{\sqrt{3}}\) or \(2U = 3V\)	M1*	2.1
\(\tan\theta = \dfrac{V}{U} \Rightarrow \tan\theta = \dfrac{\frac{2}{\sqrt{3}}\left(\sqrt{ga}\right)}{\sqrt{3ga}}\)	M1dep*	3.1b
\(\tan\theta = \dfrac{2}{3} \Rightarrow \theta = 33.7^\circ\) (3 sf)	A1	2.2a
	[5]

Notes

B1: Substituting \(\left(a, \frac{1}{2}a\right)\) into given result from (a)

B1: Substituting \((4a, 0)\) into given result from (a) – note that using \((3a, 0)\) is not a MR

M1*: Solve simultaneously (oe) to find either \(U\) or \(V\) (or their squares) in terms of \(a\) and \(g\) only, or for a linear equation (oe) in \(V\) and \(U\) only, if correct \(2U = 3V\)
oe e.g., if correct \(V^2 = \frac{4}{3}ga\) and \(U^2 = 3ga\)

M1dep*: Using \(\tan\theta = \dfrac{V}{U}\) with their \(U\) and their \(V\)

A1: awrt 33.7 (an answer of 36.9 from using \((3a, 0)\) scores (if from correct working) B1 B0 M1 M1 A0)

Mark scheme (c)

Scheme	Marks	AO
\(\sqrt{3ga + \dfrac{4}{3}ga} = 54.6\)	M1	3.4
\(a = 70.2\)	A1	1.1
	[2]

Notes

M1: Using \(\sqrt{U^2 + V^2} = 54.6\) to set up an equation in \(a\) (and possibly \(g\))

A1: www awrt 70.2
an answer of 97.344 (awrt 97.3) from using \((3a, 0)\) in (b) scores M1 A0

Alternative 1

Scheme	Marks
\(54.6\cos(33.7\ldots) = \sqrt{3ga}\) or \(54.6\sin(33.7\ldots) = \sqrt{\tfrac{4}{3}ga}\)	M1
\(a = 70.2\)	A1

M1: Setting either the horizontal component equal to their \(U\) (from (b)) or the vertical component equal to their \(V\) (from (b)). Allow sin/cos confusion
M0 for an unsupported value of \(\theta\) (if used)

A1: www awrt 70.2

Alternative 2

Scheme	Marks
\(a = \dfrac{UV}{2g} = \dfrac{54.6\cos(33.7\ldots) \times 54.6\sin(33.7\ldots)}{2g}\)	M1
\(a = 70.2\)	A1

M1: Using their expression for \(a\) (possibly seen in (b)) in terms of \(U\) and \(V\) with 54.6 and their \(\theta\). M0 for an unsupported value of \(\theta\)
Allow sin/cos confusion

A1: www awrt 70.2

Mark scheme (d)

Scheme	Marks	AO
\(0 = V^2 - 2gH\) \(\left(\Rightarrow H = \dfrac{V^2}{2g}\right)\)	M1*	3.3
\(H = \dfrac{1}{2g}\left(\dfrac{4ga}{3}\right) = \dfrac{2}{3}a\)	M1dep*	3.4
\(H = \dfrac{2}{3}(70.2) = 46.8\) (m)	A1	2.2a
	[3]

Notes

M1*: Setting up the model using \(v^2 = u^2 + 2as\) with \(v = 0\) and \(a = -g\)
\(H\) is the maximum height of \(B\)

M1dep*: Using their expression for \(V\) from (b) to get an expression for \(H\) in terms of \(a\) (oe)
e.g., \(H = \dfrac{V^2}{2g}\) where \(V = 54.6\sin\theta\) (allow \(\cos\theta\)) with their value of \(\theta\)
M0 for an unsupported value of \(\theta\)

A1: awrt 46.8
an answer of 54.756 (awrt 54.8) from using \((3a, 0)\) in (b) scores M1M1A0

Alternative 1

Scheme	Marks
\(0 = V - gt \Rightarrow t = \dfrac{54.6\sin(33.7\ldots)}{g}\)	M1*
\(H = (54.6\sin(33.7\ldots))t - \frac{1}{2}gt^2\) \(\phantom{H} = (54.6\sin(33.7\ldots))(3.09\ldots) - \frac{1}{2}g(3.09\ldots)^2\)	M1dep*
\(H = 46.8\) (m)	A1

M1*: Find \(t\) at maximum height with \(v = 0\), \(a = -g\) and \(u =\) their \(V \ne 54.6\) (allow sin/cos confusion). M0 for an unsupported value of \(\theta\)
\(t = 3.090472522\ldots\)

M1dep*: Substituting their \(t\) into \(s = Vt - \frac{1}{2}gt^2\)
\(V \ne 54.6\) (allow sin/cos confusion)

A1: awrt 46.8

Alternative 2

Scheme	Marks
\(x = 2a \Rightarrow 2U^2y = 4UVa - 4ga^2\)	M1*
\(2(54.6\cos(33.7))^2H =\) \(4(54.6\cos(33.7))(54.6\sin(33.7))(70.2) - 4g(70.2)^2\)	M1dep*
\(H = 46.8\) (m)	A1

M1*: Setting \(x = 2a\) or \(1.5a\) and substituting into path equation from (a)

M1dep*: Substituting their \(U\), \(V\) and \(a\) to form an equation in \(H\) (and possibly \(g\)) only. M0 for an unsupported value of \(\theta\)
\(U\) and \(V \ne 54.6\) (allow sin/cos confusion)

A1: awrt 46.8

Mark scheme (e)

Scheme	Marks	AO
examples of possible refinements include taking into account the size of \(B\) taking into account that \(B\) is not a particle taking into account the spin of \(B\) taking into account the dimensions of \(B\) taking into account the wind/weather taking into account the thickness of the wall (which is assumed to be ‘thin’) taking into account the clearance of the ball at the top of the wall	B1	3.5c
	[1]

Notes

B1: Allow any correct refinement, including use of a more accurate value of \(g\) rather than the assumed 9.8

B0 if referring to

the mass or weight or shape of \(B\)
the ground is unlikely to be horizontal
modelling the problem as three dimensional rather than two dimensional (unless specific detail given)
air resistance (only)

If multiple refinements given, then all must be valid to score B1

October 2021 Paper 3 Q11

OCR ACurrent spec10 marksProjectiles

A ball is projected from point A on level ground with speed 25 m/s at 15 degrees above the horizontal; the ground slopes down beyond A to a lower level, and the ball's path lands at B, which is 4 m below the level of A

A golfer hits a ball from a point \(A\) with a speed of \(25\,\mathrm{m\,s^{-1}}\) at an angle of \(15^\circ\) above the horizontal. While the ball is in the air, it is modelled as a particle moving under the influence of gravity. Take the acceleration due to gravity to be \(10\,\mathrm{m\,s^{-2}}\).

The ball first lands at a point \(B\) which is \(4\,\mathrm{m}\) below the level of \(A\) (see diagram).

(a) Determine the time taken for the ball to travel from \(A\) to \(B\). [3]

(b) Determine the horizontal distance of \(B\) from \(A\). [2]

The horizontal distance from \(A\) to \(B\) is found to be greater than the answer to part (b).

(d) State one factor that could account for this difference. [1]

Mark scheme (a)

Scheme	Marks	AO
\(-4 = (25\sin 15)t - \frac{1}{2}(10)t^2\)	M1 A1	3.3 1.1
\(t = 1.75\ (\mathrm{s})\)	A1	2.2a
	[3]

Notes

M1: Use of \(s = ut + \frac{1}{2}at^2\) with \(a = \pm g\) and \(s = \pm 4\)
Allow sin/cos confusion

A1: BC (1.750981765…) 1.75 only
For reference: 1.779296952… (if \(g = 9.8\) used)
Penalise \(g = 9.8\) only once in the question

Alternative method

Scheme	Marks
\(0 = (25\sin 15)^2 + 2(-10)s_1\) and \(0 = 25\sin 15 + (-10)t_1\)	M1*
\(4 + s_1 = \frac{1}{2}\ 10\ t_2^{\,2}\) and \(t = t_1 + t_2\)	M1dep*
\(t = 0.6470476\ldots + 1.1039341\ldots = 1.75\,(\mathrm{s})\)	A1

Notes

M1*: Finding the maximum height \(s_1\,(= 2.093353\ldots)\) above \(A\) and corresponding time \(t_1\,(= 0.647047\ldots)\)
Using \(v = 0\) and \(a = \pm 10\)

M1dep*: Complete correct method to find \(t\)
Using \(u = 0\) and where \(t_2\,(= 1.1039341\ldots)\) is the time from the maximum height to the ground

Mark scheme (b)

Scheme	Marks	AO
\((25\cos 15)t\)	M1	3.4
42.3 (m)	A1FT	1.1
	[2]

Notes

M1: Use of \(s = ut\) with their \(t\) from (a)
Allow sin/cos confusion

A1FT: 42.2829627… - ft their positive value of \(t\) from (a) but must be using \((25\cos 15)t\)
For reference: 42.96672196… (if \(g = 9.8\) used)

Mark scheme (c)

Scheme	Marks	AO
\(v_h = 25\cos 15\)	B1	1.2
\(v_v = 25\sin 15 - 10(1.5)\)	B1	3.3
\(\tan\theta = \dfrac{v_v}{v_h}\)	M1	3.1b
\(19.5^\circ\) below the horizontal	A1	3.2a
	[4]

Notes

B1: Correct expression for horizontal velocity component (soi)
24.14814…

B1: Correct expression for vertical velocity component at \(t = 1.5\) (condone positive value)
\(-8.529523\ldots\)

M1: Use of tan to find angle (allow reciprocal) – dependent on one B mark earned
M0 if using expressions for displacements

A1: oe (e.g., \(70.5^\circ\) to the downward vertical)
For reference: \(18.8^\circ\) (if \(g = 9.8\) used)

Mark scheme (d)

Scheme	Marks	AO
e.g., a less accurate value of \(g\) was used e.g., no consideration of the wind e.g., no consideration of (back)spin on the ball (but not topspin)	B1	3.5a
	[1]

Notes

B1: Any valid reason (do not accept mention of resistance e.g., air/wind resistance)

June 2025 Paper 1 Q12

OCR MEICurrent spec8 marksProjectiles SUVAT

12 In this question \(x\) and \(y\) are the horizontal and upwards vertical directions respectively.

An astronaut is standing on the surface of the moon exploring the motion of a ball.

(a) The astronaut drops a ball from rest from 1 m above the surface. It takes 1.1 s to hit the surface.

Calculate the value of the acceleration due to the moon’s gravity. Give your answer correct to 3 significant figures. [2]

The astronaut stands in a crater of the moon and hits the ball with a golf club from the moon’s surface. The initial velocity of the ball is \(25\,\mathrm{m\,s^{-1}}\) at an angle of \(40^\circ\) above the horizontal in the \(x\)-direction.

(b) Taking the origin to be the point of projection, determine the equation of the trajectory of the ball. Give your answer in the form \(y = \mathrm{f}(x)\), with each of the coefficients correct to 3 significant figures. [4]

(c) The edge of the crater is 40 m away from the point of projection and 15 m above it.

Determine whether the ball goes over the crater’s edge. [2]

Mark scheme (b)

Scheme	Marks	AO
\(x = 25\cos 40^\circ\, t\)	B1	1.1
\(y = 25\sin 40^\circ\, t + \frac{1}{2}at^2\)	M1	3.3
\(y = 25\sin 40^\circ\left(\dfrac{x}{25\cos 40^\circ}\right) + \dfrac{1}{2}a\left(\dfrac{x}{25\cos 40^\circ}\right)^2\)	M1	2.1
\(y = 0.839x - 0.00225x^2\)	A1	1.1
	[4]

Notes

B1: Oe eg \(t = \dfrac{x}{25\cos 40^\circ}\), \(t = 0.0522x\) or \(\dfrac{x}{19.15}\) or \(\dfrac{x}{19.2}\)

M1: Forms equation of motion in the vertical direction
Soi. Allow sign errors.
Do not allow if \(a = \pm 9.8\) used here or subsequently

M1: Substitutes expression for \(t\) in their \(y\) equation

A1: Cao. Coefficients must be 3sf.

Mark scheme (c)

Scheme	Marks	AO
When \(x = 40,\ y = 30.0\ [\gt 15]\)	M1	3.4
So the ball goes over the crater’s edge	A1	3.2a
	[2]

Notes

M1: Use their model and \(x = 40\)
(29.72 if exact values used)

A1: Conclusion based on correct working.
No FT from wrong (b)
Possible interpretation of the question gives \(x = \sqrt{40^2 - 15^2} = 5\sqrt{55} = 37.1\) giving \(y = 28.0\)

Alternative method

Scheme	Marks
When \(y = 15,\ x = 18.8,\ 354\)	M1
So the ball is above 15m as 40 m is between these values	A1

M1: Use the model with \(y = 15\)

A1: Conclusion based on correct working
No FT for wrong (b)
Possible interpretation \(x = \sqrt{1375} = 37.1\) which is between these values.

When \(x = 40,\ t = 2.09\) when \(y = 15,\ t = 0.983,\ 18.5\)

June 2024 Paper 1 Q14

OCR MEICurrent spec7 marksProjectiles

14 A man runs at a constant speed of \(4\ \text{m s}^{-1}\) along a straight horizontal road. A woman is standing on a bridge that spans the road. At the instant that the man passes directly below the woman she throws a ball with initial speed \(u\ \text{m s}^{-1}\) at \(\alpha^\circ\) above the horizontal. The path of the ball is directly above the road. The man catches the ball 2.4 s after it is thrown. At the instant the man catches it, the ball is 3.6 m below the level of the point of projection.

(a) Explain what it means that the ball is modelled as a particle. [1]

(b) Find the vertical component of the ball’s initial velocity. [2]

The value of \(u\)
The value of \(\alpha\)

[4]

Mark scheme (a)

Scheme	Marks	AO
The size and shape of the ball are neglected	B1	2.4
	[1]

Notes

B1: Sensible comment equivalent to “the object is modelled as a point mass”. Do not allow for a statement that includes correct and incorrect ideas. (see appendix)

Appendix: exemplar responses for Q14(a)

Response	Mark
The mass is concentrated at the centre	B0
The object has no mass	B0
There is no air resistance	B0
The weight acts at the centre [of mass]	B0
There is no spin	B1
The ball’s size and shape do not matter	B1
Only its mass is taken into account so it doesn’t spin	B1
The size of the ball is negligible and there are no external forces acting on the ball	B0
Point mass means there is no air resistance	B0
There is no air resistance because the object has no size	B0

Mark scheme (b)

Scheme	Marks	AO
Vertical motion \(s = -3.6, a = -9.8, t = 2.4\) \(s = u_y t + \tfrac{1}{2}at^2\) \(-3.6 = 2.4u_y - 4.9 \times 2.4^2\)	M1	3.1b
Giving \(u_y = 10.26\ \text{m s}^{-1}\)	A1	1.1
	[2]

Notes

M1: Use of suvat equation(s) with \(s = \pm 3.6\) and \(v \neq 0\) leading to a value for \(u_y\) Allow sign errors

A1: cao

Allow SC1 if M0 awarded and \(u\sin\alpha\) seen.
Or award SC2 for \(u\sin\alpha\) in (b) if 10.26 seen in (c)

Mark scheme (c)

Scheme	Marks	AO
The horizontal velocity of the ball must be the same as the velocity of the man	B1	3.1b
\(u = \sqrt{4^2 + 10.26^2} = 11.01\)	B1	1.1
\(\alpha = \tan^{-1}\dfrac{10.26}{4} = 68.7^\circ\)	M1 A1	1.1 1.1
	[4]

Notes

B1: Soi eg \(u\cos\alpha = 4\)

B1: FT their (b) and their horizontal velocity

M1: Also allow from solving \(u\cos\alpha = 4\) or \(u\sin\alpha = 10.26\)

A1: FT their (b) and their horizontal velocity

Alternative solution for last 3 marks

Scheme	Marks	AO
\(-3.6 = 2.4\dfrac{4}{\cos\alpha}\sin\alpha - 4.9 \times 2.4^2\)	M1
\(\alpha = 68.7^\circ\)	A1
So \(u = \left[\dfrac{4}{\cos 68.7}\right] = 11.01\)	B1

M1: Substitute for \(u\) in terms of \(\cos\alpha\) leading to a value for \(\tan\alpha\)

A1: FT their horizontal velocity

B1: FT their value for \(\alpha\)

June 2023 Paper 1 Q15

OCR MEICurrent spec8 marksProjectiles

15 A projectile is launched from a point on level ground with an initial velocity \(u\) at an angle \(\theta\) above the horizontal.

(a) Show that the range of the projectile is given by \(\dfrac{2u^2\sin\theta\cos\theta}{g}\). [3]

(b) Determine the set of values of \(\theta\) for which the maximum height of the projectile is greater than the range, where \(\theta\) is an acute angle. Give your answer in degrees. [5]

Mark scheme (a)

Scheme	Marks	AO
vertical motion with \(u\sin\theta,\ a = -g,\ s = 0\) \(0 = u\sin\theta\, t - \frac{1}{2}gt^2\) \(t = 0,\ \dfrac{2u\sin\theta}{g}\)	M1	2.1
horizontal distance \(R\) given by \(t = \dfrac{2u\sin\theta}{g},\ R = u\cos\theta \times \dfrac{2u\sin\theta}{g}\)	M1	2.1
\(R = \dfrac{2u^2\sin\theta\cos\theta}{g}\)	A1	2.1
	[3]

Notes

M1: suvat equations used with \(u\sin\theta\) leading to an expression for the time of flight. Do not allow for time to the top unless subsequently doubled.

M1: horizontal motion with \(u\cos\theta,\ a = 0\) and their expression for \(t\) soi
Allow sin/cos interchange if consistent with their vertical equation

A1: Convincing argument AG
Note M0M1A0 for sin/cos interchange

Alternative method

Scheme	Marks
Substitute \(t = \dfrac{x}{u\cos\theta}\) to form equation of the trajectory \(y = u\sin\theta \times \dfrac{x}{u\cos\theta} - \dfrac{1}{2}g\left(\dfrac{x}{u\cos\theta}\right)^2\)	M1
Equate \(y\) to zero and attempt to rearrange	M1
\(R = \dfrac{2u^2\sin\theta\cos\theta}{g}\)	A1

M1: Allow equivalent formula quoted

M1: Allow sin/cos interchange if consistent with their horizontal equation

A1: Convincing argument AG

Mark scheme (b)

Scheme	Marks	AO
Max height \(H\) when \(v_y = 0\) \(0 = (u\sin\theta)^2 - 2gH\)	M1	3.1b
\(H = \dfrac{u^2\sin^2\theta}{2g}\)	A1	1.1b
Max height exceeds range when \(\dfrac{u^2\sin^2\theta}{2g} > \dfrac{2u^2\sin\theta\cos\theta}{g}\)	M1	1.1a
\(\tan\theta > 4\)	M1	1.1a
\(76.0^\circ < \theta\ [< 90^\circ]\)	A1	1.1b
	[5]

Notes

M1: suvat equation(s) with \(v_y = 0\) leading to an equation for \(H\) not involving \(t\). Allow sin/cos interchange if consistent with their (a)

A1: correct expression for \(H\)

M1: Compares their \(H\) with given \(R\)
Allow = used to find boundary value

M1: simplifies the inequality to an inequality for \(\tan\theta\) (or equation)

A1: must be an inequality for \(\theta\). Do not penalise for omission of \(90^\circ\)

June 2022 Paper 1 Q7

OCR MEICurrent spec6 marksProjectiles Vectors

7 In this question the \(x\)- and \(y\)-directions are horizontal and vertically upwards respectively and the origin is on horizontal ground.

A ball is thrown from a point 5 m above the origin with an initial velocity \(\begin{pmatrix}14\\7\end{pmatrix}\mathrm{m\,s^{-1}}\).

(a) Find the position vector of the ball at time \(t\) s after it is thrown. [3]

(b) Find the distance between the origin and the point at which the ball lands on the ground. [3]

Mark scheme (a)

Scheme	Marks	AO
\(x = 14t\)	B1	1.1b
\(y = 7t - \frac{1}{2}gt^2 + 5\)	M1	1.1b
So the position vector is \(\begin{pmatrix}14t\\7t - 4.9t^2 + 5\end{pmatrix}\)	A1	2.5
	[3]

Notes

B1: must be \(x = \ldots\) or seen as the first component of the vector.
Do not award for an expression that adds a vector to a scalar

M1: allow without \(+5\), or if \(-5\) seen
Do not award for an expression that adds a vector to a scalar

A1: Must be a single vector.
Accept \(\frac{1}{2}g\) in final answer
Accept \(14t\,\mathbf{i} + \left(7t - 4.9t^2 + 5\right)\mathbf{j}\)

SC1: for \(\begin{pmatrix}7t - 4.9t^2 + 5\\14t\end{pmatrix}\) or \(\begin{pmatrix}14t - 4.9t^2 + 5\\7t\end{pmatrix}\)

Mark scheme (b)

Scheme	Marks	AO
Lands when \(y = 0\) \(7t - 4.9t^2 + 5 = 0\)	M1	3.1b
\(t = 1.95\)	A1	1.1b
gives \(x = 14t = 27.3\) m	B1	1.1b
	[3]

Notes

M1: Award for correct quadratic or an attempt to find value of \(t\) when their quadratic \(y = 0\)

A1: cao

B1: FT their \(t\) and their linear expression for \(x\)
ISW where candidates find the distance from the point of projection

October 2021 Paper 1 Q10

OCR MEICurrent spec11 marksProjectiles SUVAT

10 A ball is thrown upwards with a velocity of \(29.4\,\text{m}\,\text{s}^{-1}\).

(a) Show that the ball reaches its maximum height after 3 s. [1]

(b) Sketch a velocity-time graph for the first 5 s of motion. [2]

Axes printed in the Printed Answer Booklet for part (b):

Blank axes: v (m s⁻¹) vertically, extending above and below the t-axis, and t (s) horizontally marked 0 to 5

A second ball is thrown at \(u\,\text{m}\,\text{s}^{-1}\) at an angle of \(\alpha^\circ\) above the horizontal. It reaches the same maximum height as the first ball.

(d) Use this information to write down

the vertical component of the second ball’s initial velocity,
the time taken for the second ball to reach its greatest height.

[2]

This second ball reaches its greatest height at a point which is 48 m horizontally from the point of projection.

(e) Calculate the values of \(u\) and \(\alpha\). [3]

Mark scheme (a)

Scheme	Marks	AO
Time when \(v = 0\) given by \(0 = 29.4 - 9.8t\), so \(t = 3\) s	E1	2.1
	[1]

Notes

E1: Using suvat equation(s) leading to correct value for \(t\) with \(v = 0\)
Allow for verifying that \(t = 3\) gives \(v = 0\) if identified as the maximum point oe

Velocity-time graph: a straight line from (0, 29.4) crossing the t-axis at t = 3 and continuing to about −19.6 at t = 5 — Mark scheme (b)

Mark scheme (c)

Scheme	Marks	AO
When \(t = 5\), \(v = 29.4 - 9.8 \times 5\)	M1	1.1a
\(v = -19.6\)	A1	1.1b
Speed is \(19.6\,\text{m}\,\text{s}^{-1}\)	A1	1.1b
	[3]

Notes

M1: Using suvat equation(s) leading to a value for \(v\) with \(t = 5\). Allow sign errors

A1: May be implied by 19.6 seen

A1: FT their negative velocity

If motion from the highest point considered \(u = 0\), \(t = 2\), \(g = +9.8\) then \(v = 19.6\) is fully correct.
Allow M1A1A0 if \(29.4 - 9.8 \times 5 = 19.6\) seen

Mark scheme (e)

Scheme	Marks	AO
\(u_x \times 3 = 48\)	M1	1.1a
\(u = \sqrt{u_x^2 + u_y^2} = \sqrt{16^2 + 29.4^2} = 33.5\)	M1	1.1b
\(\tan\alpha = \dfrac{u_y}{u_x} = \dfrac{29.4}{16}\) giving \(\alpha = 61.4^\circ\)	A1	1.1b
	[3]

Notes

M1: Using (their) \(t = 3\) to find \(u_x\)

M1: Combining their components to find either one of \(u\) and \(\alpha\)

A1: Both values correct

October 2020 Paper 1 Q13

OCR MEICurrent spec11 marksProjectiles

13 A projectile is fired from ground level at \(35\,\mathrm{m\,s^{-1}}\) at an angle of \(\theta^\circ\) above the horizontal.

(a) State a modelling assumption that is used in the standard projectile model. [1]

(b) Find the cartesian equation of the trajectory of the projectile. [4]

The projectile travels above horizontal ground towards a wall that is 110 m away from the point of projection and 5 m high. The projectile reaches a maximum height of 22.5 m.

Mark scheme (b)

Scheme	Marks	AO
\(u_x = 35\cos\theta\) giving \(x = (35\cos\theta)t\)	B1	3.3
\(u_y = 35\sin\theta\) giving \(y = (35\sin\theta)t - \dfrac{1}{2}gt^2\)	B1	3.3
Substitute for \(t\)	M1	3.3
\(y = (35\sin\theta)\left(\dfrac{x}{35\cos\theta}\right) - \dfrac{1}{2}g\left(\dfrac{x}{35\cos\theta}\right)^2\) \(\left[y = x\tan\theta - \dfrac{x^2}{250\cos^2\theta}\right]\)	A1	1.1
	[4]

Notes

B1: Award seen in any form

B1: soi

M1: Substituting for \(t\) in their equation for \(y\)

A1: Award in any form ISW

Mark scheme (c)

Scheme	Marks	AO
EITHER Using \(s = 22.5,\quad u = 35\sin\theta,\quad v = 0,\quad a = -9.8\) \(v^2 = u^2 + 2as\)	M1	3.1b
\(0 = (35\sin\theta)^2 - 2 \times 9.8 \times 22.5\)	A1	1.1a
\(\sin\theta = 0.6\) [giving \(\theta = 36.9^\circ\)]	A1	1.1
Use the trajectory with \(x = 110\) \(y = 110 \times \tan\theta^\circ - \dfrac{1}{250\cos^2\theta^\circ} \times 110^2\)	M1	3.1b
\(= 6.875\)	A1	1.1
So it goes over the wall	E1	3.2a
	[6]

Notes

M1: Using any suvat in \(y\)-direction with \(v_y = 0\)

A1: Correct equation for \(\theta\) only

A1: allow \(37^\circ\)
Either \(\theta = 37^\circ\) or \(\cos\theta = \dfrac{4}{5},\ \tan\theta = \dfrac{3}{4}\) may be used.

M1: Allow in terms of \(\theta\)
FT their value for \(\theta\) if used
Allow this M mark for \(t = \dfrac{110}{35\cos\theta} = \dfrac{55}{14}\) used in a suitable equation for \(y\)

A1: Correct \(y\) value

E1: Conclusion in context from correct values

Alternative (OR)

Scheme	Marks
Using \(s = 22.5,\quad u = 35\sin\theta,\quad v = 0,\quad a = -9.8\) \(v^2 = u^2 + 2as\)	M1
\(0 = (35\sin\theta)^2 - 2 \times 9.8 \times 22.5\)	A1
\(\sin\theta = 0.6\) giving \(\theta = 36.9^\circ\)	A1
Use the trajectory with \(y = 5\) \(5 = x\tan\theta^\circ - \dfrac{1}{2450}g\sec^2\theta^\circ x^2\)	M1
\(\dfrac{1}{160}x^2 - \dfrac{3}{4}x + 5 = 0\) \(x = 7.089,\quad 112.9\)	A1
So particle is above the height of the wall between 7 m and 112.9 m away, so when \(x = 110\) m so it does not hit the wall	E1

M1: Using any suvat in \(y\)-direction with \(v_y = 0\)

A1: Correct equation for \(\theta\) only

A1: allow \(37^\circ\)
Either \(\theta = 37^\circ\) or \(\cos\theta = \dfrac{4}{5},\ \tan\theta = \dfrac{3}{4}\) may be seen.

M1: Allow in terms of \(\theta\)
FT their value for \(\theta\) if used
Allow this M mark for roots of their \(y = 5\) used in a suitable equation for \(x\).

A1: Both correct \(x\)-values

E1: Conclusion in context from correct values