June 2022 Paper 3 Q13
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.
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.
| 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
| 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)
| 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
| 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
| Scheme | Marks | AO |
|---|---|---|
examples of possible refinements include
| 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