June 2022 Paper 3 Mechanics Q5
5.
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,
The ball reaches a maximum height of 10 m above the ground.
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 \(V\)
| 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
| 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) |
Notes
B1: Clear statement isw
| 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