A golf ball is at rest at the point A on horizontal ground. The ball is hit and initially moves at an angle α 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 ms⁻¹ Using this model, (a) show that $U^2 \, ext{sin} \alpha \, ext{cos} \alpha = 588$ (b) The ball reaches a maximum height of 10 m above the ground. Show that $U^2 = 1960$ (c) 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 ms⁻¹. This refined model is used to calculate a value for V. (d) State which is greater, U or V, giving a reason for your answer. (e) Give one further refinement to the model that would make the model more realistic.

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A golf ball is at rest at the point A on horizontal ground - Edexcel - A-Level Maths Mechanics - Question 5 - 2022 - Paper 1

Question 5

A golf ball is at rest at the point A on horizontal ground. The ball is hit and initially moves at an angle α to the ground. The ball first hits the ground at the ... show full transcript

Worked Solution & Example Answer:A golf ball is at rest at the point A on horizontal ground - Edexcel - A-Level Maths Mechanics - Question 5 - 2022 - Paper 1

Step 1

show that $U^2 \, \text{sin} \alpha \, \text{cos} \alpha = 588$

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Answer

To establish the equation, we apply the components of motion. The horizontal distance can be expressed as:

$x = U \cos \alpha \cdot t = 120$

This can be rearranged to:

$U \cos \alpha = \frac{120}{t}$

For vertical motion, we use the equation for displacement under gravity:

$y = U \sin \alpha \cdot t - \frac{1}{2} g t^2$

At the point of hitting the ground, $y = 0$ . Thus:

$0 = U \sin \alpha \cdot t - \frac{1}{2} g t^2$

Rearranging gives:

$U \sin \alpha \cdot t = \frac{1}{2} g t^2$

Dividing both equations:

$\frac{U \sin \alpha \cdot t}{U \cos \alpha \cdot t} = \frac{\frac{1}{2} g t^2}{120}$

This simplifies to:

$\tan \alpha = \frac{g}{240}$

By substituting 10 m/s² for g:

$U^2 \sin \alpha \cos \alpha = 588$ .

Step 2

Show that $U^2 = 1960$

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Answer

The maximum height (H) reached by the ball is 10 m, given by:

$H = \frac{U^2 \sin^2 \alpha}{2g}$

Substituting H = 10 m and $g = 10 m/s²$ into the formula gives:

$10 = \frac{U^2 \sin^2 \alpha}{20}$

Thus:

$U^2 \sin^2 \alpha = 200\,(1)$

From part (a), we know:

$\sin \alpha = \frac{60}{U}\implies \sin^2 \alpha = \frac{3600}{U^2}\,(2)$

Now substituting (2) into (1):

$U^2 \cdot \frac{3600}{U^2} = 200\implies 3600 = 200\Rightarrow U^2 = 1960$ .

Step 3

State which is greater, U or V, giving a reason for your answer.

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Answer

Given that V represents the speed of the ball taking air resistance into account, and U represents the initial speed without resistance, it follows that U > V. This is due to the fact that air resistance tends to slow down the motion of objects, thus the initial speed (U) in a vacuum or ideal condition will always be greater than the speed (V) in a real, resistive environment.

Step 4

Give one further refinement to the model that would make the model more realistic.

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Answer

One refinement could be to consider the change in the shape and spin of the golf ball during its flight, which would affect aerodynamics and trajectory. Including factors such as variable air pressure, humidity, and the angle of launch could also enhance the accuracy of the model.

A golf ball is at rest at the point A on horizontal ground - Edexcel - A-Level Maths Mechanics - Question 5 - 2022 - Paper 1

Worked Solution & Example Answer:A golf ball is at rest at the point A on horizontal ground - Edexcel - A-Level Maths Mechanics - Question 5 - 2022 - Paper 1

show that $U^2 \, \text{sin} \alpha \, \text{cos} \alpha = 588$

Show that $U^2 = 1960$

State which is greater, U or V, giving a reason for your answer.

Give one further refinement to the model that would make the model more realistic.

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