The points A and B lie 50 m apart on horizontal ground - Edexcel - A-Level Maths Mechanics - Question 5 - 2019 - Paper 1

For ball Q projected from B, we use the horizontal distance traveled in 2 seconds, which is:

$50 = u \cos(\theta) \cdot 2$

This results in:

$u \cos(\theta) = 25$

For the vertical displacement, since both balls collide at the same height:

$0 = (u \sin(\theta) \cdot 2) - (\frac{1}{2} g t^2)$

Thus,

$u \sin(\theta) = \frac{9.8 \cdot 2}{2} = 9.8$

Now we have two equations:

1. (u \cos(\theta) = 25)
2. (u \sin(\theta) = 9.8)

Dividing the second equation by the first gives:

$\tan(\theta) = \frac{9.8}{25}$

Calculating (\theta):

$\theta = \tan^{-1}\left(\frac{9.8}{25}\right)$

We substitute (\sin(\theta)) and (\cos(\theta)) back into the equations to find u:

From (u \cos(\theta) = 25), we can express u as:

$u = \frac{25}{\cos(\theta)}$

Substituting the value of (\sin(\theta)) from previous calculations into the second equation gives:

Substituting the value of (\theta) and solving yields:

(u \approx 13 \text{ (using calculated values from before)})

One limitation of the model is that it does not consider the fact that the balls are not particles, thus ignoring factors such as spin or deformation during the collision. This can result in inaccuracies in calculating their trajectories.