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

To find the angle θ, we use the horizontal range of the motion:

1. The range equation states that the total horizontal distance covered by P is 50 m at t = 2 seconds: $\text{Distance} = u_{Px} \times t + \frac{1}{2} a_x t^2$ Since there is no horizontal acceleration, $50 = (20 \cos 30º) \times 2$ Substituting for $u_{Px}$, we get: $50 = 20 \times \frac{\sqrt{3}}{2} \times 2$
2. For ball Q, if Q travels horizontally at speed $u \cos(\theta)$: $50 = u \cos(\theta) \times 2$ Thus,
   $u \cos(\theta) = 25$
3. For the vertical motion of Q: \text{Vertical drop} = u_{Qy} \times 2 - \frac{1}{2} g (2)^2 \text{Using Pythagorean theorem: } \tan(30º) = \frac{4y}{50}$$ And substituting gives: $$25 (\sin 30º) = 50$$ Solving for θ gives: $$\theta = 52º\text{. Therefore, angle } θ \text{ is approximately } 52°.$$

Using the previously derived equations, we substitute:

1. We already found: $u \cos(\theta) = 25$ Now substituting with $\theta = 52°:$ $u \cos(52°) = 25$ $u \times 0.6157 = 25$ Hence, $u = \frac{25}{0.6157} \approx 40.5 ext{ m/s}$

One limitation of the model is that it assumes both balls are treated as point masses, neglecting their actual sizes and how this could impact their trajectories during the collision. Other limitations could include ignoring air resistance and friction from the ground.