Figure 4 shows two particles P and Q, of mass 3 kg and 2 kg respectively, connected by a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 7 - 2007 - Paper 1

Substituting the equation for Q into that for P, we have: $2g - (3g \sin(30°) + 3a) = 0$ Simplifying, we can find values: Using $g \approx 9.81 m/s²$,
We have: $2(9.81) - T = 2a$ Upon solving these interconnected equations, we can derive that: $a \approx 0.98 m/s².$

Starting with the equation for Q: $T = 2g - 2a$ Substituting the known values: $T = 2(9.81) - 2(0.98)$ which simplifies to: $T \approx 17.6 \, N.$

The information that the string is inextensible is crucial in sub-part (e) where we assume that both particles P and Q experience the same magnitude of acceleration; this is because the length of the string does not change as one particle moves while the other reacts.

Using the equation of motion: $v^2 = u^2 + 2as$ where $u = 0$, $a = 0.98 m/s²$, and $s = 0.8 m$ (the height from which Q falls): $v^2 = 0 + 2(0.98)(0.8) = 1.568$ Thus, the speed of Q as it reaches the ground: $v \approx 1.25 \, m/s.$

Using the formula: $s = ut + \frac{1}{2}at^2$ Here, $s = 0.8 m$, $u = 0$, and $a = 0.98 m/s²$: $0.8 = 0 + \frac{1}{2}(0.98)t^2$ Solving for $t$, we find: $t = \sqrt{\frac{0.8}{0.49}} \approx 0.51 \, s.$