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

Using the equations:

1. From particle Q: $T = 2g - 2a$

2. Substitute in particle P's equation: $2g - (2g - 2a) = 3a ext{sin}(30^ ext{o})$

   Since $ext{sin}(30^ ext{o}) = 0.5$, simplifying gives: $5a = 2g \ herefore a = \frac{2g}{5}$

   Substituting g = 9.81 m/s² gives: $a = \frac{2 \times 9.81}{5} = 0.98 ext{ m/s}^2$

Substituting the value of acceleration a into equation for Q gives:

$T = 2g - 2(0.98)$

Calculating it: $T = 2 \times 9.81 - 1.96 \\ = 19.62 - 1.96 = 17.66 ext{ N}$

The inextensibility of the string is crucial in deriving the equations of motion. It ensures that the acceleration of Q is equal to the acceleration of P. Therefore, we can set their accelerations equal as the same string connects them together.

Using the equation of motion:

$v^2 = u^2 + 2as$

With initial speed $u = 0$, distance $s = 0.8 m$, and acceleration $a = 0.98 m/s²$:

$v^2 = 0 + 2 \times 0.98 \times 0.8 = 1.568 \\ v = \sqrt{1.568} \approx 1.25 ext{ m/s}$

Using the equation:

$s = ut + \frac{1}{2}at^2$

Substituting $u = 0$, $s = 0.8 m$, $a = 0.98 m/s²$:

$0.8 = \frac{1}{2} \times 0.98 t^2 \\ t^2 = \frac{0.8 imes 2}{0.98} \approx 1.63265306 \\ t \approx 0.51 s$