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

From the equations derived, we can set them equal to find the acceleration:

Substituting the known values:

$2g - T = 2a \ ext{and } T - 3g ext{sin}(30^ ext{o}) = 3a$

We find that the acceleration of Q is 0.98 m s⁻².

To find the tension, we can rewrite one of the equations:

Using the first equation:
$T = 2g - 2a$
Substituting the values gives us:

$T ext{ approximately } 18 N$

I used the fact that the string is inextensible to equate the magnitudes of the accelerations of both P and Q. This means both particles must experience the same magnitude of acceleration when the system is released.

Using the equation of motion:

$v^2 = u^2 + 2as$
With initial speed $u = 0$, the total distance $s = 0.8 ext{ m}$ and $a = 0.98 ext{ m s}^{-2}$, we find:

$v = ext{approximately } 1.568 ext{ m s}^{-1}$

Using the equation:

s = ut + rac{1}{2}at^2
Where $u = 0$, $s = 0.8$ m, and $a = 0.98$ m/s².
Solving for $t$ gives:
$t ext{ approximately } 0.5 ext{ s}$