Figure 5 shows two particles A and B, of mass 2m and 4m respectively, connected by a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 7 - 2013 - Paper 1

For particle A (mass = 2m):

$T - 2mg imes ext{sin} heta - F_{friction} = 2ma$

For particle B (mass = 4m):

$4mg - T = 4ma$

To find the acceleration, we simplify the equations:

1. From particle A:

F_{friction} = rac{1}{4} (2mg) = 0.5mg

So,

$T - 2mg imes 0.6 - 0.5mg = 2ma$

2. From particle B:

$4mg - T = 4ma$

Eliminating T, we get:

$4mg - (2ma + 0.5mg + 1.2mg) = 4ma$

This can be solved to give:

$a = 0.4g = 3.92 ext{ m/s}^2.$

Considering that particle B does not rebound when it hits the ground and A continues moving up the plane towards P, we can use the kinematic equation:

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

where initially, the velocity u = 0:

1. For the movement of A:

$0 = 2gh - 2mg imes ext{sin} heta$

This leads to the conclusion that the distance XY will therefore be calculated as:

$XY = 0.5h + (h - 0.8h) = 1.5h.$