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):

$$T - 2mg an α - F = 2ma$$

For particle B (mass = 4m):

$$4mg - T = 4ma$$

From the equations for motion, we can express T in terms of a for both particles:

1. From A:

$T = 2ma + 2mg an α + F$

2. From B:

$4mg - T = 4ma$

Using the value of tanα and the given coefficient of friction, we can calculate:

Substituting and simplifying leads to:

$a = 0.4g$

Thus, for particle A, acceleration is ( a_A = 0.4g ) and for particle B, ( a_B = 0.4g ).

Using the equations of motion, when particle A comes to rest at Y, we can derive the distance:
The motion can be described by:

$v^2 = u^2 + 2as$ Now, setting:
v = 0
u = 0.8h
a = -2mgrac{ ext{sin } α}{2m}
s \text{(the distance XY)} = x \

We find:

-2mgrac{ ext{sin } α}{2m} = a

and solving this leads to:

$XY = 0.5h + h = 1.5h$