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)

Using Newton's second law:

$T - 2mg \sin \alpha - F = 2ma$

For particle B (mass 4m):

$4mg - T = 4ma$

From the equations of motion for both particles, we can substitute:

1. Rearranging the equations gives:

   • For A: $T = 2ma + 2mg \sin \alpha + F$
   • For B: $T = 4mg - 4ma$

2. Setting these equations equal gives: $4mg - 4ma = 2ma + 2mg \sin \alpha + F$

3. By eliminating T, we find: $4mg - 2ma - 2mg \sin \alpha - F = 0$

4. Substituting the known values:

   • Using ( \sin \alpha = \frac{3}{5} ) and ( F = \frac{1}{4} \times 2mg ) leads to:
   • Expanding terms yields: $a = 0.4g \approx 3.92 \, (\text{with} \, g \approx 9.81)$

For particle B:

Since B does not rebound, the motion is governed by its principle of conservation. Taking into account its fall:

1. The final velocity of A before reaching Y can be expressed as: $v^2 = 2gh$

2. The motion of A along the incline can be expressed as: $-2mg \sin \alpha = -2ma$

3. After applying kinematic equations: $d = 0.5h$

Thus the distance XY can be calculated as: $XY = 0.5h + h = 1.5h$