Two particles, A and B, have masses 2m and m respectively - Edexcel - A-Level Maths Mechanics - Question 8 - 2017 - Paper 1

For particle B, the forces acting on it include the gravitational force downwards (mg) and the tension in the string (T) acting upwards. The equation of motion for B can thus be expressed as:

$mg - T = ma$.

Using the equations derived for A and B, we can substitute for T from the second equation into the first:

Substituting $T = mg - ma$ into the equation for A gives:

$mg - ma - μ imes 2mg = 2ma$.

Rearranging gives:

$mg - μ imes 2mg = 5ma$.

Factoring out the m:

$g(1 - 2μ) = 5a$.

Thus the acceleration a of A can be simplified to:

a = rac{g}{5}(1 - 2μ).

The speed of A at the instant B hits the floor can be derived using the kinematic equation based on its acceleration. The velocity of A, denoted as v, can be calculated using:

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

where u = 0 (initial velocity), a = rac{g}{5}(1 - 2μ), and s = d - h.

Substituting these values into the equation gives:

v^2 = 0 + 2 imes rac{g}{5}(1 - 2μ)(d - h).

Thus, v = rac{g}{5}(1 - 2μ) imes ext{distance}.

If μ = rac{1}{2}$, then the frictional force is reduced. This reduction allows particle A to accelerate more easily. Thus the tension in the string may not be sufficient to balance the gravitational pull on B, leading to a scenario where either particle A does not move or both particles remain in a state of limiting equilibrium. Essentially, the dynamics of the system would change significantly with A being able to slide more freely along the table.