4. (a) Two particles A and B each of mass m are connected by a light inextensible string passing over a light, smooth, fixed pulley - Leaving Cert Applied Maths - Question 4 - 2012

To calculate the distance traveled by A after B strikes the ground, we will again utilize the equations of motion:

1. We already determined the speed when B hits the ground is rac{g}{13}.

2. With the known acceleration from part (i), we can say after B strikes the ground, the equation becomes:

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

   Setting $v = 0$ for when A comes to a stop, and u = v_{B} = rac{g}{13}, we have:

   0 = rac{g^2}{169} + 2(- rac{g}{13}) s \\ s = rac{g^2 / 169}{2g / 13} \\ s = rac{g}{13} = 11 m

   Therefore, the distance A will travel is 11m.

The tension in the string can be found by analyzing the forces acting on both masses.

1. For mass m:

   $T - μ(mg) = m a$ \ ext{(1)}

2. For mass 2m:

   $$$$

3. Adding (1) and (2):

   $2mg - μ(2mg) = 3ma \ T = mg(1 + μ) \\ \text{(Tension)}$

To find the acceleration of the 2m kg mass, we can leverage Newton's laws of motion:

1. The acceleration of the m kg mass is given as $f₁$.

2. Since the tension from the first mass affects the second, we can express the acceleration of the 2m kg mass as:

   2m(g - a) = 2m f \text{(from tension)} \rightarrow f_2 = rac{f₁}{2} \text{(Using the ratio of the masses)} \rightarrow a_2 = rac{3f₁}{5}

   So, the acceleration of the 2m kg mass in terms of $f₁$ is rac{3f₁}{5}.