Two scale pans A and B, each of mass $m$ kg, are attached to the ends of a light inextensible string which passes over a light smooth fixed pulley - Leaving Cert Applied Maths - Question 4 - 2017

To find the distance B rises, we use the kinematic equation:

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

Here,

• Initial velocity $u = 0$
• Final velocity $v = 0.4$ m s$^{-1}$
• Acceleration $a = \frac{3g}{5}$.

Substituting these values yields:

$(0.4)^2 = 0 + 2 \left(\frac{3g}{5}\right) s$

This simplifies to:

$s = \frac{0.16}{\frac{6g}{5}} = \frac{0.16 \times 5}{6g} = \frac{0.8}{6g} = 0.0136 m$.

To find the reaction force R on the mass 3m, we consider the forces acting on it:

1. The weight acting downwards is $3mg$.

2. Using Newton's second law, we get:

   $3mg - R = 3m(\frac{3g}{5})$

3. Rearranging gives us:

   $R = 6mg/5$.

For the wedge:

• The forces acting are:
  1. Normal force $R$ upward
  2. Weight $4mg$ downward.

For the particle:

• The forces acting on the particle on the wedge are:
  1. Gravitational force $mg$ downward.
  2. Normal force $R$ from the wedge inclined at angle eta.

Analyzing the movements:

1. For the particle:

   • The acceleration relative to the wedge is $kp ext{ cos } \beta$.

2. For the wedge:

   • The acceleration of the wedge is $kp ext{ sin } \beta$.

Combining the forces, we derive:

$mg \text{ cos } \beta = R = mkp \text{ sin } \beta$

From which we have: $k = \frac{1}{5}$.

Using Newton's second law and substituting the two equations:

1. For the wedge: $mg \text{ sin } \beta - R = mkp \text{ cos } \beta$
2. For the particle: $R + 5g \text{ sin } \beta = mp - p \text{ cos } \beta$

By solving these simultaneously, we simplify to find: $p = \frac{49 \text{ sin } \beta}{4 + \text{ sin } \beta}$.