Since \(3/4^{\text{th}}\) of kinetic energy is lost, the remaining kinetic energy is \(1/4^{\text{th}}\), meaning the final velocity \(v = v_0/2\). Using \(v = v_0 - at_0\) where \(a = \mu g\), we get \(v_0/2 = v_0 - \mu gt_0 ⇒ \mu = \frac{v_0}{2gt_0}\).

Retardation \(a = g(\sin\theta + \mu\cos\theta) = 10(\sin 30^\circ + \frac{1}{\sqrt{3}}\cos 30^\circ) = 10(0.5 + 0.5) = 10\text{ m/s}^2\). Using \(v^2 = u^2 - 2as ⇒ 0 = 1^2 - 2(10)s\), we get \(s = 0.05\text{ m} = 5\text{ cm}\).

The hanging part of length \( \frac{2L}{5} \) exerts a pulling force of \( \frac{2}{5} Mg \). The part on the table of length \( \frac{3L}{5} \) experiences a maximum friction force of \( \mu \frac{3}{5} Mg \). Equating the forces at the verge of sliding gives \( \mu = \frac{2}{3} \).

At the highest point, velocity is horizontal: \(v_x = u\cos(37^\circ) = 100 \times 0.8 = 80\text{ m/s}\). By conservation of linear momentum along the horizontal: \(m v_x = m_1 v_1 + m_2 v_2\). Here \(m_1 = m/4\) (comes to rest, \(v_1 = 0\)) and \(m_2 = 3m/4\). Thus, \(m(80) = \frac{3m}{4} v_2\), which gives \(v_2 = \frac{320}{3}\text{ m/s}\).

Impulse is defined as the product of the average force and the short time interval over which it acts, i.e., \(\text{Impulse} = f \times t\).

The maximum limiting frictional force is \(f_{max} = \mu m g = 0.2 \times 20 \times 10 = 40\text{ N}\) (using \(g = 10\text{ m/s}^2\)). Since the applied force of \(40\text{ N}\) is equal to the limiting friction, the net horizontal force on the block is zero, resulting in zero acceleration.

The velocity of the ball just before striking the ground is \(u = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = 14.14\text{ m/s}\). Since it rebounds to the same height, its velocity just after is \(v = 14.14\text{ m/s}\) upwards. The change in momentum is \(Delta p = m(v - (-u)) = 2mu = 2 \times 0.15 \times 14.14 \approx 4.24\text{ kg m/s}\).

Formula: \(F = v \frac{dm}{dt} = 20 \times 2 = 40\text{ N}\). Using Newton's second law, acceleration \(a = \frac{F}{m} = \frac{40}{2} = 20\text{ m/s}^2\).

By Newton's second law, \(F = \frac{\Delta p}{\Delta t} = \frac{mv}{t}\). Substituting the values, \(F = \frac{0.15\text{ kg} \times 20\text{ m/s}}{0.1\text{ s}} = 30\text{ N}\).

For complete mechanical equilibrium, a body must be in both translational equilibrium (net external force is zero) and rotational equilibrium (net external torque is zero).