To find the value of μ, we resolve forces parallel to the slopes for both particles.

For particle B on the 50° slope:

\(1.6g \sin 50^{\circ} - T - F_B = 0\)

For particle A on the 40° slope:

\(T - F_A - 1.2g \sin 40^{\circ} = 0\)

Normal reactions are:

\(R_A = 1.2g \cos 40^{\circ}\)

\(R_B = 1.6g \cos 50^{\circ}\)

Frictional forces are:

\(F_A = 1.2g \mu \cos 40^{\circ}\)

\(F_B = 1.6g \mu \cos 50^{\circ}\)

Substitute the frictional forces into the equations:

\(1.6g \sin 50^{\circ} - T - 1.6g \mu \cos 50^{\circ} = 0\)

\(T - 1.2g \mu \cos 40^{\circ} - 1.2g \sin 40^{\circ} = 0\)

Eliminate \(T\) by adding the equations:

\(1.6g \sin 50^{\circ} - 1.6g \mu \cos 50^{\circ} = 1.2g \sin 40^{\circ} + 1.2g \mu \cos 40^{\circ}\)

Solve for \(\mu\):

\(\mu = \frac{1.6g \sin 50^{\circ} - 1.2g \sin 40^{\circ}}{1.2g \cos 40^{\circ} + 1.6g \cos 50^{\circ}}\)

Calculating gives:

\(\mu = 0.233\)

(i) For particle A, using Newton's second law:

\(0.36g \sin 60^{\circ} - T = 0.36 \times 0.25\)

Solving for \(T\):

\(T = 0.36g \sin 60^{\circ} - 0.36 \times 0.25\)

\(T = 3.03 \text{ N}\)

(ii) For particle B, using Newton's second law:

\(T \pm F - 0.24g \sin 60^{\circ} = 0.24 \times 0.25\)

Solving for \(F\):

\(F = 3.03 - 0.24g \sin 60^{\circ} - 0.24 \times 0.25\)

\(F = 0.889 \text{ N}\)

Normal reaction \(R\):

\(R = 0.24g \cos 60^{\circ}\)

\(R = 1.2 \text{ N}\)

Coefficient of friction \(\mu\):

\(\mu = \frac{F}{R} = \frac{0.889}{1.2}\)

\(\mu = 0.74\)

(a) Resolve forces perpendicular to the plane for Q:

\(R = 0.1g \cos 60 = 0.5\)

Friction force \(F = 0.7 \times 0.1g \cos 60 = 0.35\)

Using Newton's second law for the whole system:

\(0.2g \sin 60 - 0.1g \sin 60 - F = 0.866... - F\)

\(0.2g \sin 60 - 0.1g \sin 60 - 0.35 = 0.516...\)

Since \(\frac{\sqrt{3}}{2} - 0.35 > 0\), the particles do move.

(b) Using Newton's second law for P:

\(0.2g \sin 60 - (\sqrt{3} - 1) = 0.2a\)

\(a = 5 \text{ m/s}^2\)

For the system:

\(0.2g \sin 60 - 0.1g \sin \theta = 0.3(5)\)

\(\theta = 13.4\)

(a) Using Newton's Second Law for particle P on the plane AB:

\(0.3g \sin 60^{\circ} - T = 0.3a\)

For particle Q on the plane BC:

\(T + 0.2g \sin 30^{\circ} = 0.2a\)

Adding the equations for the system:

\(0.3g \sin 60^{\circ} + 0.2g \sin 30^{\circ} = 0.5a\)

Solving for \(a\):

\(a = \frac{0.3g \sin 60^{\circ} + 0.2g \sin 30^{\circ}}{0.5}\)

\(a = 7.20 \text{ ms}^{-2}\)

Substitute \(a\) back to find \(T\):

\(T = 0.3g \sin 60^{\circ} - 0.3 \times 7.20\)

\(T = 0.439 \text{ N}\)

(b) For equilibrium of P:

\(0.3g \sin 60^{\circ} - T = 0\)

\(T = \frac{3\sqrt{3}}{2}\)

For equilibrium of Q on the point of moving down:

\(T + 0.2g \sin 30^{\circ} - F - 3 = 0\)

Using \(F = \mu R\) and \(R = 0.2g \cos 30^{\circ}\):

\(\frac{3\sqrt{3}}{2} + 0.2g \sin 30^{\circ} - \mu (0.2g \cos 30^{\circ}) - 3 = 0\)

Solving for \(\mu\):

\(\mu = 0.345\)

(a) Resolve forces parallel to plane P for particle A: \(T = 2g \sin 10^{\circ}\).

Resolve forces parallel to plane Q for particle B: \(3g \sin 20^{\circ} = F + T\).

Resolve forces perpendicular to plane Q for particle B: \(R = 30 \cos 20^{\circ}\).

Using \(\mu = \frac{F}{R}\), we have:

\(\mu = \frac{3g \sin 20^{\circ} - 2g \sin 10^{\circ}}{30 \cos 20^{\circ}}\).

Calculating gives \(\mu = 0.241\).

(b) Apply Newton's second law to the system: \(3g \sin 20^{\circ} - 2g \sin 10^{\circ} = 5a\).

Solve for \(a\): \(a = \frac{3g \sin 20^{\circ} - 2g \sin 10^{\circ}}{5}\).

Calculate \(a = 1.3575 \text{ m/s}^2\).

Using expressions for height change: \(h_1 = x \sin 20^{\circ}\), \(h_2 = x \sin 10^{\circ}\).

\(x \sin 20^{\circ} + x \sin 10^{\circ} = 1\).

Using \(s = ut + \frac{1}{2}at^2\) with \(s = x\), \(u = 0\), \(a = 1.3575\):

\(\frac{1}{\sin 10^{\circ} + \sin 20^{\circ}} = \frac{1}{2} \times 1.3575 \times t^2\).

Solve for \(t\): \(t = 1.69 \text{ s}\).

(i) Apply Newton's second law to the system:

For particle A: \(T - 0.8g \sin 45^{\circ} = 0.8a\)

For particle B: \(1.2g \sin 30^{\circ} - T = 1.2a\)

System equation: \(1.2g \sin 30^{\circ} - 0.8g \sin 45^{\circ} = 2a\)

Solve for \(a\):

\(a = 0.171 \text{ m/s}^2\)

Use the equation \(v^2 = u^2 + 2as\) with \(u = 0\):

\(v^2 = 2 \times 0.171 \times 0.4\)

\(v = 0.370 \text{ m/s}\)

(ii) Calculate the normal reactions:

\(R_A = 0.8g \cos 45^{\circ} = 4\sqrt{2}\)

\(R_B = 1.2g \cos 30^{\circ} = 6\sqrt{3}\)

Frictional forces:

\(F_A = 4\sqrt{2} \mu\) and \(F_B = 6\sqrt{3} \mu\)

Resolve parallel to the plane:

For A: \(0.8g \sin 45^{\circ} + F_A = T\)

For B: \(1.2g \sin 30^{\circ} - F_B = T\)

System equation: \(12 \sin 30^{\circ} - 8 \sin 45^{\circ} = F_A + F_B\)

Eliminate \(T\) and solve for \(\mu\):

\(\mu = \frac{(6 - 4\sqrt{2})}{(6\sqrt{3} + 4\sqrt{2})} = 0.0214\)

(i) Applying Newton's second law to particle A:

\(T - 0.9g \sin 15 = 0.9a\)

Applying Newton's second law to particle B:

\(2.5 + 0.4g \sin 25 - T = 0.4a\)

Solving these equations simultaneously:

\(1.3a = 1.86 \ldots\)

\(a = 1.43 \text{ m/s}^2\)

(ii) Calculating the frictional force:

\(F = 0.8 \times 0.4g \cos 25\)

Using equilibrium of forces on B:

\(2.5 + 0.4g \sin 25 - T - F = 0\)

Using equilibrium of forces on A:

\(T - 0.9g \sin \theta = 0\)

Solving for \(\theta\):

\(\theta = 8.2^\circ\)

(i) Resolve forces along the plane for particle A:

\(T - 0.8g \sin 30^{\circ} = 0.8a\)

Resolve forces along the plane for particle B:

\(1.2g \sin 60^{\circ} - T = 1.2a\)

Combine the equations:

\(1.2g \sin 60^{\circ} - 0.8g \sin 30^{\circ} = 2a\)

Solving for \(a\):

\(a = 3\sqrt{3} - 2 = 3.20 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = \frac{12}{5}(1 + \sqrt{3}) = 6.56 \text{ N}\)

(ii) Normal reactions:

\(R_A = 0.8g \cos 30^{\circ} = 4\sqrt{3}\)

\(R_B = 1.2g \cos 60^{\circ} = 6\)

Frictional forces:

\(F_A = 4\sqrt{3} \mu\)

\(F_B = 6\mu\)

Resolve parallel to the plane:

\(12 \sin 60^{\circ} - 6\mu - T = 0\)

or

\(T - 8 \sin 30^{\circ} - 4\sqrt{3} \mu = 0\)

System equation:

\(12 \sin 60^{\circ} - 8 \sin 30^{\circ} - 6\mu - 4\sqrt{3} \mu = 0\)

Solve for \(\mu\):

\(\mu = \frac{6\sqrt{3} - 4}{6 + 4\sqrt{3}} = 0.494\)

Consider the forces acting on particles P and Q. For particle P on plane A, the component of gravitational force along the plane is \(0.65 \times 10 \times \frac{63}{65}\). The tension \(T\) acts up the plane. Applying Newton's second law:

\(0.65 \times 10 \times \frac{63}{65} - T = 0.65a\)

For particle Q on plane B, the component of gravitational force along the plane is \(0.65 \times 10 \times \frac{16}{65}\). The tension \(T\) acts up the plane. Applying Newton's second law:

\(T - 0.65 \times 10 \times \frac{16}{65} = 0.65a\)

Adding these two equations to eliminate \(a\):

\(0.65 \times 10 \times \frac{63}{65} - 0.65 \times 10 \times \frac{16}{65} = 2T\)

\(0.65 \times 10 \times \frac{47}{65} = 2T\)

\(T = \frac{0.65 \times 10 \times \frac{47}{65}}{2}\)

\(T = 3.95 \text{ N}\)

(i) Using Newton's second law for particle P:

\(0.6g \times 0.8 - T = 0.6a\)

For particle Q:

\(T - 0.4g \times 0.8 = 0.4a\)

Adding the equations:

\((0.6 - 0.4)g \times 0.8 = (0.6 + 0.4)a\)

Solving for \(a\):

\(a = 1.6\, \text{m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = 3.84\, \text{N}\)

(ii) Using the equation \(v = u + at\) for P:

\(2 = 1.6t_1\)

\(t_1 = 1.25\, \text{s}\)

For Q, using \(v^2 = u^2 + 2as\) with final velocity zero:

\(0 = 2 - 0.8g t_2\)

\(t_2 = 0.25\, \text{s}\)

Total time \(= t_1 + t_2 = 1.5\, \text{s}\)

(i) To find the tension in the string, consider the forces acting on the pulley. The resultant force exerted by the string is \(3\sqrt{3} \text{ N}\). Using the cosine of the angle \(30^\circ\), we have:

\(2T \cos 30^\circ = 3\sqrt{3}\)

\(2T \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\)

\(T \sqrt{3} = 3\sqrt{3}\)

\(T = 3 \text{ N}\)

(ii) For particle Q in limiting equilibrium, resolve forces parallel and perpendicular to AC. The tension \(T = 3 \text{ N}\), and the coefficient of friction \(\mu = 0.75\).

Resolve forces parallel to AC:

\(T = F + mg \sin 30^\circ\)

Resolve forces perpendicular to AC:

\(R = mg \cos 30^\circ\)

Using \(F = \mu R\):

\(3 = 0.75 (10 \cos 30^\circ)m + 10m \sin 30^\circ\)

\(3 = 0.75 \times \frac{\sqrt{3}}{2} \times 10m + 10m \times \frac{1}{2}\)

\(3 = 3.75m \sqrt{3} + 5m\)

Solving for \(m\):

\(m = 0.261 \text{ kg}\)

(i) The tension \(T\) in the string is given by the horizontal component of the force: \(T = 4\sqrt{2} \cos 45^\circ = 4\) N. The mass of Q is found using \(T = mg\), so \(4 = 0.4g\), giving the mass of Q as 0.4 kg.

(ii) For P on the point of slipping, the frictional force \(F\) equals the tension \(T\). Using \(F = \mu R\) and \(R = m_P g\), we have \(4 = 0.8 \times m_P \times 10\), giving \(m_P = 0.5\) kg.

(iii) When a particle of mass 0.1 kg is attached to Q, the system starts to move. Applying Newton's second law to P and Q, we have: \(T - 0.8 \times 0.5g = 0.5a\) and \(0.5g - T = 0.5a\). Solving these equations gives the tension \(T = 4.5\) N.

(a) Resolve forces for the 5 kg particle: The normal reaction is given by the weight of the 5 kg particle, so \(R = 5g\). The frictional force \(F\) is the difference between the weight of the 6 kg particle and the 4 kg particle, so \(F = 6g - 4g = 2g\). The coefficient of friction \(\mu\) is given by \(\mu = \frac{F}{R} = \frac{2g}{5g} = 0.4\).

(b) Apply Newton's second law to the 4 kg and 8 kg particles: For the 4 kg particle, \(T_1 - 4g = 4a\). For the 8 kg particle, \(8g - T_2 = 8a\). The tension difference is \(T_2 - T_1 = F = 0.4 \times 5g\). Adding the equations gives \(8g - 4g - 2g = 17a\), leading to \(a = 1.18 \text{ m/s}^2\). Solving for tensions, \(T_1 = 44.7 \text{ N}\) and \(T_2 = 70.6 \text{ N}\).

(c) For the 8 kg particle, \(T - 4g = 4a\) and \(-T - F = 5a\), or \(-4g - 2g = 9a\). Solving gives \(a = \frac{60}{9}\). Use \(v^2 = u^2 + 2as\) to find \(v\) when the 8 kg particle reaches the ground: \(v^2 = 2 \times \frac{20}{17} \times 0.5 = \frac{20}{17}\), leading to \(v = 1.0846 \ldots\). Use \(v = u + at\) to find \(t\): \(0 = \frac{20}{17} - \frac{60}{9}t\), giving \(t = 0.163 \text{ s}\).

(i) Using Newton's second law for both particles:

For particle P: \(7g - T = 7a\)

For particle Q: \(T - 3g = 3a\)

Adding these equations gives: \(7g - 3g = 10a\)

Solving for acceleration, \(a = \frac{4g}{10} = 4 \text{ m/s}^2\)

Using the equation of motion for speed of P: \(v^2 = u^2 + 2as\)

Where initial speed \(u = 0\), \(a = 4 \text{ m/s}^2\), and \(s = 0.4 \text{ m}\):

\(v^2 = 0 + 2 \times 4 \times 0.4\)

\(v^2 = 3.2\)

\(v = \sqrt{3.2} = 1.79 \text{ m/s}\)

(ii) To determine if Q comes to rest, use:

\(0 = 3.2 + 2 \times (-g) \times s\)

Solving for \(s\):

\(s = 0.16 \text{ m}\)

Since \(0.16 + 0.4 = 0.56 \text{ m}\), particle Q does not come to rest before it reaches the pulley.

(i) For block B in limiting equilibrium, the frictional force is given by:

\(\text{Frictional force} = \mu \times 0.25g\)

Since the system is in equilibrium, the tension in the string due to P and Q must balance the frictional force and the weight of P:

\(0.3g = 0.2g + \mu \times 0.25g\)

Solving for \(\mu\):

\(\mu = \frac{0.3g - 0.2g}{0.25g} = 0.4\)

(ii) When Q is removed, apply Newton's second law to P and B:

For P:

\(0.2g - T = 0.2a\)

For B:

\(T - 0.4 \times 0.25g = 0.25a\)

Solving these equations simultaneously:

From the equation for B:

\(T = 0.25a + 0.4 \times 0.25g\)

Substitute into the equation for P:

\(0.2g - (0.25a + 0.4 \times 0.25g) = 0.2a\)

Simplifying:

\(0.2g - 0.1g = 0.45a\)

\(0.1g = 0.45a\)

\(a = \frac{0.1g}{0.45} = 2.22 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = 0.25 \times 2.22 + 0.1g = 1.56 \text{ N}\)

(i) Applying Newton's second law to particle A:

\(T_A - 2.5 = 0.25a\)

\(T_A = 2.5 + 0.25a\)

For particle B:

\(7.5 - T_B = 0.75a\)

\(T_B = 7.5 - 0.75a\)

(ii) For particle P, the frictional force \(F = 0.4 \times 5 = 2\) N. Using Newton's second law:

\(T_B - T_A - F = 0.5a\)

Substituting the expressions for \(T_A\) and \(T_B\):

\(7.5 - 0.75a - (2.5 + 0.25a) - 2 = 0.5a\)

\(7.5 - 2.5 - 2 = 0.75a + 0.25a + 0.5a\)

\(3 = 1.5a\)

\(a = 2\)

(iii) Using \(v^2 = 2as\) with \(s = 1 - \frac{1}{2}(5.28 - 4) = 0.36\) m:

\(v^2 = 2 \times 2 \times 0.36\)

\(v = 1.2\) m/s

(iv) Immediately after B reaches the floor, apply Newton's second law to P:

\(-T_A - 2 = 0.5a\)

Substitute \(T_A = 2.5 + 0.25a\):

\(-(2.5 + 0.25a) - 2 = 0.5a\)

\(-4.5 = 0.75a\)

\(a = -6\) m/s² (deceleration)

(i) For particle A on the horizontal plane, using Newton's second law:

\(T = 1.6a\)

For particle B on the inclined plane, using Newton's second law:

\(2.4g \sin 30^{\circ} - T = 2.4a\)

Solving these equations simultaneously:

\(2.4g \sin 30^{\circ} = 4a\)

\(a = 3 \text{ m/s}^2\)

Substituting back to find tension:

\(T = 4.8 \text{ N}\)

(ii) Friction force on A is:

\(F = 0.2 \times 1.6g = 3.2 \text{ N}\)

Using Newton's second law for A and B:

\(T - F = 1.6a\)

\(2.4g \sin 30^{\circ} - T = 2.4a\)

Solving these equations:

\(a = 2.2 \text{ m/s}^2\)

Using the equation for motion when B reaches the barrier:

\(v^2 = 2 \times 2.2 \times 1\)

Subsequent acceleration of A is:

\(-2 \text{ m/s}^2\)

Finding the distance A travels while decelerating to \(v = 0\):

\(4.4 = 2 \times x \times s\)

Total distance travelled is 2.1 m.

(i) For particle \(A\):

Using Newton's second law, \(T = 0.3a\).

For particle \(B\):

Using Newton's second law, \(1.5g \sin \theta - T = 1.5a\).

Substitute \(\sin \theta = \frac{3}{5}\):

\(1.5 \times 9.8 \times \frac{3}{5} - T = 1.5a\).

System equation: \(1.5 \times 9.8 \times \frac{3}{5} = 1.8a\).

Solving gives \(a = 5 \text{ m s}^{-2}\) and \(T = 1.5 \text{ N}\).

(ii) Given \(v = 5 \text{ m s}^{-1}\) after 3 s, use \(v = u + at\):

\(5 = 0 + 3a \Rightarrow a = \frac{5}{3} \text{ m s}^{-2}\).

For particle \(A\), friction \(F_A = \mu R_A\) where \(R_A = 3 \times 9.8 = 29.4\).

For particle \(B\), friction \(F_B = \mu R_B\) where \(R_B = 1.5 \times 9.8 \cos \theta = 12\).

Using the system equation:

\(1.5g \sin \theta - F_A - F_B = 1.8a\).

Substitute \(F_A = 3\mu \times 9.8\) and \(F_B = 1.5\mu \times 9.8 \cos \theta\).

Solving gives \(\mu = \frac{2}{5}\).

Let the tension in the string be T and the acceleration of the system be a. For particle Q (mass 0.4 kg), applying Newton's second law gives:

\(0.4g - T = 0.4a\)

For particle P (mass 0.6 kg), applying Newton's second law gives:

\(T = 0.6a\)

Adding these two equations to eliminate T gives:

\(0.4g = (0.4 + 0.6)a\)

Solving for a:

\(a = \frac{0.4g}{1} = 4 \text{ m/s}^2\)

Substituting back to find T:

\(T = 0.6a = 0.6 \times 4 = 2.4 \text{ N}\)

(i) Applying Newton's second law to both particles:

For particle B:
\(0.45a = 0.45g - T\)

For particle A:
\(0.2a = T - F\)

Where \(F = 0.3 \times 0.2g\).

Combining the equations:
\((0.45 + 0.2)a = 0.45g - F\)

Substitute \(F\) and solve for \(a\):
\(a = 6\) m s^-2.

Using the equation \(v^2 = u^2 + 2as\) for particle B:
\(v^2 = 0 + 2 \times 6 \times (2 - (2.8 - 2.1))\)
\(v = 3.95\) m s^-1.

(ii) For particle A reaching the pulley:

Using \(v^2 = 3.95^2 + 2a_2[2.1 - (2 - 0.7)]\), where \(a_2 = -0.06g\).

Calculate \(v\):
\(v = 3.29\) m s^-1.

Alternative method for (ii):

Work done against friction:
\(WD = 0.06g \times [2.1 - (2 - 0.7)]\)

Using kinetic energy loss:
\(\frac{1}{2} \times 0.2 \times 3.95^2 - \frac{1}{2} \times 0.2 \times v^2 = 0.48\)
\(v = 3.29\) m s^-1.

To solve this problem, we apply Newton's second law to both masses.

For block B on the table, the frictional force is given by:

\(F = 0.5 imes 0.6g\)

For particle A hanging vertically, the forces are:

\(0.4g - T = 0.4a\)

For block B on the table, the forces are:

\(T - F = 0.6a\)

Substituting the expression for frictional force:

\(T - 0.5 imes 0.6g = 0.6a\)

We solve these two equations simultaneously:

1. \(0.4g - T = 0.4a\)

2. \(T - 0.3g = 0.6a\)

Adding the two equations to eliminate T:

\(0.4g - 0.3g = 0.4a + 0.6a\)

\(0.1g = a\)

Thus, the acceleration \(a = 1 \text{ m/s}^2\).

Substitute \(a = 1\) into equation 1:

\(0.4g - T = 0.4 \times 1\)

\(T = 0.4g - 0.4\)

\(T = 3.6 \text{ N}\)

For part (ii), using the equation of motion:

\(s = ut + \frac{1}{2}at^2\)

Where \(s = 3 \text{ m}\), \(u = 0\), \(a = 1 \text{ m/s}^2\):

\(3 = 0 + \frac{1}{2} \times 1 \times t^2\)

\(t^2 = 6\)

\(t = \sqrt{6} \approx 2.45 \text{ s}\)

(i) For particle A in limiting equilibrium, the tension \(T\) in the string is equal to the weight of A, so \(T = 0.2g\).

For particle B, the tension \(T\) is equal to the frictional force \(F\), so \(T = F\).

The normal reaction \(R\) on B is \(R = 0.3g\).

Using the friction formula \(F = \mu R\), we have \(0.2g = \mu \times 0.3g\).

Solving for \(\mu\), we get \(\mu = \frac{2}{3}\).

(ii) With the additional force on B, the vertical component is 1.8 N upwards, so the new normal reaction is \(R = 0.3g - 1.8\).

The frictional force is \(F = \frac{2}{3}(0.3g - 1.8)\).

In limiting equilibrium, the horizontal force \(X\) is equal to the sum of the tension and the frictional force, so \(X = T + F\).

Substituting \(T = 0.2g\) and \(F = \frac{2}{3}(0.3g - 1.8)\), we find \(X = 2.8\).

Let the acceleration of the system be a and the tension in the string be T.

For particle P on the table, applying Newton's second law:

\(T = 1.7a\)

For particle Q hanging vertically, applying Newton's second law:

\(0.3g - T = 0.3a\)

Substitute the expression for T from the first equation into the second equation:

\(0.3g - 1.7a = 0.3a\)

Combine and solve for a:

\(0.3g = 2a\)

\(a = \frac{0.3g}{2}\)

Using \(g = 9.8 \text{ m/s}^2\):

\(a = \frac{0.3 \times 9.8}{2} = 1.47 \text{ m/s}^2\)

Rounding to one decimal place, \(a = 1.5 \text{ m/s}^2\).

Substitute \(a = 1.5 \text{ m/s}^2\) back into the equation for T:

\(T = 1.7 \times 1.5 = 2.55 \text{ N}\)

To solve this problem, we apply Newton's second law to both particles.

For particle A on the table:

\(T - 0.6 = 0.4a\)

For particle B hanging vertically:

\(0.1g - T = 0.1a\)

Adding these two equations to eliminate \(T\):

\(0.1g - 0.6 = 0.5a\)

Solving for \(a\):

\(a = \frac{0.1 \times 9.8 - 0.6}{0.5} = 0.8 \text{ m/s}^2\)

Substitute \(a = 0.8\) back into the first equation to find \(T\):

\(T - 0.6 = 0.4 \times 0.8\)

\(T = 0.92 \text{ N}\)

For part (ii), using \(v = at\) to find the speed of B after 1.5 s:

\(v = 0.8 \times 1.5 = 1.2 \text{ m/s}\)

(i) For the smooth table, apply Newton's second law to the system:

\(2.4g - T = 2.4a\)

\(T = 1.6a\)

Combine to solve for \(a\):

\(2.4g = (1.6 + 2.4)a\)

\(a = 6 \text{ m/s}^2\)

Using \(s = \frac{1}{2} a t^2\):

\(0.5 = \frac{1}{2} \times 6 \times t^2\)

\(t = 0.408 \text{ s}\)

(ii) For the rough table, the frictional force \(F = \frac{3}{8} \times 1.6g = 6 \text{ N}\).

Apply Newton's second law:

\(2.4g - 6 = (1.6 + 2.4)a\)

\(a = 4.5 \text{ m/s}^2\)

Find velocity \(v\) as B reaches the ground:

\(v^2 = 2 \times 4.5 \times 0.5\)

\(v = 2.12 \text{ m/s}\)

Deceleration of A:

\(-6 = 1.6a - a\)

\(a = -3.75 \text{ m/s}^2\)

Using \(v^2 = u^2 + 2as\) to find \(s\):

\(0 = 4.5 + 2 \times (-3.75) \times (s - 0.5)\)

\(s = 1.1 \text{ m}\)

(i) For a smooth surface, apply Newton's second law to the system:

For A: \(T = 0.8a\)

For B: \(2 - T = 0.2a\)

Combine: \(0.2g = (0.2 + 0.8)a\)

Solve for \(a\): \(a = 2 \, \text{ms}^{-2}\)

Use the equation \(s = \frac{1}{2} a t^2\) to find \(t\):

\(2.5 = \frac{1}{2} \times 2 \times t^2\)

\(t = 1.58 \text{ s}\)

(ii) For a rough surface, the normal force \(N = 8\) and friction \(F = 0.1 \times N = 0.8\).

Apply Newton's second law:

For A: \(T - 0.8 = 0.8a\)

For B: \(2 - T = 0.2a\)

Combine: \(0.2g - 0.8 = (0.2 + 0.8)a\)

Solve for \(a\): \(a = 1.2 \, \text{ms}^{-2}\)

Use the equation \(v^2 = u^2 + 2as\) to find \(v\):

\(v^2 = 0 + 2 \times 1.2 \times 2.5\)

\(v = \sqrt{6} = 2.45 \text{ ms}^{-1}\)

(i) After B reaches the floor, A continues at constant speed until it reaches the pulley (no tension and the surface is smooth). Thus A's speed when B reaches the floor is the same as A's speed (3 m s^-1) when it reaches the pulley. Until the instant when B reached the floor, A and B have the same speed and hence B reaches the floor with speed 3 m s^-1.

(ii) Using energy conservation:

Loss of potential energy (PE) of B is given by:

\(0.15gh\)

Gain of kinetic energy (KE) of the system is:

\(\frac{1}{2} (0.35 + 0.15) \times 3^2\)

Equating loss of PE to gain of KE:

\(0.15gh = \frac{1}{2} \times 0.5 \times 9\)

\(1.5h = 0.25 \times 9\)

\(h = 1.5\)

(i) Using Newton's second law for both particles, we have:

\(T - 0.2 \times 3 = 0.3a \quad \text{and} \quad 7 - T = 0.7a\)

Solving these equations gives the acceleration:

\(a = 6.4 \text{ m/s}^2\)

Using the equation \(v = u + at\) to find the speed when the string breaks:

\(v = 0 + 6.4 \times 0.25 = 1.6 \text{ m/s}\)

Using \(s = ut + \frac{1}{2}at^2\) to find the distance moved before the break:

\(s = 0 + \frac{1}{2} \times 6.4 \times (0.25)^2 = 0.2 \text{ m}\)

Using \(v^2 = u^2 + 2gs\) to find the speed when B hits the floor:

\(v^2 = 1.6^2 + 2 \times 9.8 \times (0.5 - 0.2)\)

Solving gives:

\(v = 2.93 \text{ m/s}\)

(ii) For finding the distance travelled by A after the break from \(v^2 = u^2 + 2as\):

Distance travelled after break:

\(= (0 - 1.6^2) \div (2 \times -2) = 0.64 \text{ m}\)

Total distance travelled:

\(= 0.2 + 0.64 = 0.84 \text{ m}\)

(i) For particle A on the table, applying Newton's second law:

\(T - \frac{2}{7} \times 1.26 \times g = 1.26 \times a\)

For particle B hanging, applying Newton's second law:

\(0.9 \times g - T = 0.9 \times a\)

Solving these equations simultaneously gives:

\(0.9g - \frac{2}{7} \times 1.26g = (0.9 + 1.26) \times a\)

\(a = 2.5 \text{ m s}^{-2}\)

Substitute \(a\) back to find tension \(T\):

\(T = \frac{2}{7} \times 1.26 \times g + 1.26 \times 2.5\)

\(T = 6.75 \text{ N}\)

(ii) Using \(v^2 = 2ah\) for particle B:

\(v^2 = 2 \times 2.5 \times 0.45\)

\(v = 1.5 \text{ m s}^{-1}\)

(iii) For particle A, using \(v^2 = v_B^2 + 2as\):

\(v^2 = 2.25 + 2 \times \left( -\frac{20}{7} \right) \times 0.03\)

\(v = 1.44 \text{ m s}^{-1}\)

(i) Applying Newton's second law to particle A:

\(0.32g - T = 0.32a\)

For particle B:

\(T = 0.48a\)

Solving these equations simultaneously:

\(0.32g = 0.32a + 0.48a\)

\(0.32g = 0.8a\)

\(a = \frac{0.32g}{0.8} = 4 \text{ m/s}^2\)

Substitute back to find tension:

\(T = 0.48 \times 4 = 1.92 \text{ N}\)

(ii) Using the equation \(s = \frac{1}{2}at^2\) for A:

\(0.98 = \frac{1}{2} \times 4 \times t^2\)

\(t^2 = \frac{0.98}{2}\)

\(t = 0.7 \text{ s}\)

(iii) For B to reach the pulley:

Velocity when A hits the floor:

\(v = at = 4 \times 0.7 = 2.8 \text{ m/s}\)

Time for B to travel remaining distance:

\(t = \frac{1.4 - 0.98}{2.8} = 0.15 \text{ s}\)

Total time:

\(0.7 + 0.15 = 0.85 \text{ s}\)

(i) For particle B, applying Newton's second law:

\(0.2g - T = 0.2 imes 1.6\)

Solving for \(T\):

\(T = 0.2g - 0.2 imes 1.6\)

\(T = 0.2 imes 9.8 - 0.32\)

\(T = 1.96 - 0.32\)

\(T = 1.68 ext{ N}\)

(ii) For particle A, applying Newton's second law:

\(T - F = 0.3 imes 1.6\)

Substitute \(T = 1.68\) N:

\(1.68 - F = 0.48\)

Solving for \(F\):

\(F = 1.68 - 0.48\)

\(F = 1.2 ext{ N}\)

(i) Resolve forces on Q vertically:

\(R + 3.2 \sin 30^{\circ} = 0.5g\)

Resolve forces on Q horizontally:

\(F + 0.2g = 3.2 \cos 30^{\circ}\)

Using \(F = \mu R\), solve for \(\mu\):

\(\mu = \frac{3.2 \cos 30^{\circ} - 2}{5 - 3.2 \sin 30^{\circ}}\)

Coefficient is 0.227.

(ii) For the motion of P and Q:

Equation for P:

\(2 - T = 0.2a\)

Equation for Q:

\(T - 0.227 \times 5 = 0.5a\)

Solve these equations simultaneously to find:

Acceleration \(a = 1.24 \text{ m/s}^2\)

Tension \(T = 1.75 \text{ N}\)

(a) Resolving forces for both particles:

For particle A on the plane:

\(T - F - 2.4g \sin 30 = 0\)

For particle B hanging vertically:

\(3.3g - T = 0\)

Where \(F = \mu R\) and \(R = 2.4g \cos 30\).

Substitute \(F = \mu \times 2.4g \cos 30\) into the equation for A:

\(3.3g - 2.4g \sin 30 - \mu \times 2.4g \cos 30 = 0\)

Solve for \(\mu\):

\(\mu = \frac{3.3g - 2.4g \sin 30}{2.4g \cos 30} \approx 1.01\)

(b) Using Newton's second law for both particles:

For particle B:

\(3.3g - T = 3.3a\)

For particle A:

\(T - F - 2.4g \sin 20 = 2.4a\)

Where \(F = 1.01 \times 2.4g \cos 20\).

Substitute and solve for \(a\):

\(a = 0.353 \text{ m/s}^2\)

Using \(v^2 = u^2 + 2as\) with \(v = 0\), \(u = 0\), solve for \(s\):

\(0 = 0 + 2 \times 0.353 \times s\)

\(s = 1.03 \text{ m}\)

(i) Applying Newton's second law to P and Q:

For P: \(0.5g \times \frac{7}{25} - T = 0.5a\)

For Q: \(T - 0.1g = 0.1a\)

Solving these equations:

\(0.5g \times \frac{7}{25} - T = 0.5a\)

\(T - 0.1g = 0.1a\)

\(1.4 - 1 = 0.6a\)

\(a = \frac{2}{3} \text{ ms}^{-2}\)

Substitute \(a\) to find \(T\):

\(T = 1.07 \text{ N}\)

(ii) Using \(v^2 = u^2 + 2as\) for the motion of P:

\([v^2 = 2 \times \left( \frac{2}{3} \right) \times 0.7]\)

\([2^2 = 2 \times \frac{2}{3} \times 0.7 + 2 \times 0.28g \times s]\)

\(Length of string = 2.5 - s = 1.95 m\)

The forces acting on block P include gravity, normal force, tension, and friction. The frictional force is given by:

\(F = 0.2 \times mg \cos 35^{\circ}\)

For equilibrium, resolve forces along the plane:

\(5g - mg \sin 35^{\circ} - 0.2 mg \cos 35^{\circ} = 0\)

This equation represents the condition when block P is on the point of moving up the plane.

For the condition when block P is on the point of moving down the plane:

\(5g - Mg \sin 35^{\circ} + 0.2 Mg \cos 35^{\circ} = 0\)

Solving these equations gives:

\(m = 6.78 \quad \text{or} \quad M = 12.2\)

Thus, the set of values for m is:

\(6.78 \leq m \leq 12.2\)

(i) Apply Newton's second law to particle Q:

\(0.49g - T = 0.49a\)

For particle P on the inclined plane:

\(T - \frac{40}{160} \times 0.76g = 0.76a\)

Solving these equations simultaneously gives:

\(0.49g - \frac{40}{160} \times 0.76g = (0.49 + 0.76)a\)

Acceleration, \(a = 2.4 \text{ m/s}^2\) and tension, \(T = 3.72 \text{ N}\).

(ii) Using \(v^2 = u^2 + 2as\) with \(u = 0\), \(a = 2.4 \text{ m/s}^2\), and \(s = 0.3 \text{ m}\):

\(v^2 = 2 \times 2.4 \times 0.3\)

Speed, \(v = 1.20 \text{ m/s}\).

(iii) Total distance travelled by P before coming to rest:

Distance while Q is on the ground:

\(\frac{2 \times 2.4 \times 0.3}{2(40 \div 160)g} = 0.288 \text{ m}\)

Total distance travelled, \(0.588 \text{ m}\).

For part (i), apply Newton's second law to both particles. For particle A on the inclined plane:

\(T - 0.26g \sin \alpha = 0.26a\)

For particle B hanging freely:

\(0.52g - T = 0.52a\)

Solving these equations simultaneously:

\(T = 0.26g \sin \alpha + 0.26a\)

\(0.52g - (0.26g \sin \alpha + 0.26a) = 0.52a\)

\(0.52g - 0.26g \sin \alpha = 0.78a\)

\(a = \frac{0.52g - 0.26g \sin \alpha}{0.78}\)

Substitute \(g = 9.8\), \(\sin \alpha = \frac{16}{65}\):

\(a = \frac{0.52 \times 9.8 - 0.26 \times 9.8 \times \frac{16}{65}}{0.78} = 5.85 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = 0.26 \times 9.8 \times \frac{16}{65} + 0.26 \times 5.85 = 2.16 \text{ N}\)

For part (ii), use the equation \(v^2 = u^2 + 2as\) for particle B:

\(v^2 = 2 \times 5.85 \times 0.6\)

\(v = \sqrt{2 \times 5.85 \times 0.6} = 2.65 \text{ m/s}\)

For particle A, use \(v^2 = u^2 - 2g \sin \alpha \times s\):

\(0 = 2.65^2 - 2 \times 9.8 \times \frac{16}{65} \times s\)

\(s = \frac{2.65^2}{2 \times 9.8 \times \frac{16}{65}} = 1.425 \text{ m}\)

Distance AP when A comes to rest:

\(2.5 - 0.6 - 1.425 = 0.475 \text{ m}\)

For part (i):

The normal reaction \(R\) on particle A is given by:

\(R = 0.26 \times \left( \frac{12}{13} \right) = 2.4 \text{ N}\)

The frictional force \(F\) is:

\(F = 0.2 \times 2.4 = 0.48 \text{ N}\)

Applying Newton's second law to particle A along the plane:

\(T - 0.26 \times \frac{5}{13} - 0.48 = 0.26a\)

For particle B, applying Newton's second law vertically:

\(0.54g - T = 0.54a\)

Solving these equations simultaneously:

\(T - 1 - 0.48 = 0.26a\)

\(5.4 - T = 0.54a\)

\((5.4 - 1 - 0.48) = (0.54 + 0.26)a\)

Acceleration \(a = 4.9 \text{ m/s}^2\)

Tension \(T = 2.75 \text{ N}\)

For part (ii):

Using the equation of motion:

\(s = \frac{1}{2} a t^2\)

\(s = \frac{1}{2} \times 4.9 \times (0.4)^2\)

\(s = 0.392 \text{ m}\)

(i) The forces acting on P are:

• Weight \(W = 0.13g\) acting vertically downwards.
• Tension \(T\) in the string acting up the plane.
• Frictional force \(F\) acting down the plane.
• Normal reaction \(R\) perpendicular to the plane.

(ii) To show \(T - F > 0.32\):

The component of weight of P down the plane is \(0.13g \sin \theta\), where \(\sin \theta = \frac{16}{65}\).

Thus, \(0.13g \sin \theta = 0.13g \left(\frac{16}{65}\right)\).

The resultant upward force (RUF) is \(T - F - 0.13g \left(\frac{16}{65}\right)\).

For P to move upwards, \(T - F - 0.13g \left(\frac{16}{65}\right) > 0\).

Therefore, \(T - F > 0.13g \left(\frac{16}{65}\right) \approx 0.32\).

(iii) To find the acceleration of P:

The normal reaction \(R = 0.13g \cos \theta\), where \(\cos \theta = \frac{63}{65}\).

Thus, \(R = 0.13g \left(\frac{63}{65}\right) \approx 1.26\).

The frictional force \(F = \mu R = 0.6 \times 1.26 = 0.756\).

Using Newton's second law for P:

\(T - 0.32 = 0.13a\) and \(0.11g - T = 0.11a\).

Solving these equations gives \(a = 0.1 \text{ m/s}^2\).

(a)(i) For particle P, using Newton's second law:

\(2g \sin 30° - F - T = 2a\)

For particle Q:

\(T - 0.25g = 0.25a\)

Solving the system:

\(2g \sin 30° - F - 0.25g = (2 + 0.25)a\)

Friction force:

\(F = 0.3 \times 2g \cos 30° = 3\sqrt{3} = 5.1961...\)

Acceleration from A to B:

\(a = 1.02 \, \text{m/s}^2\)

(a)(ii) Using SUVAT for v^2:

\(v^2 = u^2 + 2as\)

Substitute:

\(v^2 = 0 + 2 \times 1.02 \times 1.2\)

Velocity at C:

\(v = 3.1 \, \text{m/s}\)

(b) Time from A to B:

\(0.8 = \frac{1}{2} \times 1.02 \times t^2\)

Solve for t:

\(t = 1.25 \, \text{s}\)

Time from B to C:

\(1.2 = \frac{1}{2} \times (3.1 + 0) \times \frac{10}{3}\)

Solve for t:

\(t = 0.547 \, \text{s}\)

Total time:

\(1.25 + 0.547 = 1.8 \, \text{s}\)

Apply Newton's second law to the system:

For particle A: \(2g - T = 2a\)

For particle B: \(T - 3g \sin 18 = 3a\)

Solving these equations gives:

\(2g - 3g \sin 18 = 5a\)

\(a = 2.145898034\)

Use the equation \(v^2 = 2as\) to find \(v^2\) when A reaches the ground:

\(v^2 = 2 \times 2.145898034 \times 0.45 = 1.931308\)

\(v = 1.389715162\)

When the string becomes slack, find the acceleration \(a\) for the motion of B:

\(T = 0, 3g \sin 18 = 3a\)

\(a = 3.0901699\)

Use \(v^2 = 2as\) to find the further distance \(s\) travelled by B:

\(0 = 1.93 - 2 \times 3.09 \times s\)

\(s = 0.312\)

Total distance moved by B is \(0.45 + 0.312 = 0.762\) m.

(a) For A to remain at rest on a smooth plane, the tension \(T\) in the string must balance the component of the weight of A down the plane. Thus, \(T = 2mg\).

Resolving forces parallel to the plane for A, we have:

\(3mg \sin \theta - T = 0\)

Substituting \(T = 2mg\), we get:

\(3mg \sin \theta = 2mg\)

\(\sin \theta = \frac{2}{3}\)

\(\theta = \arcsin \left( \frac{2}{3} \right) \approx 41.8^\circ\)

(b) For the rough plane, the normal reaction \(R = 3mg \cos 30^\circ\).

The frictional force \(F = \mu R\).

Using Newton's second law for B and A:

For B: \(2mg - T = 0.1 \times 2m\)

For A: \(T - 3mg \sin 30^\circ - \mu \times 3mg \cos 30^\circ = 0.1 \times 3m\)

Solving these equations gives:

\(\mu = \frac{\sqrt{3}}{10}\)

(c) Using the equation \(v^2 = u^2 + 2as\) with \(v = 0\), \(u = 0.4\), \(a = -6.5\), and \(s = 0.8\):

\(0 = 0.4^2 + 2 \times (-6.5) \times 0.8\)

Solving for \(t\):

\(t = \frac{0.4}{6.5} \approx 0.0615 \text{ s}\)

(i) Using Newton's second law for P and Q:

For P: \(T - 0.3g \frac{3}{5} = 0.3a\)

For Q: \(0.2g - T = 0.2a\)

Solving these equations simultaneously:

\(0.2g - 0.3g \frac{3}{5} = (0.2 + 0.3)a\)

\(0.2g - 0.18g = 0.5a\)

\(a = 0.4 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = 1.92 \text{ N}\)

(ii) Using the equation \(s = ut + \frac{1}{2}at^2\) with \(s = 0.8\), \(u = 0\), \(a = 0.4\):

\(0.8 = \frac{1}{2} \times 0.4 \times t^2\)

\(t = 2 \text{ s}\)

(iii) Speed when Q hits the floor: \(v = 2 \times 0.4 = 0.8 \text{ m/s}\)

Using \(v = u + at\) for P with \(u = 0\), \(a = -6\):

\(0 = 0.8 + (-6)t\)

\(t = \frac{2}{15} \text{ s}\)

Total time = \(2 + \frac{2}{15} = \frac{34}{15} \text{ s}\)

(i) For particle Q, the tension \(T\) in the string is given by \(T = 0.7g\).

The normal reaction \(R\) on particle P is \(R = 0.4g \times \frac{4}{5} = 3.2\).

Resolving forces along the plane for particle P, we have:

\(X + 0.4g \times \frac{3}{5} - F - T = 0\)

Solving for \(X\), we find \(X = 6.2\).

(ii) Using Newton's second law for both particles:

For particle Q: \(0.7g - T = 0.7a\)

For particle P: \(T - 0.8 - 0.4g \times \frac{3}{5} - F = 0.4a\)

Combining the equations, we solve for \(a\):

\(0.7g - 0.8 - 0.4g \times \frac{3}{5} - F = (0.7 + 0.4)a\)

Solving gives \(a = 2 \text{ m/s}^2\).

(i) For particle P on the plane, the normal reaction \(R = mg \cos \alpha\). Given \(\tan \alpha = \frac{7}{24}\), \(\cos \alpha = \frac{24}{25}\). Thus, \(R = mg \cdot \frac{24}{25} = 0.96mg\).

In limiting equilibrium, the frictional force \(F = \mu R\) and \(F = mg \sin \alpha + mg\). Given \(\sin \alpha = \frac{7}{25}\), \(F = mg \cdot \frac{7}{25} + mg = 1.28mg\).

Equating \(F = \mu R\), we have \(1.28mg = \mu \cdot 0.96mg\). Solving for \(\mu\), \(\mu = \frac{1.28}{0.96} = \frac{4}{3}\).

(ii) Applying Newton's second law to P: \(10 - mg \sin \alpha - F - T = 2.5m\).

For Q: \(T - mg = 2.5m\).

Eliminate \(T\) and solve: \(10 - mg \sin \alpha - \mu mg \cos \alpha - mg = 5m\).

Substitute \(\mu = \frac{4}{3}\), \(\sin \alpha = \frac{7}{25}\), \(\cos \alpha = \frac{24}{25}\):

\(10 - 0.28m - 1.28m - m = 5m\).

Solving gives \(m = 0.327\) kg.

(i) Resolve forces for particle A:

The normal reaction force is given by:

\(R = mg \cos 30^{\circ}\)

The frictional force is:

\(F = 0.2 \times mg \cos 30^{\circ}\)

For particle B, the tension is:

\(T = 4g\)

For particle A, resolve parallel to the plane:

\(T = mg \sin 30^{\circ} + F\)

Substitute for T and F:

\(40 = mg \sin 30^{\circ} + 0.2 \times mg \cos 30^{\circ}\)

Solve for m:

\(m = \frac{40}{5 + \sqrt{3}} = 5.94\)

\((ii) For m = 3, resolve forces and use energy principles:\)

Resolve forces for particle A:

\(R = 3g \cos 30^{\circ}\)

\(F = 0.2 \times 3g \cos 30^{\circ}\)

Apply Newton's Second Law:

\(4g - T = 4a\)

\(T - 3g \sin 30^{\circ} - F = 3a\)

Solve for acceleration a:

\(a = \frac{25}{3 \sqrt{3}} - 7\)

Use the equation:

\(v^2 = u^2 + 2as\)

\(Since initial velocity u = 0, solve for s:\)

\(s = \frac{0.5 \times 2.83}{2} = 0.710 \text{ m}\)

First, calculate the normal reaction \(R\) on the 5 kg particle:

\(R = 5g \cos \alpha = 4g\)

The frictional force \(F\) is:

\(F = 0.5 \times 4g = 2g\)

Apply Newton's second law to the 5 kg particle on the slope:

\(T - 2g - 5g \sin \alpha = 5a\)

Since \(\sin \alpha = \frac{3}{5}\), substitute to get:

\(T - 5g = 5a\)

For the 10 kg particle hanging vertically:

\(10g - T = 10a\)

Combine the equations:

\(10g - 5g = 15a\)

Solve for \(a\):

\(a = \frac{g}{3} = 3.33 \text{ m/s}^2\)

Substitute \(a\) back to find \(T\):

\(T = 10g - 10 \left( \frac{g}{3} \right) = \frac{20g}{3} = 66.7 \text{ N}\)

Using Newton's second law for particle P:

\(10 - 0.5g - T = 0.5a\)

For particle Q:

\(T - 0.3g = 0.3a\)

For the system of both particles:

\(10 - 0.5g - 0.3g = 0.8a\)

Solving for acceleration \(a\):

\(10 - 0.8g = 0.8a\)

\(a = \frac{10 - 0.8 \times 9.8}{0.8} = 2.5 \text{ m/s}^2\)

Substitute \(a = 2.5\) into the equation for Q:

\(T - 0.3 \times 9.8 = 0.3 \times 2.5\)

\(T = 0.3 \times 9.8 + 0.75 = 3.75 \text{ N}\)

(i) For block B in equilibrium, the tension in S₂ must balance the weight of both blocks. Therefore, the tension in S₂ is given by:

\(T_{S_2} = (3 + 2)g = 5g = 50 \text{ N}\)

For block A, the tension in S₁ must balance the weight of block A alone. Therefore, the tension in S₁ is:

\(T_{S_1} = 3g = 30 \text{ N}\)

(ii) After S₂ breaks, apply Newton's second law to each block. For block A:

\(3g - T - 1.6 = 3a\)

For block B:

\(2g + T - 4 = 2a\)

Adding these equations:

\((3g + 2g) - (1.6 + 4) = (3 + 2)a\)

\(5g - 5.6 = 5a\)

\(a = \frac{5g - 5.6}{5} = 8.88 \text{ m/s}^2\)

Substitute \(a\) back into one of the equations to find \(T\):

\(3g - T - 1.6 = 3 \times 8.88\)

\(T = 3g - 1.6 - 26.64\)

\(T = 1.76 \text{ N}\)

(i) In equilibrium, the forces on each particle must balance. For particle A, the tension in S₁ must balance the weight of A and the tension in S₂. Therefore, we have:

\(T_1 = W_A + T_2\)

For particle B, the tension in S₂ must balance the weight of B:

\(T_2 = W_B\)

Given that the weight of each particle is \(W = mg = 0.2 \times 9.8 = 1.96 \text{ N}\), we find:

\(T_2 = 1.96 \approx 2 \text{ N}\)

Substituting back, we find:

\(T_1 = 1.96 + 1.96 = 3.92 \approx 4 \text{ N}\)

(ii) When S₁ is cut, both particles fall with acceleration. Using Newton's second law for particle A:

\(T_2 + 2 - 0.4 = 0.2a\)

For particle B:

\(2 - T_2 - 0.2 = 0.2a\)

Solving these equations simultaneously, we find:

\(a = 8.5 \text{ m/s}^2\)

\(T_2 = 0.1 \text{ N}\)

(a) Applying Newton's second law to particle A:

\(2.4g - T = 2.4a\)

For particle B:

\(T - 1.2g = 1.2a\)

Adding these equations:

\(2.4g - 1.2g = (2.4 + 1.2)a\)

\(1.2g = 3.6a\)

\(a = \frac{1.2g}{3.6} = \frac{10}{3}\) m/s²

Substituting \(a\) back into one of the equations:

\(T = 1.2g + 1.2 \times \frac{10}{3}\)

\(T = 12 + 4 = 16\) N

(b) Using energy conservation for particle B:

Initial potential energy = \(1.2g \times 1.5\)

Final potential energy = \(1.2g \times h\)

Using \(v^2 = u^2 + 2as\) with \(v^2 = 2 \times \frac{10}{3} \times 2.1\)

\(v^2 = 14\)

Using \(v^2 = u^2 + 2gh\) with \(u = 0\):

\(0 = 14 - 2g \times s\)

\(s = 4.3\) m

Greatest height = initial height + additional height = \(1.5 + 2.1 + 0.7 = 4.3\) m

(i) To find the speed of A at the instant B reaches the ground, we use Newton's Second Law and energy conservation.

Using Newton's Second Law for the system:

\(T - 0.35g = 0.35a\)

\(0.45g - T = 0.45a\)

Solving these equations gives:

\(0.45g - 0.35g = (0.35 + 0.45)a\)

\(0.1g = 0.8a\)

\(a = \frac{0.1g}{0.8} = 1.25 \text{ m/s}^2\)

Using the equation of motion:

\(v^2 = u^2 + 2as\)

where initial velocity \(u = 0\), \(a = 1.25 \text{ m/s}^2\), and \(s = 0.64 \text{ m}\):

\(v^2 = 2 \times 1.25 \times 0.64 = 1.6\)

\(v = \sqrt{1.6} = 1.26 \text{ m/s}\)

(ii) To find the total distance travelled by A after B reaches the ground, we consider the motion of A under gravity alone.

Using the equation of motion:

\(v^2 = u^2 + 2gs\)

where \(v = 0\), \(u = 1.26 \text{ m/s}\), and \(g = 9.8 \text{ m/s}^2\):

\(0 = 1.26^2 - 2 \times 9.8 \times s\)

\(s = \frac{1.26^2}{2 \times 9.8} = 0.08 \text{ m}\)

The total distance travelled by A is twice this distance (as it moves up and then down):

\(\text{Total distance} = 2 \times 0.08 = 0.16 \text{ m}\)

(i) Apply Newton's second law to the particles. For the 1.2 kg particle:

\(12 - T = 1.2a\)

For the 0.8 kg particle:

\(T - 8 = 0.8a\)

Solving these equations simultaneously gives:

\(a = 2 ext{ m/s}^2, \, T = 9.6 ext{ N}\)

(ii) First, find the time taken for the 1.2 kg particle to reach the ground:

\(0.64 = \frac{1}{2} \times 2 \times t_1^2\)

Solving gives:

\(t_1 = 0.8 ext{ s}\)

The speed of the 0.8 kg particle when the 1.2 kg particle hits the ground is:

\(v = 2 \times 0.8 = 1.6 ext{ m/s}\)

Next, find the time and distance for the 0.8 kg particle to come to rest:

\(0 = 1.6 - 10t_2\)

Solving gives:

\(t_2 = 0.16 ext{ s}\)

The distance travelled during this time is:

\(s_2 = \frac{1}{2} \times 1.6 \times 0.16 = 0.128 ext{ m}\)

Finally, find the distance travelled in the remaining time:

\(t_3 = 1 - 0.8 - 0.16 = 0.04 ext{ s}\)

\(s_3 = \frac{1}{2} \times 10 \times 0.04^2 = 0.008 ext{ m}\)

Total distance travelled:

\(0.64 + 0.128 + 0.008 = 0.776 ext{ m}\)

(i) (a) Using Newton's Second Law for the system, we have:

\(1.3g - T = 1.3a\)

\(T - 0.7g = 0.7a\)

Adding these equations gives:

\(1.3g - 0.7g = (1.3 + 0.7)a\)

\(0.6g = 2a\)

\(a = 3 \text{ m/s}^2\)

Substituting back to find tension:

\(1.3g - T = 1.3 \times 3\)

\(T = 1.3g - 3.9\)

\(T = 9.1 \text{ N}\)

(b) Using the equation of motion:

\(s = \frac{1}{2} a t^2\)

\(2 = \frac{1}{2} \times 3 \times t^2\)

\(t^2 = \frac{4}{3}\)

\(t = \sqrt{\frac{4}{3}} \approx 1.15 \text{ seconds}\)

(ii) Using energy conservation or equations of motion:

At the point when the 1.3 kg particle reaches the plane, the speed \(v\) is given by:

\(v^2 = 2 \times 3 \times 2\)

\(v = \sqrt{12}\)

The 0.7 kg particle continues upwards:

\(0 = 12 - 2gs\)

\(s = \frac{12}{2g}\)

Greatest height above the plane is:

\(2 + \frac{12}{2g} = 4.6 \text{ m}\)

(i) Applying Newton's second law to particle A:

\(0.35g - T = 0.35a\)

For particle B:

\(T - 0.15g = 0.15a\)

Adding the two equations:

\((0.35 - 0.15)g = (0.35 + 0.15)a\)

Solving for acceleration \(a\):

\(0.2g = 0.5a\)

\(a = \frac{0.2g}{0.5} = 4\, \text{m/s}^2\)

Substituting \(a\) back to find tension \(T\):

\(T = 0.15g + 0.15 \times 4 = 2.1\, \text{N}\)

(ii) Using the equation \(v_1^2 = 0 + 2a \times 1.6\):

\(v_1^2 = 8 \times 1.6 = 12.8\)

Using the formula for greatest height \(H\):

\(H = 1.6 + \frac{0 - v_1^2}{-2g}\)

\(H = 1.6 + \frac{-12.8}{-20} = 2.24\, \text{m}\)

Since the system is in equilibrium, the tension \(T\) in the string is equal to the weight of the particle \(P\). Therefore, \(T = 3g\), where \(g\) is the acceleration due to gravity. Assuming \(g = 10 \text{ m/s}^2\), we have:

\(T = 3 \times 10 = 30 \text{ N}\).

For the normal component of the contact force on \(B\), resolve the forces perpendicular to the plane. The normal force \(R\) is given by:

\(R = (4g - T) \cos \alpha\).

Substitute \(T = 30 \text{ N}\) and \(\cos \alpha = 0.8\):

\(R = (4 \times 10 - 30) \times 0.8 = 10 \times 0.8 = 8 \text{ N}\).

(i) The area under the velocity-time graph represents the distance traveled. The graph shows a triangle with base 0.5 s and height 2 m/s. The area is given by:

\(h = \frac{1}{2} \times 0.5 \times 2 = 0.5\)

(ii) The acceleration \(a\) can be found using the gradient of the velocity-time graph:

\(a = \frac{2}{0.5} = 4 \text{ m/s}^2\)

Using Newton's second law for both particles:

For P: \(T - mg = ma\)

For Q: \((1-m)g - T = (1-m)a\)

Solving these equations simultaneously gives:

\(a = \frac{(1-2m)}{(1+m)}g\)

Substituting \(a = 4\) and \(g = 10\):

\(4 = \frac{(1-2m)}{(1+m)} \times 10\)

Solving for \(m\):

\(m = 0.3\)

Substitute \(m = 0.3\) back to find tension \(T\):

\(T = 0.3 \times 10 + 4 \times 0.3 = 4.2 \text{ N}\)

(iii) The string is slack while Q is at rest on the ground. The graph shows that P moves from 2 m/s to -2 m/s, taking 0.5 s to reach 0 m/s. The total time is:

\(T = 0.5 + 0.4 = 0.9 \text{ s}\)

(i) From the velocity-time graph, the acceleration can be found using the gradient. The acceleration is 4 m/s². Using Newton's second law for both particles, we have:

\(T - mg = 4m\)

\((1 - m)g - T = 4(1 - m)\)

Solving these equations gives:

\(4 = (1 - m - m)g\)

Thus, P has mass 0.3 kg and Q has mass 0.7 kg.

(ii) Using the area under the velocity-time graph or the formula \(h = \frac{1}{2} a t^2\), we find \(h = 2\) m.

(iii) The distance travelled upwards by P is given by the area under the graph:

\(\text{Distance} = \frac{1}{2} \times 1.4 \times 4 = 2.8 \text{ m}\)

Adding the initial height, the greatest height is 4.8 m.

(i) Using Newton's second law for both particles and eliminating tension, we have:

\((M + m)a = (M - m)g\)

where \(M = 0.75\) kg, \(m = 0.25\) kg, and \(g = 9.8\) m/s².

Solving gives:

\(a = \frac{(0.75 - 0.25) \times 9.8}{0.75 + 0.25} = 5\) m/s².

Using the equation \(s = ut + \frac{1}{2}at^2\) with \(u = 0\), \(t = 0.6\) s:

\(s = 0 + \frac{1}{2} \times 5 \times (0.6)^2 = 0.9\) m.

(ii) From the velocity-time graph, the area under the graph from 0 to 0.6 s is:

\(\frac{1}{2} \times 0.6 \times V = 0.9\)

Solving gives:

\(V = 3\) m/s.

Using the equation \(0 = V - g(T - 0.6)\):

\(0 = 3 - 9.8(T - 0.6)\)

Solving gives:

\(T = 0.9\) s.

(iii) The distance travelled upwards is given by:

\(s_{\text{up}} = \frac{1}{2} \times 0.9 \times 3 = 1.35\) m.

The distance travelled downwards is given by:

\(s_{\text{down}} = 0 + \frac{1}{2} \times 9.8 \times (1.6 - 0.9)^2 = 2.45\) m.

Thus, the height \(h\) is:

\(h = s_{\text{down}} - s_{\text{up}} = 2.45 - 1.35 = 1.1\) m.

Let the acceleration of the system be \(a\). Using Newton's second law for both particles:

For particle A: \(T - 0.2g = 0.2a\)

For particle B: \(0.6g - T = 0.6a\)

Adding these equations gives:

\(0.6g - 0.2g = 0.6a + 0.2a\)

\(0.4g = 0.8a\)

\(a = \frac{0.4g}{0.8} = 5 \, \text{m/s}^2\)

When B reaches the floor, using \(v^2 = u^2 + 2as\) with \(u = 0\), \(a = 5\), and \(s = 1.6\):

\(v^2 = 2 \times 5 \times 1.6\)

\(v^2 = 16\)

\(v = 4 \, \text{m/s}\)

For A to reach 3 m above the floor, the total distance moved by A is \(3 - h\). Using energy conservation:

\(Potential energy gain = Kinetic energy loss\)

\(0.2g(3 - h) = \frac{1}{2} \times 0.2 \times 4^2\)

\(0.2g(3 - h) = 1.6\)

\(2g(3 - h) = 16\)

\(3 - h = 0.8\)

\(h = 3 - 0.8 = 0.6\)

(i) Applying Newton's second law to particle B gives:

\(0.7g - T = 0.7a\)

For particle A:

\(T - 0.3g = 0.3a\)

Adding these equations:

\(0.7g - 0.3g = 0.7a + 0.3a\)

\(0.4g = a\)

Substituting \(g = 9.8 \text{ m/s}^2\):

\(a = 4 \text{ m/s}^2\)

Substitute \(a\) back into one of the equations:

\(0.7g - T = 0.7 \times 4\)

\(T = 0.7g - 2.8\)

\(T = 6.86 - 2.8 = 4.06 \text{ N}\)

Rounding gives \(T = 4.2 \text{ N}\).

(ii) When the string breaks, both particles have a speed of 1.6 m/s. The distance A falls is 0.52 m. Using the kinematic equation:

\(s = ut + \frac{1}{2}gt^2\)

\(0.52 = 1.6t + \frac{1}{2} \times 9.8 \times t^2\)

\(0.52 = 1.6t + 4.9t^2\)

Solving the quadratic equation:

\(4.9t^2 + 1.6t - 0.52 = 0\)

Using the quadratic formula:

\(t = \frac{-1.6 \pm \sqrt{(1.6)^2 - 4 \times 4.9 \times (-0.52)}}{2 \times 4.9}\)

\(t = \frac{-1.6 \pm \sqrt{2.56 + 10.192}}{9.8}\)

\(t = \frac{-1.6 \pm \sqrt{12.752}}{9.8}\)

\(t = \frac{-1.6 \pm 3.57}{9.8}\)

\(t = 0.6 \text{ s}\) (taking the positive root).

(a) Consider the forces on the 0.1 kg particle:

\(T - 0.1g = 0.1a\)

For the m kg particle:

\(mg - T = ma\)

Adding these equations for the system gives:

\(mg - 0.1g = (m + 0.1)a\)

Substitute \(T = 1.5\) N and solve for \(m\):

\(a = 5\)

\(m = 0.3\)

(b) Use the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 0.9\), and \(a = 5\):

\(v^2 = 0 + 2 \times 5 \times 0.9\)

\(v = 3 \text{ m s}^{-1}\)

(i) Using the equation of motion:

\(v = u + at\)

where \(v = 0.6 \text{ m s}^{-1}\), \(u = 0 \text{ m s}^{-1}\), and \(t = 0.3 \text{ s}\).

Substitute the values:

\(0.6 = 0 + a \times 0.3\)

Solve for \(a\):

\(a = \frac{0.6}{0.3} = 2 \text{ m s}^{-2}\)

(ii) Applying Newton's second law to particle A:

\(mg - T = ma\)

For particle B:

\(T - (1 - m)g = (1 - m)a\)

Equating the two expressions for tension \(T\):

\(mg - ma = (1 - m)g + (1 - m)a\)

Substitute \(a = 2 \text{ m s}^{-2}\) and \(g = 10 \text{ m s}^{-2}\):

\(10m - 2m = 10 - 10m + 2 - 2m\)

Simplify and solve for \(m\):

\(8m = 12 - 12m\)

\(20m = 12\)

\(m = 0.6\)

Substitute \(m = 0.6\) back to find \(T\):

\(T = 8m = 8 \times 0.6 = 4.8 \text{ N}\)

(i) Using Newton's second law, the equations for the forces on A and B are:

\(T - 0.12g = 0.12a\)

\(0.38g - T = 0.38a\)

Adding these equations gives:

\(0.38g - 0.12g = 0.12a + 0.38a\)

\(a = \frac{0.38 - 0.12}{0.38 + 0.12}g\)

Thus, the acceleration \(a = 5.2 \text{ m/s}^2\).

(ii) Using \(v^2 = 2as\) for B:

\(v^2 = 2 \times 5.2 \times 0.65\)

\(v = 2.6 \text{ m/s}\)

Time \(T_B\) is given by:

\(s = \frac{1}{2} a T_B^2\)

\(0.65 = \frac{1}{2} \times 5.2 \times T_B^2\)

\(T_B = 0.5 \text{ s}\)

(iii) For A, using the equation:

\(-2.6 = 2.6 - 10(T - 0.5)\)

Solving gives:

\(T = 1.02 \text{ s}\)

The velocity-time graph is a straight line increasing to 2.6 m/s at 0.5 s, then decreasing back to 0 m/s at 1.02 s.

(iv) Total distance travelled by A:

\(0.65 + 0.5(1.02 - 0.5) \times 2.6\)

\(= 1.326 \text{ m}\)

Using Newton's second law for particle P:

\(0.9g - 7.2 = 0.9a\)

Solving for acceleration \(a\):

\(a = 2\)

Using the equation of motion \(v^2 = u^2 + 2ah\) with initial velocity \(u = 0\):

\(v^2 = 2 \times (0.9g - 7.2)/0.9 \times 2\)

\(v = \sqrt{8}\)

When the string becomes slack, the velocity of Q is:

\(u_{\text{slack}} = v_{\text{taut}} = 2\sqrt{g - 8}\)

Using the equation \(0 = u^2 - 2gh\) to find the distance:

\(\text{distance} = 2h\)

The total distance travelled by Q is 0.8 m.

(i) Applying Newton's second law to particle A:

\(0.9g - T = 0.9a\)

For particle B:

\(T - 0.6g = 0.6a\)

Adding the two equations:

\(0.9g - 0.6g = 0.9a + 0.6a\)

\(0.3g = 1.5a\)

Solving for \(a\):

\(a = \frac{0.3g}{1.5} = 2 \text{ m/s}^2\)

Substituting \(a\) back to find \(T\):

\(T = 0.6g + 0.6 \times 2 = 0.6 \times 9.8 + 1.2 = 7.2 \text{ N}\)

(ii) After A hits the floor, B moves upwards for 0.3 s with initial velocity \(u = 3 \text{ m/s}\) (from previous motion):

Using \(v = u - gt\):

\(0 = 3 - 9.8 \times 0.3\)

Solving for \(t\):

\(t = 0.3 \text{ s}\)

Using \(v^2 = u^2 + 2ah\) to find height \(h\):

\(0 = 3^2 - 2 \times 9.8 \times h\)

\(9 = 19.6h\)

\(h = \frac{9}{19.6} = 0.459 \text{ m}\)

Total height from initial position:

\(h = 2.25 \text{ m}\)

Let the tension in the string be denoted by \(T\) and the acceleration of the system be \(a\).

For particle A (mass 0.65 kg), applying Newton's second law:

\(0.65g - T = 0.65a\)

For particle B (mass 0.35 kg), applying Newton's second law:

\(T - 0.35g = 0.35a\)

Adding these two equations to eliminate \(T\):

\(0.65g - 0.35g = 0.65a + 0.35a\)

\(0.30g = a\)

Substitute \(a = 0.30g\) back into one of the equations to find \(T\):

\(0.65g - T = 0.65(0.30g)\)

\(T = 0.65g - 0.195g\)

\(T = 0.455g\)

Using \(g = 9.8 \text{ m/s}^2\), \(T = 0.455 \times 9.8 = 4.55 \text{ N}\).

The resultant force exerted on the pulley by the string is twice the tension:

\(2T = 2 \times 4.55 = 9.1 \text{ N}\).

(i) Using Newton's second law for both particles:

For particle A: \(T - 0.3g = 0.3a\)

For particle B: \(0.7g - T = 0.7a\)

Adding these equations gives:

\(0.7g - 0.3g = (0.7 + 0.3)a\)

\(0.4g = a\)

\(a = 4 \text{ m s}^{-2}\)

(ii) Using the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(v = 1.6 \text{ m s}^{-1}\), and \(a = -4 \text{ m s}^{-2}\):

\(0 = 1.6^2 - 2 \times 4 \times s\)

\(s = \frac{1.6^2}{2 \times 4} = 0.32 \text{ m}\)

For the total height, \(s_1 + s_2 = 0.32 + 0.128 = 0.448 \text{ m}\)

(iii) Using \(v = u + at\) with \(u = 0\), \(v = 1.6 \text{ m s}^{-1}\), and \(a = 4 \text{ m s}^{-2}\):

\(1.6 = 4t_1\)

\(t_1 = \frac{1.6}{4} = 0.4 \text{ s}\)

For the descent, \(t_2 = \frac{1.6}{10} = 0.16 \text{ s}\)

Total time \(t_1 + t_2 = 0.4 + 0.16 = 0.56 \text{ s}\)

(i) Using Newton's second law for the system, the net force is given by the difference in weights:

\(0.55g - T = 0.55a\)

\(T - 0.45g = 0.45a\)

Adding these equations, we eliminate tension \(T\):

\(0.55g - 0.45g = 0.55a + 0.45a\)

\(0.10g = 1.00a\)

Thus, the acceleration \(a\) is:

\(a = \frac{0.10g}{1.00} = 1 \text{ m/s}^2\)

(ii) (a) Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with \(u = 0\), \(a = 1 \text{ m/s}^2\), and \(t = 2 \text{ s}\):

For P:

\(s_P = 5 - \frac{1}{2} \times 1 \times (2)^2 = 5 - 2 = 3 \text{ m}\)

For Q:

\(s_Q = 5 + \frac{1}{2} \times 1 \times (2)^2 = 5 + 2 = 7 \text{ m}\)

(b) The speed \(v\) of the particles after 2 s is given by:

\(v = u + at = 0 + 1 \times 2 = 2 \text{ m/s}\)

(iii) After the string breaks, both particles move under gravity. For P, using \(s = ut + \frac{1}{2}gt^2\) with \(u = 2 \text{ m/s}\), \(g = 9.8 \text{ m/s}^2\), and \(s = 3 \text{ m}\):

\(3 = 2t_P + \frac{1}{2} \times 9.8 \times t_P^2\)

Solving gives \(t_P = 0.6 \text{ s}\).

For Q, using \(s = ut + \frac{1}{2}gt^2\) with \(u = 2 \text{ m/s}\), \(s = 7 \text{ m}\):

\(7 = -2t_Q + \frac{1}{2} \times 9.8 \times t_Q^2\)

Solving gives \(t_Q = 1.4 \text{ s}\).

Thus, Q reaches the ground 0.8 s later than P.

(i) Using the equation of motion: \(s = ut + \frac{1}{2} a t^2\), where \(s = 0.36\) m, \(u = 0\), and \(t = 0.6\) s. Substituting, we get:

\(0.36 = 0 + \frac{1}{2} a (0.6)^2\)

\(0.36 = 0.18a\)

\(a = \frac{0.36}{0.18} = 2 \text{ m/s}^2\)

(ii) Applying Newton's second law to particle A: \(0.45g - T = 0.45a\)

\(0.45 \times 10 - T = 0.45 \times 2\)

\(4.5 - T = 0.9\)

\(T = 4.5 - 0.9 = 3.6 \text{ N}\)

(iii) Applying Newton's second law to particle B: \(T - mg = ma\)

\(3.6 - m \times 10 = m \times 2\)

\(3.6 = 12m\)

\(m = \frac{3.6}{12} = 0.3 \text{ kg}\)

(iv) Using the equation \(v^2 = u^2 + 2as\) for particle B when it reaches maximum height (\(v = 0\)):

\(0 = (0.6 \times 2)^2 - 2 \times 10 \times s\)

\(0 = 1.44 - 20s\)

\(s = \frac{1.44}{20} = 0.072 \text{ m}\)

Maximum height above the floor = initial height + \(s = 0.72 + 0.072 = 0.792 \text{ m}\)

Since the system is in equilibrium, the tension in the string is equal to the weight of the particle P. The weight of P is given by:

\(T = m_P imes g = 4 imes 10 = 40 \text{ N}\)

Thus, the tension in the string is 40 N.

For block B, the forces acting vertically are its weight, the tension in the string, and the normal reaction from the ground. Since B is in equilibrium, we have:

\(R + T = W_B\)

where \(R\) is the normal reaction, \(T\) is the tension, and \(W_B\) is the weight of block B.

The weight of block B is:

\(W_B = m_B imes g = 5 imes 10 = 50 \text{ N}\)

Substituting the known values:

\(R + 40 = 50\)

Solving for \(R\):

\(R = 50 - 40 = 10 \text{ N}\)

Therefore, the force exerted on B by the ground is 10 N.

(i) (a) To find the acceleration of A, use the equation of motion:

\(v = u + at\)

where \(v = 5 \text{ m s}^{-1}\), \(u = 0\), and \(t = 2 \text{ s}\).

Substitute the values:

\(5 = 0 + 2a\)

\(a = \frac{5}{2} = 2.5 \text{ m s}^{-2}\)

(b) To find the tension in the string, apply Newton's second law to particle A:

\(0.5g - T = 0.5 \times 2.5\)

\(T = 0.5g - 0.5 \times 2.5\)

\(T = 0.5 \times 9.8 - 1.25\)

\(T = 4.9 - 1.25 = 3.75 \text{ N}\)

(ii) To find the value of m, apply Newton's second law to particle B:

\(T - mg = 2.5m\)

Substitute \(T = 3.75\) and solve for \(m\):

\(3.75 - mg = 2.5m\)

\(3.75 = mg + 2.5m\)

\(3.75 = m(g + 2.5)\)

\(m = \frac{3.75}{g + 2.5}\)

\(m = \frac{3.75}{9.8 + 2.5}\)

\(m = \frac{3.75}{12.3} \approx 0.3 \text{ kg}\)

(a) Apply Newton's second law to each particle:

For the 0.8 kg particle:

\(0.8g - T = 0.8a\)

For the 0.2 kg particle:

\(T - 0.2g = 0.2a\)

For the system:

\(0.8g - 0.2g = (0.8 + 0.2)a\)

Solve for \(a\):

\(a = 6 \text{ m s}^{-2}\)

Solve for \(T\):

\(T = 3.2 \text{ N}\)

(b) Use the equation \(v^2 = u^2 + 2as\) to find the velocity \(v\) of the 0.8 kg particle as it reaches the ground:

\(v^2 = 2 \times 6 \times 0.5\)

Find the extra height reached by the 0.2 kg particle using \(v^2\):

\(0 = 6 - 20s\)

\(Greatest height = 0.5 + 0.5 + 0.3 = 1.3 m\)

(i) Apply Newton's second law to particle P:

\(0.6g - T = 0.6a\)

Apply Newton's second law to particle Q:

\(T - 0.2g = 0.2a\)

Add the two equations:

\(0.6g - T + T - 0.2g = 0.6a + 0.2a\)

\(0.4g = 0.8a\)

Solve for acceleration \(a\):

\(a = \frac{0.4g}{0.8} = 5 \text{ m/s}^2\)

Substitute \(a = 5\) into the equation for P:

\(0.6g - T = 0.6 \times 5\)

\(T = 0.6g - 3\)

\(T = 3 \text{ N}\)

(ii) Use the equation of motion \(s = ut + \frac{1}{2}at^2\) with \(s = 0.9\), \(u = 0\), \(a = 5\):

\(0.9 = \frac{1}{2} \times 5 \times t^2\)

\(t^2 = \frac{0.9 \times 2}{5}\)

\(t^2 = 0.36\)

\(t = 0.6 \text{ s}\)

(i) Using the equation of motion \(s = ut + \frac{1}{2}at^2\), we have:

\(0.81 = 0 + \frac{1}{2}a \times 0.9^2\)

Solving for \(a\), we get \(a = 2 \text{ m s}^{-2}\).

Using Newton's Second Law for particle A: \(T - mg = ma\) and for particle B: \(kmg - T = kma\).

Solving these equations simultaneously gives:

\(a = \frac{(kg - g)}{k+1}\)

Setting \(a = 2\), we solve for \(k\):

\(\frac{(kg - g)}{k+1} = 2\)

\(k = 1.5\).

The tension \(T = 10m + 2m = 12m \text{ N}\).

(ii) The velocity of A when the string breaks is \(2 \times 0.9 = 1.8 \text{ m s}^{-1}\) upwards.

Using \(v^2 = u^2 + 2gs\) to find \(v\) at the ground:

\(v^2 = 1.8^2 + 2g \times 1.62\)

Speed is \(5.97 \text{ m s}^{-1}\).

Time taken:

\(\frac{(1.8 + 5.97)}{g} = 0.777 \text{ s}\)

(iii) The velocity-time graph is a straight line from (0, 0) to (0.9, 1.8), then from (0.9, 1.8) to approximately (1.7, -6).

(i) Using Newton's Second Law for particle A:

\(1.3g - T = 1.3a\)

For particle B:

\(T - 0.7g = 0.7a\)

Adding these equations gives:

\(1.3g - 0.7g = 1.3a + 0.7a\)

\(0.6g = 2a\)

Thus,

\(a = \frac{0.6g}{2} = 3 ext{ m s}^{-2}\)

Substitute back to find tension:

\(1.3g - T = 1.3 \times 3\)

\(T = 1.3g - 3.9 = 9.1 ext{ N}\)

(ii) The distance while connected is 0.375 m. Using the equation

\(v^2 = u^2 + 2as\)

where initial velocity,

\(u = 0,\)

we find

\(v = 1.5 ext{ m s}^{-1}\)

For particle A after the string breaks:

\(1.375 = 1.5t + \frac{1}{2}gt^2\)

Solving gives

\(t = 0.395 ext{ s}\)

For particle B:

\(-1.375 = 1.5t - \frac{1}{2}gt^2\)

Solving gives

\(t = 0.695 ext{ s}\)

\(Difference in times = 0.3 s.\)

(i) Apply Newton's second law to particle A:

\(4 - T = 0.4a\)

Apply Newton's second law to particle B:

\(T - 2 = 0.2a\)

For the system:

\(4 - 2 = (0.4 + 0.2)a\)

Solving the system of equations gives:

\(a = \frac{10}{3} \text{ m s}^{-2}, \; T = \frac{8}{3} \text{ N}\)

(ii) Use \(v^2 = u^2 + 2as\) for particle B with \(a = \frac{10}{3}\):

\(v^2 = 0 + 2 \times \frac{10}{3} \times 0.5\)

\(v^2 = \frac{10}{3}\)

After A hits the ground, apply \(v^2 = u^2 + 2as\) to find \(s\):

\(0 = \frac{10}{3} - 2 \times 10 \times s\)

\(s = \frac{1}{6} \text{ m}\)

Maximum height = \(\frac{1}{2} + \frac{1}{2} + \frac{1}{6} = \frac{7}{6} = 1.17 \text{ m}\)

(i) Using Newton's second law for P and Q:

\(T - 0.3g = 0.3a\)

\(0.5g - T = 0.5a\)

Adding these equations gives:

\(0.5g - 0.3g = 0.8a\)

Solving for \(a\):

\(a = \frac{0.2g}{0.8} = 2.5 \text{ m/s}^2\)

Using \(s = ut + \frac{1}{2}at^2\) to find \(h\):

\(h = 0 + \frac{1}{2} \times 2.5 \times 0.6^2 = 0.45 \text{ m}\)

(ii) Velocity of P when Q reaches the floor:

\(v = 0 + 0.6 \times 2.5 = 1.5 \text{ m/s}\)

Using \(v = u + at\) to find time to highest point:

\(0 = 1.5 - 9.8t\)

\(t = \frac{1.5}{9.8} \approx 0.15 \text{ s}\)

Total time:

\(2 \times 0.15 + 0.6 = 0.9 \text{ s}\)

To solve this problem, we use the equations of motion and Newton's second law.

(i) Finding the tension in the string:

First, we find the acceleration of the system. Using the equation of motion:

\(v^2 = u^2 + 2as\)

where \(v = 0.6 \text{ m/s}\), \(u = 0 \text{ m/s}\), and \(s = 0.8 \text{ m}\).

Substituting the values, we get:

\(0.6^2 = 0 + 2a \times 0.8\)

\(a = 0.225 \text{ m/s}^2\)

Now, applying Newton's second law to the 0.3 kg particle:

\(T - 0.3g = 0.3a\)

Substituting \(g = 9.8 \text{ m/s}^2\) and \(a = 0.225 \text{ m/s}^2\):

\(T - 0.3 \times 9.8 = 0.3 \times 0.225\)

\(T = 3.07 \text{ N}\)

(ii) Finding the value of \(m\):

Applying Newton's second law to the \(m\) kg particle:

\(mg - T = ma\)

Substituting \(T = 3.07 \text{ N}\) and \(a = 0.225 \text{ m/s}^2\):

\(mg - 3.07 = m \times 0.225\)

Solving for \(m\):

\(m(9.8 - 0.225) = 3.07\)

\(m = 0.314 \text{ kg}\)

Let the tension in the string be denoted by T and the acceleration of the system by a. For particle A (mass 0.8 kg), applying Newton's second law gives:

\(8 - T = 0.8a\)

For particle B (mass 0.2 kg), applying Newton's second law gives:

\(T - 2 = 0.2a\)

Adding these two equations to eliminate T:

\((8 - T) + (T - 2) = 0.8a + 0.2a\)

\(6 = a\)

Thus, the acceleration a is 6 m s^-2.

\(Substitute a = 6 into the equation for particle B:\)

\(T - 2 = 0.2(6)\)

\(T - 2 = 1.2\)

\(T = 3.2\)

Therefore, the tension in the string is 3.2 N.