Past Exam Questions

(i) To find the value of C, we use the triangle of forces. The horizontal component of the applied force is given by:

\(F = 4 \cos 30^{\circ}\)

The vertical component of the applied force is:

\(R = 10 - 4 \sin 30^{\circ}\)

Using the cosine rule for the triangle of forces:

\(C^2 = (4 \cos 30^{\circ})^2 + (10 - 4 \sin 30^{\circ})^2\)

Calculating gives:

\(C = 8.72\)

(ii) The coefficient of friction \(\mu\) is given by:

\(\mu = \frac{4 \cos 30^{\circ}}{10 - 4 \sin 30^{\circ}}\)

Calculating gives:

Coefficient is 0.433 (accept 0.43)

To find the coefficient of friction, we resolve the forces vertically and horizontally.

Vertically, the normal reaction force is given by:

\(R + 2000 \cos 15^{\circ} = 400g\)

where \(g\) is the acceleration due to gravity \((9.8 \text{ m/s}^2)\).

Horizontally, the frictional force \(F\) is:

\(F = 2000 \sin 15^{\circ}\)

In limiting equilibrium, the frictional force \(F\) is equal to \(\mu R\), where \(\mu\) is the coefficient of friction:

\(2000 \sin 15^{\circ} = \mu (400g - 2000 \cos 15^{\circ})\)

Solving for \(\mu\):

\(\mu = \frac{2000 \sin 15^{\circ}}{400g - 2000 \cos 15^{\circ}}\)

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

\(\mu = \frac{2000 \times 0.2588}{400 \times 9.8 - 2000 \times 0.9659}\)

\(\mu = \frac{517.6}{3920 - 1931.8}\)

\(\mu = \frac{517.6}{1988.2}\)

\(\mu \approx 0.26\)

Rounding to 2 significant figures, the coefficient of friction is \(0.25\).

(i) Resolve horizontally:

\(T \cos 60^\circ = 75 \cos 30^\circ\)

\(T \times \frac{1}{2} = 75 \times \frac{\sqrt{3}}{2}\)

\(T = 75 \sqrt{3} \approx 130 \text{ N}\)

Resolve vertically:

\(T \sin 60^\circ + 75 \sin 30^\circ + R = 20g\)

\(130 \times \frac{\sqrt{3}}{2} + 75 \times \frac{1}{2} + R = 200\)

\(R = 200 - 130 \times \frac{\sqrt{3}}{2} - 75 \times \frac{1}{2}\)

\(R \approx 50 \text{ N}\)

(ii) Resolve horizontally:

\(T \cos 60^\circ + 25 = 75 \cos 30^\circ\)

\(T \times \frac{1}{2} + 25 = 75 \times \frac{\sqrt{3}}{2}\)

\(T = 79.9 \text{ N}\)

Resolve vertically:

\(79.9 \sin 60^\circ + 75 \sin 30^\circ + R = 200\)

\(R = 200 - 79.9 \times \frac{\sqrt{3}}{2} - 75 \times \frac{1}{2}\)

\(R \approx 93.3 \text{ N}\)

Coefficient of friction:

\(\mu = \frac{25}{93.3} \approx 0.268\)

(a) In limiting equilibrium, the frictional force is equal to the applied force. The normal reaction \(R\) is given by \(R = 0.2g\).

The frictional force is \(\mu R\), so \(1.2 = \mu \times 0.2g\).

Solving for \(\mu\), we get \(\mu = \frac{1.2}{0.2g} = 0.6\).

(b) Using Newton's Second Law, the net force is \(1.2 - 0.3 \times 0.2g = 0.2a\).

Solving for \(a\), we get \(a = 3 \, \text{m/s}^2\).

To find the distance travelled in the third second, use the equation \(s = ut + \frac{1}{2}at^2\).

For the first 3 seconds, \(s_3 = 0 + \frac{1}{2} \times 3 \times 3^2 = 13.5 \, \text{m}\).

For the first 2 seconds, \(s_2 = 0 + \frac{1}{2} \times 3 \times 2^2 = 6 \, \text{m}\).

Thus, the distance travelled in the third second is \(13.5 - 6 = 7.5 \, \text{m}\).

(i) Using the equation of motion, the acceleration is given by:

\(v = u + at\)

\(3.5 = 0 + 10a\)

\(a = 0.35 \text{ m/s}^2\)

Applying Newton's second law horizontally:

\(10\cos(15^{\circ}) - F = 2 \times 0.35\)

\(F = 10\cos(15^{\circ}) - 0.7\)

\(F = 8.96 \text{ N}\)

(ii) Resolving forces vertically, the normal reaction \(R\) is:

\(R = 2g - 10\sin(15^{\circ})\)

The coefficient of friction \(\mu\) is given by:

\(F = \mu R\)

\(\mu = \frac{8.96}{2g - 10\sin(15^{\circ})}\)

\(\mu = 0.515\)

(i) Resolve the forces horizontally: \(5 \cos \alpha = F\). Given \(\tan \alpha = \frac{3}{4}\), \(\cos \alpha = \frac{4}{5}\). Thus, \(F = 5 \times \frac{4}{5} = 4 \text{ N}\).

Resolve the forces vertically: \(R + 5 \sin \alpha = 8\). Since \(\sin \alpha = \frac{3}{5}\), \(R + 5 \times \frac{3}{5} = 8\), so \(R = 8 - 3 = 5 \text{ N}\).

Using \(F = \mu R\), we have \(4 = 5 \mu\). Therefore, \(\mu = 0.8\).

(ii) Resolve the forces vertically with the new force: \(R + 10 \sin \alpha = 8\). \(R + 10 \times \frac{3}{5} = 8\), so \(R = 8 - 6 = 2 \text{ N}\).

Using \(F = \mu R\), \(F = 0.8 \times 2 = 1.6 \text{ N}\).

Resolve horizontally: \(10 \cos \alpha - F = 0.8a\). \(10 \times \frac{4}{5} - 1.6 = 0.8a\), so \(8 - 1.6 = 0.8a\).

Thus, \(6.4 = 0.8a\) and \(a = 8 \text{ ms}^{-2}\).

(i) Using Newton's second law, the frictional force is given by:

\(F = \mu R = \mu mg\)

where \(\mu\) is the coefficient of friction, \(R\) is the normal reaction, and \(m\) is the mass of the particle. The deceleration \(a\) is given by:

\(\mu mg = ma\)

Thus, \(a = \mu g\).

For particle P, \(\mu = 0.2\):

\(a_P = 0.2 \times 9.8 = 2 \text{ m s}^{-2}\)

For particle Q, \(\mu = 0.25\):

\(a_Q = 0.25 \times 9.8 = 2.5 \text{ m s}^{-2}\)

(ii) Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where \(s_P = s_Q + 5\):

For P:

\(s_P = 8t - \frac{1}{2} \times 2 \times t^2 = 8t - t^2\)

For Q:

\(s_Q = 3t - \frac{1}{2} \times 2.5 \times t^2 = 3t - 1.25t^2\)

Equating and solving:

\(8t - t^2 = 3t - 1.25t^2 + 5\)

\(5t = 0.25t^2 + 5\)

\(t^2 - 5t + 5 = 0\)

Solving for \(t\):

\(t = \sqrt{120} - 10 \approx 0.95445\)

Using \(v = u + at\) for both P and Q:

For P:

\(v_P = 8 - 2 \times 0.95445 = 6.09 \text{ m s}^{-1}\)

For Q:

\(v_Q = 3 - 2.5 \times 0.95445 = 0.614 \text{ m s}^{-1}\)

(i) Resolve the forces in the x-direction:

\(100 \cos 30^{\circ} + 120 \cos 60^{\circ} - F \cos \alpha = 136\)

\(F \cos \alpha = 10.6025\)

Resolve the forces in the y-direction:

\(100 \sin 30^{\circ} - 120 \sin 60^{\circ} + F \sin \alpha = 0\)

\(F \sin \alpha = 53.9230\)

Using the equations:

\(F^2 = (F \cos \alpha)^2 + (F \sin \alpha)^2\)

Calculate \(F\) and \(\alpha\):

\(F = 55.0\) N, \(\alpha = 78.9^{\circ}\)

(ii) The magnitude of the frictional force is equal to the resultant force, which is 136 N. The normal reaction \(R = 40g\), where \(g\) is the acceleration due to gravity.

The coefficient of friction \(\mu = \frac{136}{40g} = 0.34\).

(i) In limiting equilibrium, the frictional force is equal to the tension in the string. Therefore, we have:

\(T = \mu R\)

Given that the tension \(T = 24 \text{ N}\) and the weight \(W = 30 \text{ N}\), the normal reaction \(R = W = 30 \text{ N}\). Thus:

\(24 = \mu \times 30\)

Solving for \(\mu\):

\(\mu = \frac{24}{30} = 0.8\)

(ii) When the block is in motion, resolve the forces vertically and horizontally. The vertical component of the tension is \(25 \sin 30^\circ\) and the horizontal component is \(25 \cos 30^\circ\).

The normal reaction \(R\) is given by:

\(R = 30 - 25 \sin 30^\circ\)

The frictional force \(F\) is:

\(F = \mu R = 0.8 \times (30 - 25 \sin 30^\circ) = 14 \text{ N}\)

Using Newton's second law horizontally:

\(25 \cos 30^\circ - F = \frac{30}{g} a\)

Substitute \(F = 14 \text{ N}\) and solve for \(a\):

\(25 \cos 30^\circ - 14 = \frac{30}{g} a\)

\(a = 2.55 \text{ m/s}^2\)

(i) The frictional force \(F\) required to move the block horizontally is given by \(F = \mu W\), where \(\mu = 1.25\) and \(W\) is the weight of the block. Therefore, \(F = 1.25W\). Since the vertical force required to lift the block is \(W\), and \(F = 1.25W\), the vertical force \(W\) is less than the horizontal force \(F\).

(ii) Using Newton's second law, the net force \(F_{\text{net}} = ma\), where \(m\) is the mass and \(a\) is the acceleration. The mass \(m = \frac{W}{g} = \frac{60}{10} = 6\) kg (assuming \(g = 10\) m/s\(^2\)).

The net force is also given by \(P - F = ma\), where \(F = 1.25 \times 60 = 75\) N is the frictional force.

Thus, \(P - 75 = 6 \times 4\).

Solving for \(P\), we get \(P = 24 + 75 = 99\) N.

(i) Resolve forces in the x-direction:

\(F_x = 6.1 - 5 \times 0.28 = 0\)

Resolve forces in the y-direction:

\(F_y = 4.8 - 5 \times 0.96 = 0\)

The frictional force acts parallel to the x-axis and to the right. Therefore, \(F_y = 0 \Rightarrow F = F_x\). The frictional force has magnitude 7.5 N.

(ii) Using \(F = \mu R\) and \(R = mg\), we have:

\(\mu = \frac{7.5}{1.25 \times 10}\)

The coefficient of friction is 0.6.

(iii) Applying Newton's second law:

\([7.5 - 8.6 - 1.4 = 1.25a \Rightarrow a = -2]\)

The magnitude of acceleration is 2 m/s² and the direction is parallel to the x-axis and to the left.

(i) Use the equation of motion: \(v^2 = u^2 + 2as\).

Substitute \(v = 25 \text{ m s}^{-1}\), \(u = 10 \text{ m s}^{-1}\), and \(s = 525 \text{ m}\):

\(25^2 = 10^2 + 2a \times 525\)

\(625 = 100 + 1050a\)

\(525 = 1050a\)

\(a = \frac{525}{1050} = 0.5 \text{ m s}^{-2}\)

(ii) Apply Newton's second law: \(F = ma\).

The net force \(F\) is given by the driving force minus the resistance: \(900 - R = 1200 \times 0.5\).

\(900 - R = 600\)

\(R = 900 - 600 = 300 \text{ N}\)

(i) Resolve the forces vertically. The normal force \(N\) and the vertical component of the force \(X \cos \theta\) balance the weight of the slab:

\(N + X \cos \theta = mg\)

Given \(\tan \theta = \frac{7}{24}\), we find \(\cos \theta = \frac{24}{25}\).

Substitute \(m = 320 \text{ kg}\) and \(g = 10 \text{ m/s}^2\):

\(N + X \left( \frac{24}{25} \right) = 320 \times 10\)

\(N = 3200 - \frac{24}{25}X\)

(ii) Resolve the forces horizontally. The frictional force \(F\) is given by \(F = X \sin \theta\).

Given \(\sin \theta = \frac{7}{25}\), the frictional force is:

\(F = X \left( \frac{7}{25} \right)\)

The slab is about to slip when \(F = \mu N\), where \(\mu = \frac{3}{8}\):

\(\frac{7}{25}X = \frac{3}{8} \left( 3200 - \frac{24}{25}X \right)\)

Solving for \(X\):

\(\frac{7}{25}X = \frac{3}{8} \times 3200 - \frac{3}{8} \times \frac{24}{25}X\)

\(\frac{7}{25}X + \frac{72}{200}X = 1200\)

\(\frac{175}{625}X + \frac{72}{200}X = 1200\)

\(\frac{187}{625}X = 1200\)

\(X = 1875\)

(i) The frictional force is given by the formula:

\(F = \mu R\)

where \(R = mg\) is the normal reaction force. Thus,

\(F = 0.025 \times 0.15 \times 9.8 = 0.0375 \text{ N}\).

(ii) Using \(F = ma\), we have:

\(-0.0375 = 0.15a\)

Solving for \(a\), we get:

\(a = -0.25 \text{ m/s}^2\).

(iii) Using the equation of motion:

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

\(s = 5.5 \times 4 + \frac{1}{2}(-0.25) \times 16\)

\(s = 20 \text{ m}\).

(iv) Using \(v^2 = u^2 + 2as\), we have:

\(v^2 = 3.5^2 - 2 \times 0.25 \times 20\)

\(v = 1.5 \text{ m/s}\).

(v) The return distance is given by:

\(\frac{3.5^2}{2 \times 0.25}\)

\(Total distance = 24.5 + 20 = 44.5 m.\)

Using Newton's second law for the trailer, we have:

\(300 - 200 = m \times 1.25\)

\(100 = m \times 1.25\)

\(m = \frac{100}{1.25} = 80\)

For the car, using Newton's second law:

\(3200 - F - 300 = 1500 \times 1.25\)

\(2900 - F = 1875\)

\(F = 2900 - 1875 = 1025\)

Thus, the mass of the trailer \(m\) is 80 kg and the resistance force \(F\) is 1025 N.

(a) Apply Newton's second law to the van:

\(2500 - 700 - T = 3600a\)

For the trailer:

\(T - 300 = 1200a\)

For the system of van and trailer:

\(2500 - 700 - 300 = (3600 + 1200)a\)

Simplifying gives:

\(1500 = 4800a\)

Thus, the acceleration \(a\) is:

\(a = \frac{5}{16} = 0.3125\)

Substitute \(a\) back into the equation for the trailer:

\(T - 300 = 1200 \times 0.3125\)

\(T - 300 = 375\)

\(T = 675 \text{ N}\)

(b) For the van with braking force \(F\):

\(-F - 700 = 3600a\)

For the trailer:

\(-300 = 1200a\)

For the system:

\(-F - 700 - 300 = (3600 + 1200)a\)

Solving gives:

\(F = 200 \text{ N}\)

(a) Use the equation of motion: \(s = ut + \frac{1}{2} a t^2\). Given \(u = 0\), \(s = 2\), and \(t = 5\), we have:

\(2 = \frac{1}{2} \times a \times 25\)

Solving for \(a\), we get \(a = 0.16 \text{ ms}^{-2}\).

(b) The normal reaction \(R\) is given by:

\(R = 5g - X \sin 30\)

Using Newton's second law horizontally:

\(X \cos 30 - F = 5a\)

Where \(F = 0.4R\). Substituting \(R\) and \(F\):

\(X \cos 30 - 0.4(5g - X \sin 30) = 5 \times 0.16\)

Solving for \(X\), we find \(X = 19.5 \text{ N (3sf)}\).

(c) For limiting equilibrium, \(R = 5g - 25 \sin 30\). Calculating \(R\):

\(R = 37.5\)

The frictional force \(F\) is:

\(F = 25 \cos 30 = \frac{25 \sqrt{3}}{2}\)

The coefficient of friction \(\mu\) is:

\(\mu = \frac{F}{R} = 0.577 \text{ (3sf)}\)

(a) Using the equation of motion \(s = ut + \frac{1}{2} a t^2\), where \(u = 0\), \(s = 1.44\) m, and \(a = 2\) m/s\(^2\):

\(1.44 = 0 + \frac{1}{2} \times 2 \times t^2\)

\(1.44 = t^2\)

\(t = \sqrt{1.44} = 1.2 \text{ s}\)

(b) Resolve forces vertically to find the normal reaction \(R\):

\(R = 0.4g - 3 \times \frac{3}{5} = 0.4g - 3 \sin 36.9^\circ \approx 2.2 \text{ N}\)

Resolve forces horizontally using Newton's second law:

\(3 \times \frac{4}{5} - F = 3 \cos 36.9^\circ - F = 0.4 \times 2\)

\(F = 1.6 \text{ N}\)

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

\(\mu = \frac{1.6}{2.2} \approx 0.727\)

The crate is moving at a constant speed, so the net force acting on it is zero. The pulling force is balanced by the frictional force.

The frictional force \(F\) is given by \(F = \mu R\), where \(R\) is the normal reaction force.

Since the crate is on horizontal ground, \(R = mg\), where \(m = 500 \text{ kg}\) and \(g = 9.8 \text{ m/s}^2\).

Thus, \(R = 500 \times 9.8 = 4900 \text{ N}\).

The pulling force is 2500 N, so \(2500 = \mu \times 4900\).

Solving for \(\mu\), we get \(\mu = \frac{2500}{4900} = 0.5\).

(i) Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where \(u = 0\), \(s = 4.5\), and \(t = 5\), we have:

\(4.5 = 0 + \frac{1}{2} \times a \times 5^2\)

Solving for \(a\), we get \(a = 0.36 \text{ m/s}^2\).

Resolving horizontally, the net force is \(6 \times \frac{24}{25} - F = 3 \times 0.36\).

Solving for \(F\), we find \(F = 4.68 \text{ N}\).

(ii) The normal reaction \(R\) is given by:

\(R = 3g - 6 \sin 16.3 = 3g - 6 \times \frac{7}{25} = 28.32 \text{ N}\)

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

\(4.68 = \mu \times 28.32\)

Solving for \(\mu\), we get \(\mu = 0.165\).

(iii) The velocity at \(t = 5\) is \(v = 5 \times 0.36 = 1.8 \text{ m/s}\).

Using \(3a = -0.165 \times 3g\), we solve for \(t\) when \(v = 0\):

\(0 = 1.8 - 0.165g t\)

Solving for \(t\), we find \(t = 1.09 \text{ s}\).

The total time is \(5 + 1.09 = 6.09 \text{ s}\).