Past Exam Questions

First, calculate the change in potential energy (PE) as the block moves from A to B:

\(\text{PE change} = 60g \times 17.5 = 10500 \text{ J}\)

Next, calculate the change in kinetic energy (KE):

\(\text{KE change} = \frac{1}{2} \times 60 \times (8.5^2 - 3.5^2) = 1800 \text{ J}\)

The work done (WD) against resistance is:

\(\text{WD against resistance} = 6 \times 250 = 1500 \text{ J}\)

The total work done by the pulling force is the sum of the changes in PE, KE, and the work done against resistance:

\(\text{WD by pulling force} = 10500 + 1800 + 1500 = 13800 \text{ J}\)

Using the formula for work done by the pulling force:

\(\text{WD by pulling force} = 50 \cos \alpha \times 250\)

Equating the two expressions for work done:

\(50 \cos \alpha \times 250 = 13800\)

Solve for \(\cos \alpha\):

\(\cos \alpha = \frac{13800}{12500} = \frac{102}{125}\)

Finally, calculate \(\alpha\):

\(\alpha = \cos^{-1} \left( \frac{102}{125} \right) \approx 35.3^\circ\)

(i) Using Newton's second law and the equation for friction \(F = \mu R\), we have:

\(-\left(\frac{1}{3}\right)(W \cos \alpha) - W \sin \alpha = \left(\frac{W}{g}\right)a\)

Substituting \(\sin \alpha = 0.28\) and \(\cos \alpha = \sqrt{1 - \sin^2 \alpha} \approx 0.96\), we get:

\(-(0.32 - 0.28)g = a\)

\(a = -6\).

(ii) Using the equation \(0 = u^2 + 2as\) where \(u = 5.4 \text{ m s}^{-1}\) and \(a = -6 \text{ m s}^{-2}\):

\(0 = (5.4)^2 + 2(-6)s\)

Solving for \(s\), we find:

\(s = 2.43 \text{ m}\)

(i) Apply Newton's second law to the cyclist and bicycle system. The net force equation along the incline is:

\(F - 780 \times \frac{36}{325} - 32 = 78 \times (-0.2)\)

Solving for \(F\):

\(F = 78 \times (-0.2) + 780 \times \frac{36}{325} + 32\)

\(F = -15.6 + 86.4 + 32\)

\(F = 102.8\)

Rounding to the nearest whole number, \(F = 103\) N.

(ii) Use the equation of motion:

\(0 = 7^2 + 2(-0.2)s\)

\(0 = 49 - 0.4s\)

\(0.4s = 49\)

\(s = \frac{49}{0.4}\)

\(s = 122.5 \text{ m}\)

(i) Resolve forces parallel to the plane: The force equation is \(F = 5.9 - 6.1 \sin \alpha\). The normal reaction is \(R = 6.1 \cos \alpha\). For equilibrium, \(5.9 - 6.1 \sin \alpha \leq \mu (6.1 \cos \alpha)\). Given \(\tan \alpha = \frac{11}{60}\), \(\sin \alpha = \frac{11}{61}\) and \(\cos \alpha = \frac{60}{61}\). Substituting, \(5.9 - 6.1 \times \frac{11}{61} \leq \mu \times 6.1 \times \frac{60}{61}\). Simplifying gives \(\mu \geq \frac{4}{5}\).

(ii) When the force acts down the plane, the net force is \(6.1 \times \frac{11}{61} + 5.9 - \mu \times 6.1 \times \frac{60}{61} > 0\). Simplifying gives \(\mu < \frac{7}{6}\).

(iii) Using Newton's second law, \(6.1 \times \frac{11}{61} + 5.9 - \mu \times 6.1 \times \frac{60}{61} = 6.1 \times 1.7\). Solving gives \(\mu = 0.994\).

(i) Resolve the forces normal to the plane: \(R = mg \cos \alpha\).

Given \(R = 1.2\) N and \(\cos \alpha = 0.96\), we have:

\(1.2 = mg \times 0.96\)

\(mg = \frac{1.2}{0.96} = 1.25\)

\(m = \frac{1.25}{10} = 0.125\) kg

(ii) Use Newton's second law along the plane:

\(-mg \sin \alpha - F = ma\)

\(-0.125 \times 10 \times 0.28 - 0.4 = 0.125a\)

\(-0.35 - 0.4 = 0.125a\)

\(a = -6\)

Deceleration is 6 m/s²

(iii) Compare magnitudes of \(\mu R\) and \(mg \sin \alpha\):

\(\mu R = 0.4\) and \(mg \sin \alpha = 0.35\)

Since \(\mu R > mg \sin \alpha\), the particle remains at rest.

(i) Using Newton's second law along the plane, the equation is:

\(-F - 0.6g\sin 18^{\circ} = 0.6(-4)\)

Solving for the frictional force \(F\):

\(F = 0.6g\sin 18^{\circ} - 0.6 \times 4\)

\(F = 0.546\, \text{N}\)

For the normal component, resolve forces perpendicular to the plane:

\(R = 0.6g\cos 18^{\circ}\)

\(R = 5.71\, \text{N}\)

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

\(\mu = \frac{F}{R} = \frac{0.546}{5.71} = 0.096\)

(ii) When the particle moves down the plane, using Newton's second law:

\(0.6g\sin 18^{\circ} - 0.546 = 0.6a\)

Solving for \(a\):

\(a = \frac{0.6g\sin 18^{\circ} - 0.546}{0.6}\)

\(a = 2.18\, \text{m s}^{-2}\)

(i) The normal force R is given by:

\(R = mg \cos 21^{\circ} = 9.336m\)

Using Newton's second law for motion up the plane:

\(a = -5\)

The equation of motion is:

\(-mg \sin 21^{\circ} - F = -5m\)

Solving for F gives:

\(F = 1.416m\)

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

\(\mu = \frac{F}{R} = \frac{1.416m}{9.336m} = 0.152\)

(iii) Using the equation of motion for the particle moving down the plane:

\(m g \sin 21^{\circ} - 1.416m = ma\)

Solving for acceleration \(a\):

\(a = 2.167 \text{ m/s}^2\)

Using the equation \(v^2 = 2as\) with \(s = 10\):

\(v^2 = 2(2.167)(10)\)

Solving for \(v\):

\(v = 6.58 \text{ m/s}\)

(a) In limiting equilibrium, the frictional force \(F\) is equal to \(\mu R\), where \(R\) is the normal reaction force. The forces acting on the particle are its weight \(0.4g\) and the normal reaction \(R\). The component of the weight perpendicular to the plane is \(0.4g \cos 30^{\circ}\), and the component parallel to the plane is \(0.4g \sin 30^{\circ}\).

Since the particle is in limiting equilibrium, \(F = \mu R = 0.4g \sin 30^{\circ}\).

The normal reaction \(R = 0.4g \cos 30^{\circ}\).

Thus, \(\mu = \frac{0.4g \sin 30^{\circ}}{0.4g \cos 30^{\circ}} = \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = \tan 30^{\circ} = \frac{1}{3} \sqrt{3}\).

(b) Applying Newton's second law along the plane: \(7.2 - 0.4g \sin 30^{\circ} - F = 0.4a\).

Substituting \(F = \mu R = \frac{1}{3} \sqrt{3} \times 0.4g \cos 30^{\circ}\), we find \(a = 8 \text{ m/s}^2\).

Using the equation of motion \(s = ut + \frac{1}{2}at^2\), where \(s = 1 \text{ m}\), \(u = 0\), and \(a = 8 \text{ m/s}^2\), we solve for \(t\):

\(1 = 0 + \frac{1}{2} \times 8 \times t^2\)

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

\(t = 0.5 \text{ s}\).

First, use the equation of motion to find the acceleration:

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

\(2.25 = \frac{1}{2} a (1.5^2)\)

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

Next, calculate the normal reaction force \(R\):

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

Apply Newton's second law along the plane:

\(mg \sin 30^{\circ} - \mu mg \cos 30^{\circ} = ma\)

\(mg \sin 30^{\circ} - \mu mg \cos 30^{\circ} = 2m\)

Solving for \(\mu\):

\(\mu = \frac{mg \sin 30^{\circ} - 2m}{mg \cos 30^{\circ}}\)

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

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

\(\mu = \frac{9.8 \times 0.5 - 2}{9.8 \times \frac{\sqrt{3}}{2}}\)

\(\mu = 0.346\)

(i) The potential energy gain is given by:

\(PE = mg(2.5 \sin 60^{\circ})\)

Using kinetic energy: \(KE = \frac{1}{2} mv^2\)

By conservation of energy:

\(\frac{1}{2} m (8^2 - v^2) = mg(2.5 \sin 60^{\circ})\)

Solving for \(v\), we find \(v = 4.55 \text{ m s}^{-1}\).

(ii) Using \(\frac{1}{2} mu^2 > mgh_{\text{max}}\),

\(\frac{1}{2} \times 8^2 > 10h_{\text{max}}\)

This shows \(h_{\text{max}} < 3.2 \text{ m}\).

(iii) Since \(BC\) is smooth and \(B\) and \(C\) are at the same height, energy is conserved, so speed at \(C\) is the same as at \(B\).

(iv) Work done against friction is \(1.4 \times 5.2\).

Using work-energy principle:

\(1.4 \times 5.2 = \frac{1}{2} \times 0.4 (v^2 - 4.55^2)\)

Solving for \(v\), we find \(v = 5.25 \text{ m s}^{-1}\).

(a) Resolve forces parallel to the plane using Newton's second law:

\(T \cos 26^{\circ} - 8g \sin 18^{\circ} - F = 8 \times 0.2\)

Resolve forces perpendicular to the plane:

\(R + T \sin 26^{\circ} = 8g \cos 18^{\circ}\)

Use the friction equation \(F = 0.65R\) to find an equation in \(T\) only.

Solve for \(T\):

\(T = 64 \text{ N}\)

(b) Use the constant acceleration formula to find the distance \(s\) during the fourth second:

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

or

\(s = \frac{1}{2} (0 + 0.2 \times 4) \times 4 - \frac{1}{2} (0 + 0.2 \times 3) \times 3\)

\(Distance = 0.7 m\)

(a) Resolve forces perpendicular to the plane: \(R = 0.3g \cos \theta + 4 \sin \theta = 3 \times \frac{24}{25} + 4 \times \frac{7}{25}\).

Resolve forces parallel to the plane: \(F = 4 \cos \theta - 0.3g \sin \theta = 4 \times \frac{24}{25} - 3 \times \frac{7}{25}\).

Using \(F = \mu R\), solve for \(\mu\): \(3 = \mu \times 4\), hence \(\mu = \frac{3}{4}\).

(b) Use Newton's second law: \(F = \mu \times 0.3g \cos \theta = \frac{3}{4} \times 0.3 \times \frac{24}{25}\).

Net force: \(4 - \frac{3}{4} \times 0.3 \times \frac{24}{25} - 0.3 \times g \times \frac{7}{25} = 0.3a\).

Solve for \(a\): \(a = \frac{10}{3} \text{ m/s}^2\).

(c) Distance in 3 seconds: \(s_1 = \frac{1}{2} \times \frac{10}{3} \times 3^2 = 15 \text{ m}\).

Velocity after 3 seconds: \(v = \frac{10}{3} \times 3 = 10 \text{ m/s}\).

After force is removed, apply Newton's second law: \(-0.3g \sin \theta - \mu \times 0.3g \cos \theta = 0.3a\), leading to \(a = -10 \text{ m/s}^2\).

Use \(v^2 = u^2 + 2as\) to find extra distance \(s_2\): \(0 = 10^2 + 2 \times (-10) \times s_2\), so \(s_2 = 5 \text{ m}\).

Total distance: \(s_1 + s_2 = 15 + 5 = 20 \text{ m}\).

For the car moving up the hill, apply Newton's second law:

\(1200 - 350 - 1250 \times 10 \times 0.05 = 1250a\)

\(a = \frac{225}{1250} = 0.18 \text{ m/s}^2\)

Using the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 100\):

\(v^2 = 0 + 2 \times 0.18 \times 100\)

\(v = 6 \text{ m/s}\)

For the car moving down the hill, apply Newton's second law:

\(1200 - 350 + 1250 \times 10 \times 0.05 = 1250a\)

\(a = \frac{1475}{1250} = 1.18 \text{ m/s}^2\)

Using the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(s = 100\):

\(v^2 = 0 + 2 \times 1.18 \times 100\)

\(v = 15.4 \text{ m/s}\)

First, calculate the normal reaction force, \(R\), on the block. Since the block is on an inclined plane, \(R = 5g \cos 6^{\circ}\).

Next, calculate the frictional force, \(F\), using the formula \(F = \mu R\), where \(\mu = 0.3\). Thus, \(F = 0.3 \times 5g \cos 6^{\circ}\).

Since the block is moving at constant speed, the tension \(T\) in the rope must balance the component of gravitational force down the plane and the frictional force. Therefore, \(T = 5g \sin 6^{\circ} + F\).

Substitute the values to find \(T\):

\(T = 5g \sin 6^{\circ} + 0.3 \times 5g \cos 6^{\circ}\).

Calculating this gives \(T \approx 20.1 \text{ N}\).

(i) To find the friction force, we consider the component of the gravitational force acting down the slope. The gravitational force component is given by:

\(F = mg \sin \theta\)

where \(m = 0.2\) kg, \(g = 9.8\) m/s², and \(\theta = 20^{\circ}\).

Substituting the values, we get:

\(F = 0.2 \times 9.8 \times \sin 20^{\circ} = 0.684\) N

Thus, the friction force is 0.684 N.

(ii) To find the acceleration, we first calculate the normal reaction force \(R\):

\(R = mg \cos \theta = 0.2 \times 9.8 \times \cos 20^{\circ}\)

\(R = 0.2 \times 9.8 \times \cos 20^{\circ} \approx 1.84\) N

The frictional force \(F_f\) is given by:

\(F_f = \mu R = 0.6 \times 1.84 = 1.104\) N

Using Newton's second law along the plane:

\(0.9 + 0.2 \times 9.8 \times \sin 20^{\circ} - F_f = 0.2a\)

\(0.9 + 0.684 - 1.104 = 0.2a\)

\(0.48 = 0.2a\)

\(a = \frac{0.48}{0.2} = 2.4\) m/s²

However, according to the mark scheme, the correct acceleration is 2.28 m/s².

(i) First, calculate the normal reaction force, \(R\), using:

\(R = 0.8g \cos 10^{\circ} \approx 7.88 \text{ N}\)

The frictional force, \(F\), is given by:

\(F = \mu R = 0.4 \times 0.8g \cos 10^{\circ} \approx 3.15 \text{ N}\)

Using Newton's second law along the plane:

\(-0.8g \sin 10^{\circ} - F = 0.8a\)

\(-0.8g \sin 10^{\circ} - 3.15 = 0.8a\)

Solving for \(a\):

\(a = \frac{-0.8g \sin 10^{\circ} - 3.15}{0.8} \approx -5.68 \text{ m s}^{-2}\)

(ii) Using the equation of motion \(v^2 = u^2 + 2as\) with final velocity \(v = 0\), initial velocity \(u = 12 \text{ m s}^{-1}\), and acceleration \(a = -5.68 \text{ m s}^{-2}\):

\(0 = 12^2 + 2(-5.68)s\)

\(s = \frac{144}{2 \times 5.68} \approx 12.7 \text{ m}\)

(i) Using Newton's second law for the system:

\(2500 - 2000g \times 0.1 - 250 = 2000a\)

\(a = \frac{1}{8} = 0.125 \text{ m/s}^2\)

For tension \(T\):

\(2500 - T - 100 - 1200g \times 0.1 = 1200 \times 0.125\)

or

\(T - 150 - 800g \times 0.1 = 800 \times 0.125\)

Solving gives \(T = 1050 \text{ N}\).

(ii) With no driving force, using Newton's second law:

\(-2000g \times 0.1 - 250 = 2000a\)

\(a = -1.125 \text{ m/s}^2\)

Using \(v = u + at\):

\(0 = 30 - 1.125t\)

\(t = 26.7 \text{ s}\)

(a) The diagram should show three forces: the gravitational force acting downwards, the normal force perpendicular to the plane, and the frictional force opposing the motion.

(b) Resolve forces perpendicular to the plane: \(R = 8g \cos 30^{\circ} = 40\sqrt{3} \approx 69.282\). Resolve forces parallel to the plane and apply Newton's second law: \(8g \sin 30^{\circ} - F = 8 \times 2.4\). This gives \(F = 20.8\). Use \(F = \mu R\) to find \(\mu\):

\(8g \sin 30^{\circ} - 8g \mu \cos 30^{\circ} = 8 \times 2.4\)

\(40 - 40\sqrt{3} \mu = 19.2\)

\(\mu = \frac{20.8}{40\sqrt{3}} \approx 0.3\)

(c) Use the equation \(v^2 = u^2 + 2as\) with \(u = 0\), \(a = 2.4\), and \(s = 3\):

\(v^2 = 2 \times 2.4 \times 3\)

\(v = \sqrt{14.4} \approx 3.79 \text{ m/s}\)

(i) The acceleration \(a\) down the plane is given by \(a = g \sin \alpha = g \times \frac{16}{65}\).

Using the equation \(v^2 = u^2 + 2as\), where \(u = 0\), \(v = 8\), and \(s = S\), we have:

\(8^2 = 2 \times \frac{16}{65}g \times S\)

\(64 = \frac{32}{65}gS\)

\(S = \frac{64 \times 65}{32g} = 13\) m

To find the speed when \(s = \frac{1}{2}S = 6.5\) m, use \(v^2 = 2as\):

\(v^2 = 2 \times \frac{16}{65}g \times 6.5\)

\(v = \sqrt{\frac{32}{65}g \times 6.5} = 5.66\) m s^-1

(ii) The time \(T\) to travel distance \(S\) is given by \(v = u + at\), so \(8 = 0 + \frac{16}{65}gT\).

\(T = \frac{8 \times 65}{16g}\)

For \(\frac{1}{2}T\), the distance \(s\) is given by \(s = \frac{1}{2}a(\frac{T}{2})^2\):

\(s = \frac{1}{2} \times \frac{16}{65}g \times (\frac{8 \times 65}{32g})^2\)

\(s = 3.25\) m

(i) The acceleration for \(t < 0.8\) is calculated using the formula \(a = \frac{v}{t} = \frac{4}{0.8} = 5 \text{ m s}^{-2}\). Since the plane is smooth between A and B, the acceleration is due to gravity: \(a = g \sin \theta\). Therefore, \(5 = 10 \sin \theta\), giving \(\sin \theta = 0.5\). Thus, \(\theta = 30^\circ\).

(ii) For \(0.8 < t < 4.8\), the acceleration is \(\frac{-4}{4.8 - 0.8} = -1 \text{ m s}^{-2}\). Using Newton's second law, \(mg \sin 30^\circ - F = m(-1)\), where \(F\) is the frictional force. The frictional force \(F = \mu R\), where \(R = mg \cos 30^\circ\). Therefore, \(\mu = \frac{mg \sin 30^\circ + m}{mg \cos 30^\circ}\). Simplifying gives \(\mu = 0.693\).