Past Exam Questions

(i) For particle P, using the equation of motion under gravity:

\(0 = V - gt\)

\(At t = 2 s, the velocity is zero, so:\)

\(0 = V - g \times 2\)

Solving for V gives:

\(V = 2g\)

\(Assuming g = 10 m s^-2, we find:\)

\(V = 20 \text{ m s}^{-1}\)

(ii) For particle Q, using the equation of motion:

\(v = u + gt\)

\(Since Q starts from rest, u = 0, and at t = 4 s:\)

\(v = 0 + 10 \times 4 = 40 \text{ m s}^{-1}\)

(iii) The height h from which Q was dropped can be found using:

\(h = \frac{1}{2}gt^2\)

\(Substituting t = 4 s:\)

\(h = \frac{1}{2} \times 10 \times 4^2 = 80 \text{ m}\)

(i) Using the equation of motion \(v^2 = u^2 + 2as\) where initial velocity \(u = 0\), acceleration \(a = 10 \text{ m/s}^2\), and distance \(s = 1.25 \text{ m}\):

\(v^2 = 2 \times 10 \times 1.25\)

\(v^2 = 25\)

\(v = 5 \text{ m/s}\)

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

\(1.25 = \frac{1}{2} \times 10 \times t^2\)

\(t^2 = 0.25\)

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

(ii) The velocity after passing through A is given by integrating the acceleration:

\(v = \int (10 - 0.3t) \, dt = 10t - 0.15t^2 + C\)

At \(t = 0\), \(v = 5 \text{ m/s}\), so \(C = 5\).

\(v = 10t - 0.15t^2 + 5\)

Integrating to find the distance:

\(x = \int (10t - 0.15t^2 + 5) \, dt = 5t^2 - 0.05t^3 + 5t\)

Substitute \(t = 2.5\) (since \(t = 3 - 0.5\)):

\(x = 5 \times 2.5^2 - 0.05 \times 2.5^3 + 5 \times 2.5\)

\(x = 42.97 \text{ m}\)

Total distance \(OP = 1.25 + 42.97 = 44.2 \text{ m}\)

Let the upward direction be positive. For the first particle projected upwards:

Using the equation of motion: \(s_1 = ut - \frac{1}{2}gt^2\)

\(s_1 = 12.5t - \frac{1}{2}gt^2\)

For the second particle released from rest:

Using the equation of motion: \(s_2 = \frac{1}{2}gt^2\)

Since the total distance between the particles is 10 m, we have:

\(s_1 + s_2 = 10\)

Substituting the expressions for \(s_1\) and \(s_2\):

\(12.5t - \frac{1}{2}gt^2 + \frac{1}{2}gt^2 = 10\)

\(12.5t = 10\)

\(t = 0.8 \text{ s}\)

Substitute \(t = 0.8\) back into the equation for \(s_1\):

\(s_1 = 12.5 \times 0.8 - \frac{1}{2} \times 9.8 \times (0.8)^2\)

\(s_1 = 10 - 3.2\)

\(s_1 = 6.8 \text{ m}\)

Therefore, the height above O at which the particles meet is 6.8 m.

(i) For \(P_1\), using \(s = ut - \frac{1}{2}gt^2\), we have:

\(25 = 30t - 5t^2\)

Solving \(5t^2 - 30t + 25 = 0\) gives \((t-1)(t-5) = 0\), so \(t = 1\) or \(t = 5\).

Thus, \(P_1\) is higher than the tower for \(1 < t < 5\).

(ii) For \(P_1\) and \(P_2\), using \(s_1 = 30t - 5t^2\) and \(s_2 = 10t - 5t^2\), and \(s_1 = s_2 + 25\):

\(30t - 10t = 25\)

\(t = 1.25\)

Velocities: \(v_1 = 30 - 10 \times 1.25 = 17.5\) m s\(^{-1}\), \(v_2 = 10 - 10 \times 1.25 = -2.5\) m s\(^{-1}\).

(iii) Using \(t_{up} = 3\) and \(t_{equal} = 1.25\), the time is \(1.75\) s or \(1.25 < t < 3\).

1. To find the speed after 2 s, use the formula for velocity under constant acceleration: \(v = u + gt\). Since the stone is released from rest, \(u = 0\). Thus, \(v = 0 + 10 \times 2 = 20 \text{ m s}^{-1}\).
2. To find the time taken to fall 45 m, use the formula \(s = \frac{1}{2}gt^2\). Solving \(45 = \frac{1}{2} \times 10 \times t^2\) gives \(t^2 = 9\), so \(t = 3 \text{ s}\).
3. To find the distance fallen from 30 m s^-1 to 40 m s^-1, use \(v^2 = u^2 + 2gs\). Substituting \(40^2 = 30^2 + 2 \times 10 \times s\), we get \(1600 = 900 + 20s\). Solving gives \(s = 35 \text{ m}\).

(i) To find the maximum height of A, use the equation for vertical motion:

\(v = u - gt\)

where \(v = 0\) at maximum height, \(u = 5 \text{ m/s}\), and \(g = 9.8 \text{ m/s}^2\). Solving \(0 = 5 - 9.8t\) gives \(t = \frac{5}{9.8}\).

The height \(h\) is given by:

\(h = ut - \frac{1}{2}gt^2\)

Substitute \(t = \frac{5}{9.8}\) to find \(h_A\).

For B, use the same formula with \(u = 8 \text{ m/s}\) and the same \(t\). Calculate \(h_B\).

The difference in heights is \(h_B - h_A = 1.5 \text{ m}\).

(ii) To find when B is 0.9 m above A, solve:

\(h_B - h_A = 0.9\)

Using \(h = ut - \frac{1}{2}gt^2\) for both particles, set up the equation:

\(8t - \frac{1}{2}gt^2 - (5t - \frac{1}{2}gt^2) = 0.9\)

Simplify to \(3t = 0.9\), giving \(t = 0.3 \text{ s}\).

Substitute \(t = 0.3\) into \(h_A = 5t - \frac{1}{2}gt^2\) to find \(h_A = 1.05 \text{ m}\).

(a) To find the initial speed, use the equation of motion:

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

At the maximum height, \(v = 0\), \(a = -g\), and \(s = 45\) m. Thus,

\(0 = u^2 - 2g \times 45\)

Solving for \(u\), we get:

\(u^2 = 2g \times 45\)

\(u = \sqrt{2 \times 9.8 \times 45} = 30 \text{ m s}^{-1}\)

(b) To find the total time for which the speed is at least 10 m s^-1, use the equation:

\(v = u + at\)

Set \(v = 10 \text{ m s}^{-1}\), \(u = 30 \text{ m s}^{-1}\), and \(a = -g\):

\(10 = 30 - gt\)

\(gt = 20\)

\(t = \frac{20}{9.8} \approx 2 \text{ s}\)

The particle takes 2 s to reach 10 m s^-1 on the way up and another 2 s on the way down, so the total time is:

\(2 \times 2 = 4 \text{ s}\)

(a) To find the initial speed u, use the equation for maximum height:

\(0 = u^2 - 2 \times 10 \times 20\)

Solving for \(u\), we get:

\(u^2 = 400\)

\(u = 20 \text{ m/s}\)

(b) To find the total time for which P is at least 15 m above the ground, use the equation:

\(15 = 20t - \frac{1}{2} \times 10 \times t^2\)

Rearranging gives:

\(5t^2 - 20t + 15 = 0\)

Solving this quadratic equation, we find:

\(t = 1 \text{ or } t = 3\)

The total time is \(3 - 1 = 2 \text{ s}\).

(a) Using the equation of motion: \(s = ut + \frac{1}{2} a t^2\), where \(s = 24 \text{ m}\), \(t = 2 \text{ s}\), and \(a = -g = -9.8 \text{ m s}^{-2}\).

\(24 = u \times 2 - \frac{1}{2} \times 9.8 \times 2^2\)

\(24 = 2u - 19.6\)

\(2u = 43.6\)

\(u = 22 \text{ m s}^{-1}\)

(b) At maximum height, \(v = 0\), using \(v^2 = u^2 + 2as\):

\(0 = 22^2 - 2 \times 9.8 \times s\)

\(s = 24.2 \text{ m}\)

Height down in 1.8 s: \(s = \frac{1}{2} \times 9.8 \times 1.8^2 = 16.2 \text{ m}\)

\(h = 24.2 - 16.2 = 8 \text{ m}\)

Alternative method: Using \(v = u - gt\) with \(u = 22\) and \(v = 0\):

\(0 = 22 - 9.8t\)

\(t = 2.2 \text{ s}\)

\(h = 22 \times (2.2 - 1.8) - \frac{1}{2} \times 9.8 \times (2.2 - 1.8)^2\)

\(h = 8 \text{ m}\)

(a) To find the initial velocity v, we use the equation of motion:

\(v = u + at\)

where \(v = 0\) (velocity at the greatest height), \(a = -g = -10 \text{ m/s}^2\) (acceleration due to gravity), and \(t = 3 \text{ s}\).

Substituting the values, we get:

\(0 = v - 10 \times 3\)

\(v = 30 \text{ m/s}\)

(b) To find the greatest height, we use the equation:

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

where \(v = 0\), \(u = 30 \text{ m/s}\), and \(a = -10 \text{ m/s}^2\).

Substituting the values, we get:

\(0 = 30^2 + 2(-10)s\)

\(0 = 900 - 20s\)

\(20s = 900\)

\(s = 45 \text{ m}\)

Thus, the greatest height is 45 m.

(a) To find the height h of the building, we use the equation of motion for the first particle:

Using the equation for velocity, \(v = u + at\), with \(u = 40\), \(v = 0\), and \(a = -g\), we find the time to reach the highest point:

\(40 - gt = 0\) gives \(t = 4\) seconds.

The time to reach the top of the building is \(4 - \frac{1}{2} \times 10 \times 2^2 = 2\) seconds.

Using the equation for displacement, \(s = ut + \frac{1}{2} at^2\), with \(u = 40\), \(a = -g\), and \(t = 2\), we find:

\(h = 40 \times 2 - \frac{1}{2} \times 10 \times 2^2 = 60\) m.

(b) For the first particle, the height above ground is given by:

\(40t - \frac{1}{2} \times 10t^2\).

For the second particle, the height above the top of the building is:

\(20(t - 1) - \frac{1}{2} \times 10 \times (t - 1)^2\).

Setting the heights equal gives:

\(60 + 20(t - 1) - \frac{1}{2} \times 10 \times (t - 1)^2 = 40t - \frac{1}{2} \times 10t^2\).

Solving this equation, we find \(t = 3.5\) seconds.

(a) To find the greatest height above the ground reached by P, we use the equation for vertical motion:

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

where \(v = 0\) m/s (at the highest point), \(u = 5\) m/s, and \(g = 9.8\) m/s².

\(0 = 5^2 - 2 \times 9.8 \times s\)

\(s = \frac{25}{19.6} = 1.25\) m

The height above the ground is \(2.8 + 1.25 = 4.05\) m.

(b) To find the length of time for which P is at a height of more than 3.6 m above the ground, we use the equation:

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

Let \(s = 3.6 - 2.8 = 0.8\) m, \(u = 5\) m/s, \(a = -9.8\) m/s².

\(0.8 = 5t - 4.9t^2\)

Solving the quadratic equation \(4.9t^2 - 5t + 0.8 = 0\), we find \(t = 0.2\) s and \(t = 0.8\) s.

The length of time for which P is above 3.6 m is \(0.8 - 0.2 = 0.6\) s.

(a) Resolve the forces horizontally and vertically. For equilibrium:

Horizontal: \(T \sin \theta = 32\)

Vertical: \(T \cos \theta = 80\)

Divide the horizontal equation by the vertical equation:

\(\tan \theta = \frac{32}{80}\)

\(\theta = \arctan \left( \frac{32}{80} \right) \approx 21.8^{\circ}\)

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

\(T = \frac{80}{\cos \theta} \approx 86.2 \text{ N}\)

(b) Given \(T = 120 \text{ N}\), resolve the forces:

Horizontal: \(120 \sin \theta = P\)

Vertical: \(120 \cos \theta = 80\)

Divide the vertical equation by \(T\):

\(\cos \theta = \frac{80}{120}\)

\(\theta = \cos^{-1} \left( \frac{80}{120} \right) \approx 48.2^{\circ}\)

Substitute \(\theta\) back to find \(P\):

\(P = 120 \sin \theta \approx 89.4 \text{ N}\)

To solve for θ and T, we resolve the forces vertically and horizontally.

Vertically:

\(T \sin \theta + 120 \sin 45^{\circ} = 15g\)

Horizontally:

\(T \cos \theta = 120 \cos 45^{\circ}\)

Using the vertical equation:

\(T \sin \theta = 15g - 120 \sin 45^{\circ}\)

Using the horizontal equation:

\(T = \frac{120 \cos 45^{\circ}}{\cos \theta}\)

Substitute for \(T\) in the vertical equation:

\(\frac{120 \cos 45^{\circ}}{\cos \theta} \sin \theta = 15g - 120 \sin 45^{\circ}\)

Using trigonometric identities and solving for \(\theta\):

\(\tan \theta = \frac{15g - 120 \sin 45^{\circ}}{120 \cos 45^{\circ}}\)

Calculate \(\theta\):

\(\theta = 37.5^{\circ}\)

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

\(T = \sqrt{(15g \sin \theta)^2 + (120 \cos 45^{\circ})^2}\)

Calculate \(T\):

\(T = 107 \text{ N}\)

To solve for \(A\) and \(B\), we can resolve the forces horizontally and vertically at the 25 N weight.

1. Resolve vertically: \(A \cos 30^\circ + B \cos 40^\circ = 25\)

2. Resolve horizontally: \(A \sin 30^\circ = B \sin 40^\circ\)

From the horizontal resolution, we have:

\(A \sin 30^\circ = B \sin 40^\circ\)

\(A \cdot 0.5 = B \cdot 0.6428\)

\(A = \frac{B \cdot 0.6428}{0.5}\)

\(A = 1.2856B\)

Substitute \(A = 1.2856B\) into the vertical resolution equation:

\(1.2856B \cdot 0.8660 + B \cdot 0.7660 = 25\)

\(1.112B + 0.766B = 25\)

\(1.878B = 25\)

\(B = \frac{25}{1.878} \approx 13.3\)

Substitute \(B = 13.3\) back to find \(A\):

\(A = 1.2856 \times 13.3 \approx 17.1\)

Thus, \(A = 17.1\) and \(B = 13.3\).

Resolve the forces vertically and horizontally. Let \(T_A\) and \(T_B\) be the tensions in the strings at angles 20° and 40° respectively.

Vertically, the sum of the vertical components of the tensions must equal the weight of the particle:

\(T_A \sin 20^{\circ} + T_B \sin 40^{\circ} = 16\)

Horizontally, the horizontal components must balance:

\(T_A \cos 20^{\circ} = T_B \cos 40^{\circ}\)

Solving these equations simultaneously:

From the horizontal equation, express \(T_A\) in terms of \(T_B\):

\(T_A = T_B \frac{\cos 40^{\circ}}{\cos 20^{\circ}}\)

Substitute into the vertical equation:

\(T_B \frac{\cos 40^{\circ}}{\cos 20^{\circ}} \sin 20^{\circ} + T_B \sin 40^{\circ} = 16\)

Simplify and solve for \(T_B\):

\(T_B (\frac{\cos 40^{\circ} \sin 20^{\circ}}{\cos 20^{\circ}} + \sin 40^{\circ}) = 16\)

\(T_B = \frac{16}{\frac{\cos 40^{\circ} \sin 20^{\circ}}{\cos 20^{\circ}} + \sin 40^{\circ}} \approx 17.4 \text{ N}\)

Substitute \(T_B\) back to find \(T_A\):

\(T_A = 17.4 \frac{\cos 40^{\circ}}{\cos 20^{\circ}} \approx 14.2 \text{ N}\)

To solve for the tensions in the ropes, we resolve the forces horizontally and vertically.

Let the tension in rope PA be \(T_A\) and in rope PB be \(T_B\).

Resolving horizontally:

\(T_A \cos 50^{\circ} - T_B \cos 10^{\circ} = 0\)

Resolving vertically:

\(T_A \sin 50^{\circ} + T_B \sin 10^{\circ} = 20g\)

Solving these equations simultaneously:

From the horizontal resolution, \(T_A \cos 50^{\circ} = T_B \cos 10^{\circ}\).

Substitute \(T_A = \frac{T_B \cos 10^{\circ}}{\cos 50^{\circ}}\) into the vertical resolution:

\(\frac{T_B \cos 10^{\circ}}{\cos 50^{\circ}} \sin 50^{\circ} + T_B \sin 10^{\circ} = 20g\)

\(T_B \left( \frac{\cos 10^{\circ} \sin 50^{\circ}}{\cos 50^{\circ}} + \sin 10^{\circ} \right) = 20g\)

Solving for \(T_B\):

\(T_B = \frac{20g}{\frac{\cos 10^{\circ} \sin 50^{\circ}}{\cos 50^{\circ}} + \sin 10^{\circ}}\)

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

Substitute \(T_B\) back to find \(T_A\):

\(T_A = \frac{200 \cos 10^{\circ}}{\cos 50^{\circ}}\)

\(T_A = 306 \text{ N}\)

Alternatively, using Lami's Theorem:

\(\frac{T_A}{\sin 80^{\circ}} = \frac{T_B}{\sin 140^{\circ}} = \frac{20g}{\sin 140^{\circ}}\)

Solving gives the same tensions: \(T_A = 306 \text{ N}\) and \(T_B = 200 \text{ N}\).

(i) Resolve the forces acting at O vertically:

\(W \cos \alpha + 7 \times 0.6 = 8\)

\(W \cos \alpha = 3.8\)

Using the Pythagorean identity, \(W^2 = (W \sin \alpha)^2 + (W \cos \alpha)^2\), we find:

\(W \sin \alpha = 5.6\)

(ii) Using \(W^2 = (W \sin \alpha)^2 + (W \cos \alpha)^2\):

\(W^2 = 5.6^2 + 3.8^2\)

\(W = \sqrt{31.36 + 14.44} = \sqrt{45.8} \approx 6.77\)

Using \(\tan \alpha = \frac{W \sin \alpha}{W \cos \alpha}\):

\(\tan \alpha = \frac{5.6}{3.8}\)

\(\alpha = \arctan\left(\frac{5.6}{3.8}\right) \approx 55.8^\circ\)

To find the tensions in the strings, we resolve the forces horizontally and vertically.

For horizontal equilibrium:

\(0.8T_1 + \frac{12}{13}T_2 = 2.24\)

For vertical equilibrium:

\(0.6T_1 - \frac{5}{13}T_2 = 1.4\)

Solving these two equations simultaneously, we find:

\(T_1 = 2.5 \text{ N}\)

\(T_2 = 0.26 \text{ N}\)

To solve for the tensions in the strings, we resolve the forces acting on P both horizontally and vertically.

1. **Horizontal Resolution:**

The horizontal component of the tension in the string attached to A is \(T_1 \cos 36.9^ ext{o}\) and for the string attached to B is \(T_2 \cos 16.3^ ext{o}\). The horizontal force is 10 N.

Thus, the equation is:

\(T_1 \cos 36.9^ ext{o} + T_2 \cos 16.3^ ext{o} = 10\)

2. **Vertical Resolution:**

The vertical component of the tension in the string attached to A is \(T_1 \sin 36.9^ ext{o}\) and for the string attached to B is \(T_2 \sin 16.3^ ext{o}\). The weight of P is \(0.7g\) N.

Thus, the equation is:

\(T_1 \sin 36.9^ ext{o} - T_2 \sin 16.3^ ext{o} = 0.7g\)

3. **Solving the Equations:**

Using the given values:

\(0.8T_1 + 0.96T_2 = 10\)

\(0.6T_1 - 0.28T_2 = 0.7g\)

Solving these simultaneous equations gives:

\(T_1 = 11.9\) N and \(T_2 = 0.5\) N.