Past Exam Questions

(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.

To find the tensions in the strings, we resolve the forces acting on the particle P vertically and horizontally.

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

\(T_A \frac{1}{2.6} + T_B \frac{1}{1.25} = 10.5\)

Horizontally, the horizontal components of the tensions must balance each other:

\(T_A \frac{2.4}{2.6} = T_B \frac{0.75}{1.25}\)

Solving these equations simultaneously gives:

\(T_A = 6.5 \text{ N}\)

\(T_B = 10 \text{ N}\)

Given that \(\tan \alpha = \frac{8}{15}\), we can find \(\cos \alpha\) and \(\sin \alpha\) using the identity \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\).

Let \(\sin \alpha = \frac{8}{17}\) and \(\cos \alpha = \frac{15}{17}\) based on the Pythagorean identity \(\sin^2 \alpha + \cos^2 \alpha = 1\).

Resolve forces vertically: \(T \cos \alpha = mg\).

Substitute \(\cos \alpha = \frac{15}{17}\) and \(mg = 0.3 \times 9.8 = 2.94\) N:

\(T \times \frac{15}{17} = 2.94\).

\(T = \frac{2.94 \times 17}{15} = 3.4\) N.

Resolve forces horizontally: \(F = T \sin \alpha\).

Substitute \(T = 3.4\) N and \(\sin \alpha = \frac{8}{17}\):

\(F = 3.4 \times \frac{8}{17} = 1.6\) N.

To solve for the tensions in the strings, we resolve the forces acting on P horizontally and vertically. Let the tension in PA be T_A and in PB be T_B.

Using the geometry of the triangle formed by the strings and the vertical line, we have:

\(T_A \times \frac{40}{50} + T_B \times \frac{40}{104} = 21\)

Solving for T_A and T_B:

\(T_A \times \frac{30}{50} = T_B \times \frac{96}{104}\)

From the first equation, we find:

\(T_A = 20 \text{ N}\)

Substituting T_A into the second equation gives:

\(T_B = 13 \text{ N}\)

Thus, the tension in AP is 20 N and the tension in BP is 13 N.

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

1. Resolve horizontally:

\(T_1 \cos 35^{\circ} = T_2 \cos 40^{\circ}\)

2. Resolve vertically:

\(T_1 \sin 35^{\circ} + T_2 \sin 40^{\circ} = 2.4g\)

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

3. Solve the system of equations:

From the horizontal resolution, express \(T_2\) in terms of \(T_1\):

\(T_2 = \frac{T_1 \cos 35^{\circ}}{\cos 40^{\circ}}\)

Substitute \(T_2\) in the vertical resolution equation:

\(T_1 \sin 35^{\circ} + \left(\frac{T_1 \cos 35^{\circ}}{\cos 40^{\circ}}\right) \sin 40^{\circ} = 2.4 \times 9.8\)

Simplify and solve for \(T_1\):

\(T_1 (\sin 35^{\circ} + \frac{\cos 35^{\circ} \sin 40^{\circ}}{\cos 40^{\circ}}) = 23.52\)

\(T_1 = \frac{23.52}{\sin 35^{\circ} + \frac{\cos 35^{\circ} \sin 40^{\circ}}{\cos 40^{\circ}}} \approx 20.4 \text{ N}\)

4. Substitute \(T_1\) back to find \(T_2\):

\(T_2 = \frac{20.4 \cos 35^{\circ}}{\cos 40^{\circ}} \approx 19.0 \text{ N}\)

To solve for the tensions in the strings, we can use the equilibrium conditions for the forces acting on the particle P. The forces involved are the tensions in the strings S₁ and S₂, and the weight of the particle.

Let the tension in S₁ be T₁ and in S₂ be T₂. The weight of the particle is 21 N acting vertically downward.

Using trigonometry, we can resolve the tensions into horizontal and vertical components. The horizontal components must balance each other, and the sum of the vertical components must balance the weight of the particle.

From the diagram, we can determine the angles using the given lengths:

• For S₁: \\(an^{-1}(0.2/0.52) = 21.8^\circ\\)
• For S₂: \\(an^{-1}(0.2/0.25) = 38.7^\circ\\)

Resolving horizontally:

\(T_1 \cos(21.8^\circ) = T_2 \cos(38.7^\circ)\)

Resolving vertically:

\(T_1 \sin(21.8^\circ) + T_2 \sin(38.7^\circ) = 21\)

Solving these equations simultaneously gives:

\(T_1 = 13\) N

\(T_2 = 20\) N

To find the angle AP₁X, we use the cosine rule in triangle XP₁A:

\(\cos \alpha = \frac{5.5}{7.3}\)

Solving for \(\alpha\), we get \(\alpha = 41.1^\circ\).

For the value of W, we use the sine rule in triangle XP₁C:

\(\frac{W}{\sin(180^\circ - 41.1^\circ)} = \frac{7.3}{\sin 90^\circ}\)

Solving for W, we find \(W = 4.8 \text{ N}\).

To solve for \(W_1\) and \(W_2\), we use the triangle of forces and the sine rule. The angles in the triangle are 40°, 60°, and 80°.

Using the sine rule:

\(\frac{W_1}{\sin 60^{\circ}} = \frac{5}{\sin 80^{\circ}}\)

Solving for \(W_1\):

\(W_1 = \frac{5 \cdot \sin 60^{\circ}}{\sin 80^{\circ}} = 4.40 \text{ N}\)

Similarly, for \(W_2\):

\(\frac{W_2}{\sin 40^{\circ}} = \frac{5}{\sin 80^{\circ}}\)

Solving for \(W_2\):

\(W_2 = \frac{5 \cdot \sin 40^{\circ}}{\sin 80^{\circ}} = 3.26 \text{ N}\)