Past Exam Questions

(i) To find the greatest height, use the equation of motion: \(v^2 = u^2 + 2as\). Here, \(v = 0\) at the greatest height, \(u = 6\) m/s, \(a = -g = -9.8\) m/s², and \(s = 0.5\) m.

First, find the initial velocity \(u\) when the particle is at 0.5 m:

\(6^2 = u^2 - 2 \times 9.8 \times 0.5\)

\(36 = u^2 - 9.8\)

\(u^2 = 46\)

Now, find the total height:

\(0 = 46 - 2 \times 9.8 \times s\)

\(2 \times 9.8 \times s = 46\)

\(s = \frac{46}{19.6} = 2.3\) m

(ii) To find the time to return to O, use \(s = ut + 0.5gt^2\) with \(s = 2.3\) m, \(u = 0\), and \(g = 9.8\) m/s²:

\(2.3 = 0 + 0.5 \times 9.8 \times t^2\)

\(t^2 = \frac{2.3}{4.9}\)

\(t = \sqrt{\frac{2.3}{4.9}} = 0.678\) s

Total time for the round trip is \(2 \times 0.678 = 1.36\) s

(a) To find the speed of P when it is 10 m above the ground, use the equation of motion:

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

where \(u = 15 \text{ m s}^{-1}\), \(a = -g = -9.8 \text{ m s}^{-2}\), and \(s = 10 \text{ m}\).

Substitute the values:

\(v^2 = 15^2 - 2 \times 9.8 \times 10\)

\(v^2 = 225 - 196\)

\(v^2 = 29\)

\(v = \sqrt{29} \approx 5 \text{ m s}^{-1}\)

(b) For the collision, use the equations of motion for both particles:

For P: \(s_P = 15t - \frac{1}{2}gt^2\)

For Q: \(s_Q = 18 - \frac{1}{2}gt^2\)

Set \(s_P + s_Q = 18\) and solve for \(t\):

\(15t - \frac{1}{2}gt^2 + 18 - \frac{1}{2}gt^2 = 18\)

\(15t = gt^2\)

\(t = \frac{15}{g}\)

Substitute \(t\) back into \(s_P\):

\(s_P = 15 \times \frac{15}{g} - \frac{1}{2}g \left(\frac{15}{g}\right)^2\)

\(s_P = 10.8 \text{ m}\)

To solve for V and H, we use the equations of motion under gravity. Assume acceleration due to gravity is \(g\).

(i) Finding \(V\):

Using the equation \(v^2 = u^2 + 2gs\), where \(u\) is the initial velocity, \(v\) is the final velocity, and \(s\) is the displacement:

For the section from 35 m above the ground to the ground:

\(V^2 = (V - 10)^2 + 2g \times 35\)

Expanding and simplifying:

\(V^2 = V^2 - 20V + 100 + 70g\)

\(20V = 100 + 70g\)

Assuming \(g = 10 \text{ m s}^{-2}\):

\(20V = 100 + 700\)

\(20V = 800\)

\(V = 40\)

(ii) Finding \(H\):

Using the equation \(v^2 = u^2 + 2gs\) for the entire fall from H to the ground:

\(40^2 = 0^2 + 2gH\)

\(1600 = 20H\)

\(H = 80\)

(i) To find the time after projection at which P hits the ground, we use the equation of motion:

\(v = u - gt\)

where \(v = -11\) m/s (final velocity when hitting the ground), \(u = 11\) m/s (initial velocity), and \(g = 10\) m/s² (acceleration due to gravity).

Substitute the values:

\(-11 = 11 - 10t\)

Solve for \(t\):

\(-11 = 11 - 10t\)

\(-22 = -10t\)

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

(ii) To find the values of \(h\) and \(V\):

For particle Q, using the equation:

\(h = 0 + \frac{1}{2} g t^2\)

Substitute \(g = 10\) m/s² and \(t = 2.2\) s:

\(h = \frac{1}{2} \times 10 \times (2.2)^2 = 24.2 \text{ m}\)

For the speed \(V\) of Q when it hits the ground:

\(V = 0 + gt\)

Substitute \(g = 10\) m/s² and \(t = 2.2\) s:

\(V = 10 \times 2.2 = 22 \text{ m/s}\)

(i) First, calculate the speed of P just before entering the liquid using the equation:

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

where \(u = 0\), \(a = g = 9.8 \text{ m s}^{-2}\), and \(s = 7.2 \text{ m}\).

\(v^2 = 2 \times 9.8 \times 7.2\)

\(v = 12 \text{ m s}^{-1}\)

Now, use the equation \(v^2 = u^2 + 2as\) for the motion in the liquid:

\(6^2 = 12^2 + 2a \times 0.8\)

\(a = -67.5 \text{ m s}^{-2}\)

Using Newton's second law, \(0.2g - R = 0.2 \times 67.5\)

\(R = 15.5 \text{ N}\)

(ii) For the upward motion, use \(s = \frac{1}{2}at^2\) to find \(a\):

\(3.6 = \frac{1}{2}a \times 4^2\)

\(a = 0.45 \text{ m s}^{-2}\)

Using Newton's second law, \(T - R - 0.2g = 0.2 \times 0.45\)

\(T = 17.6 \text{ N}\)

First, calculate the maximum height reached by the particle. Using the equation for upward motion:

\(0 = u^2 - 2gs\)

where \(u = 9 \text{ m/s}\) and \(g = 9.8 \text{ m/s}^2\). Solving for \(s\):

\(0 = 9^2 - 2 \times 9.8 \times s\)

\(s = \frac{81}{19.6} = 4.05 \text{ m}\)

The total distance travelled is the sum of the distance upwards and downwards:

\(\text{Total distance} = 4.05 + (3.15 + 4.05) = 11.25 \text{ m}\)

Next, calculate the time taken to reach the maximum height:

\(v = u - gt\)

Setting \(v = 0\):

\(0 = 9 - 9.8t\)

\(t = \frac{9}{9.8} = 0.9 \text{ s}\)

For the downward motion, use:

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

where \(s = 3.15 + 4.05 = 7.2 \text{ m}\):

\(7.2 = \frac{1}{2} \times 9.8 \times t^2\)

\(t^2 = \frac{14.4}{9.8}\)

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

The total time taken is:

\(\text{Total time} = 0.9 + 1.2 = 2.1 \text{ s}\)

(i) The time for particle P to reach the ground is given by solving the equation for vertical motion:

\(17t - 5t^2 = 0\)

Solving gives \(t = 0\) or \(t = 3.4\) seconds. The time for particle Q to reach the ground is given by:

\(7(t - T) - 5(t - T)^2 = 0\)

Solving gives \(t - T = 0\) or \(t - T = 1.4\) seconds. Since both particles reach the ground at the same time, \(t = 3.4\) and \(t - T = 1.4\), so \(T = 2\) seconds.

(ii) At the instant when P is 5 m higher than Q, we have:

\(17(t + 2) - 5(t + 2)^2 - (7t - 5t^2) = 5\)

Solving gives \(t = 0.9\) or \(t = 2.9\) seconds. Using \(t = 0.9\), the velocities are:

For P: \(v_P = 17 - 10(t + 2) = 12 \text{ m s}^{-1}\)

For Q: \(v_Q = 7 - 10t = 2 \text{ m s}^{-1}\)

Both velocities are directed vertically downwards.

(i) Using the equation of motion: \(0 = u^2 - 2gs\), where \(s = 45 \text{ m}\) (40 m + 5 m) and \(g = 10 \text{ ms}^{-2}\), we have:

\(u^2 = 2 \times 10 \times 45\)

\(u = 30 \text{ ms}^{-1}\)

(ii) Using \(s = ut - \frac{1}{2}gt^2\) with \(s = 40\), \(u = 30\), and \(g = 10\):

\(40 = 30t - 5t^2\)

Solving \(5t^2 - 30t + 40 = 0\) gives \(t = 2\) and \(t = 4\).

The time above the top of the cliff is \(4 - 2 = 2 \text{ s}\).

(iii) For the second signal, the maximum height above the top of the cliff is \(\frac{1}{2}g(17 \div 4) = 21.25 \text{ m}\).

Using \(0 = V^2 - 2g(40 + 21.25)\):

\(V^2 = 2 \times 10 \times 61.25\)

\(V = 35 \text{ ms}^{-1}\)

To solve this problem, we use the equations of motion under constant acceleration due to gravity, where the acceleration is \(g = 9.8 \text{ m/s}^2\).

(i) Speed at the mid-point of OA:

The maximum height A is found using the equation: \(0 = u^2 - 2gs\), where \(u = 8 \text{ m/s}\) and \(v = 0 \text{ m/s}\) at the highest point.

\(0 = 8^2 - 2g \cdot s\)

\(s = \frac{64}{2 \times 9.8} = 3.2 \text{ m}\)

The mid-point of OA is at \(s = 1.6 \text{ m}\).

Using \(v^2 = u^2 - 2gs\) to find the speed at the mid-point:

\(v^2 = 8^2 - 2 \times 9.8 \times 1.6\)

\(v^2 = 64 - 31.36\)

\(v = \sqrt{32.64} \approx 5.66 \text{ m/s}\)

(ii) Time to reach the mid-point of OA:

Using \(v = u - gt\), where \(v = 5.66 \text{ m/s}\) at the mid-point:

\(5.66 = 8 - 9.8t\)

\(9.8t = 8 - 5.66\)

\(t = \frac{2.34}{9.8} \approx 0.234 \text{ s}\)

First, calculate the time \(t\) it takes for the object to fall 125 m under gravity. Using the formula for free fall:

\(125 = \frac{1}{2} \times 10 \times t^2\)

\(t^2 = 25\)

\(t = 5 \text{ s}\)

Next, calculate the distance \(s\) that \(P\) travels in this time. \(P\) starts with an initial speed of \(5 \, \text{m/s}\) and accelerates at \(0.8 \, \text{m/s}^2\):

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

\(s = 5 \times 5 + \frac{1}{2} \times 0.8 \times 5^2\)

\(s = 25 + 10 = 35 \text{ m}\)

(i) To find when the particles are travelling in opposite directions, we need to determine when one particle is moving upwards and the other downwards. The time to reach maximum height for P is given by setting the velocity to zero:

\(0 = 12 - gt \Rightarrow t = \frac{12}{g} \approx 1.2 \text{ s}\)

For Q, \(0 = 7 - gt \Rightarrow t = \frac{7}{g} \approx 0.7 \text{ s}\)

Thus, the particles are travelling in opposite directions for \(0.7 < t < 1.2\).

(ii) Given \(3h_P = 8h_Q\), we use the equations for height:

\(h_P = 12t - \frac{1}{2}gt^2\)

\(h_Q = 7t - \frac{1}{2}gt^2\)

Substitute into the given condition:

\(3(12t - \frac{1}{2}gt^2) = 8(7t - \frac{1}{2}gt^2)\)

Simplifying gives:

\(36t - 1.5gt^2 = 56t - 4gt^2\)

\(2.5gt^2 = 20t\)

\(t = \frac{8}{g}\)

Using \(v = u - gt\) for velocities:

\(v_P = 12 - gt\)

\(v_Q = 7 - gt\)

Substitute \(t = \frac{8}{g}\):

\(v_P = 12 - 8 = 4 \text{ m s}^{-1}\)

\(v_Q = 7 - 8 = -1 \text{ m s}^{-1}\)

(i) To find the time for which P's height is greater than 15 m, use the equation for vertical motion:

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

where \(u = 20\) m s^-1 and \(g = 10\) m s^-2. Set \(h = 15\) m:

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

Solve the quadratic equation:

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

Divide by 5:

\(t^2 - 4t + 3 = 0\)

Factorize:

\((t - 1)(t - 3) = 0\)

So, \(t = 1\) or \(t = 3\). The duration is \(3 - 1 = 2\) seconds.

(ii) To find the velocities when P and Q are at the same height, equate their height equations:

\(20t - 5t^2 = 25(t - 0.4) - 5(t - 0.4)^2\)

Expand and simplify:

\(20t - 5t^2 = 25t - 10 - 5(t^2 - 0.8t + 0.16)\)

\(20t - 5t^2 = 25t - 10 - 5t^2 + 4t - 0.8\)

Combine terms:

\(20t = 29t - 10.8\)

\(9t = 10.8\)

\(t = 1.2\) (or \(t = 0.8\))

Use \(v = u - gt\) to find velocities:

For P:

\(v_P = 20 - 10 \times 1.2 = 8 \text{ m s}^{-1}\)

For Q:

\(v_Q = 25 - 10 \times (1.2 - 0.4) = 17 \text{ m s}^{-1}\)

(a) To find the initial speed u, we use the equation for velocity in uniformly accelerated motion:

\(v = u + at\)

At the greatest height, the final velocity \(v = 0\), the acceleration \(a = -10 \text{ m/s}^2\) (due to gravity), and the time \(t = 3 \text{ s}\). Substituting these values, we get:

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

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

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

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

Substituting \(v = 0\), \(u = 30 \text{ m/s}\), \(a = -10 \text{ m/s}^2\), we get:

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

\(0 = 900 - 20s\)

\(20s = 900\)

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

Therefore, the greatest height is 45 m.

(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:

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