Past Exam Questions

Using the SUVAT equation for vertical motion: \(s = ut - \frac{1}{2}gt^2\), where \(g = 10 \text{ m s}^{-2}\).

At \(t = 3\) seconds, \(s = 3u - 5 \times 9\).

At \(t = 4\) seconds, \(s = 4u - 5 \times 16\).

Equating the two expressions for \(s\):

\(3u - 45 = 4u - 80\)

Solving for \(u\):

\(3u - 4u = -80 + 45\)

\(-u = -35\)

\(u = 35\)

Substitute \(u = 35\) back into one of the equations for \(s\):

\(s = 3(35) - 45 = 105 - 45 = 60\)

Thus, \(u = 35\) and \(s = 60\).

(i) For particle A, the height at time t is given by:

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

For particle B, the height at time t is:

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

Setting \(h_A = h_B\) to find the collision point:

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

Simplifying gives:

\(10t - 35 = 0\)

Solving for \(t\):

\(t = 3.5\)

Substitute \(t = 3.5\) into \(h_A\) or \(h_B\):

\(h = 20 \times 3.5 - \frac{1}{2} \times 10 \times (3.5)^2 = 8.75 \text{ m}\)

(ii) The speed of particle A at collision:

\(v_A = 20 - gt = 20 - 10 \times 3.5 = -15 \text{ m s}^{-1}\)

The speed of particle B at collision:

\(v_B = -g(t-1) = -10 \times 2.5 = -25 \text{ m s}^{-1}\)

Difference in speeds:

\(|v_A - v_B| = |-15 - (-25)| = 10 \text{ m s}^{-1}\)

(i) To find the time taken for P to reach its greatest height, use the equation of motion:

\(v = u + at\)

where \(v = 0\) (at the greatest height), \(u = 25\) m/s, and \(a = -g = -10\) m/s². Solving for \(t\):

\(0 = 25 - 10t\)

\(t = 2.5 \text{ s}\)

(ii) To find the length of time for which P is higher than 23 m above the ground, use the equation:

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

where \(s = 20\) m (since 23 m above the ground is 20 m above the projection point), \(u = 25\) m/s, and \(a = -10\) m/s². Solving the quadratic equation:

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

Rearranging gives:

\(5t^2 - 25t + 20 = 0\)

Solving this quadratic equation gives \(t = 1\) s and \(t = 4\) s. The time interval is:

\(4 - 1 = 3 \text{ s}\)

(iii) To find h such that P is higher than h m above the ground for 1 second, note that the maximum height is reached at 2.5 s. Therefore, P reaches h after 2 s. Using:

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

where \(u = 25\) m/s, \(a = -10\) m/s², and \(t = 2\) s:

\(h - 3 = 25 \times 2 - 5 \times 2^2\)

\(h - 3 = 50 - 20\)

\(h = 33 \text{ m}\)

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

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

where \(v = 0\) (at maximum height), \(u = 30\) m/s, and \(a = -g\) (acceleration due to gravity, \(g = 9.8\) m/s²).

\(0 = 30^2 + 2(-9.8)s\)

\(s = \frac{900}{19.6} = 45\) m

(ii) To find the time to reach 33.75 m, use:

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

\(33.75 = 30t - \frac{1}{2}gt^2\)

\(33.75 = 30t - 4.9t^2\)

\(4.9t^2 - 30t + 33.75 = 0\)

Solving this quadratic equation gives \(t = 1.5\) s (rejecting \(t = 4.5\) s as it is the second time reaching this height).

To find the speed at this time, use:

\(v = u + at\)

\(v = 30 - 9.8 \times 1.5\)

\(v = 15\) m/s

First, use the equation of motion to find \(V\):

\(V^2 = 5^2 + 2 \times g \times 7.2\)

Assuming \(g = 9.8 \text{ m s}^{-2}\), solve for \(V\):

\(V^2 = 25 + 2 \times 9.8 \times 7.2\)

\(V^2 = 25 + 141.12\)

\(V^2 = 166.12\)

\(V = \sqrt{166.12} \approx 12.89 \text{ m s}^{-1}\)

Rounding gives \(V = 13 \text{ m s}^{-1}\).

Next, find the time taken for the ball to reach the ground from A:

Using \(v = u + gt\), where \(v = 13 \text{ m s}^{-1}\), \(u = 5 \text{ m s}^{-1}\), and \(g = 9.8 \text{ m s}^{-2}\):

\(13 = 5 + 9.8t\)

\(8 = 9.8t\)

\(t = \frac{8}{9.8} \approx 0.82 \text{ s}\)

For the rebound, the initial speed is \(\frac{1}{2} V = 6.5 \text{ m s}^{-1}\):

Using \(v = u - gt\), where \(v = 0 \text{ m s}^{-1}\) at the highest point:

\(0 = 6.5 - 9.8t\)

\(9.8t = 6.5\)

\(t = \frac{6.5}{9.8} \approx 0.66 \text{ s}\)

\(Total time = 0.82 s + 0.66 s = 1.48 s\)

Rounding gives the total time as 1.45 s.

We use the equation of motion:

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

where \(s = -5\) m (since it ends 5 m below the starting point), \(u = 24\) m/s, and \(a = -10\) m/s² (acceleration due to gravity).

Substituting the values, we get:

\(-5 = 24t - 5t^2\)

Rearranging gives the quadratic equation:

\(5t^2 - 24t - 5 = 0\)

Solving this quadratic equation using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 5\), \(b = -24\), and \(c = -5\), we find:

\(t = 5\)

Thus, the time from projection until P reaches the ground is 5 seconds.

(i) Using the equation of motion:

\(s = \frac{1}{2} (u + v) t\)

where \(s = 200\) m, \(u = 0\), \(t = 10\) s, we have:

\(200 = \frac{1}{2} (0 + v) \times 10\)

\(v = 40 \text{ m s}^{-1}\)

To find acceleration \(a\), use:

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

\(200 = 0 \times 10 + \frac{1}{2} a \times 10^2\)

\(a = 4 \text{ m s}^{-2}\)

(ii) After 10 s, the rocket moves under gravity alone. Using:

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

where \(v = 0\), \(u = 40 \text{ m s}^{-1}\), \(a = -g\), we have:

\(0 = 40^2 - 2 \times 9.8 \times s\)

\(s = 80 \text{ m}\)

\(Total height = 200 + 80 = 280 m\)

(iii) Time to reach maximum height:

\(0 = 40 - gt_1\)

\(t_1 = 4 \text{ s}\)

Time to fall from maximum height:

\(280 = \frac{1}{2} g t_2^2\)

\(t_2 = \sqrt{\frac{560}{9.8}} = 7.48 \text{ s}\)

\(Total time = 10 + 4 + 7.48 = 21.5 s\)

(i) To find the time taken for particle P to return to the ground, we use the equation of motion:

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

where \(s = 0\) (displacement is zero when it returns to the ground), \(u = 12\) m/s (initial velocity), and \(a = -g\) (acceleration due to gravity, \(g = 9.8\) m/s²).

Substituting the values, we get:

\(0 = 12t - \frac{1}{2} \times 9.8 \times t^2\)

Solving for \(t\), we find:

\(t = \frac{2 \times 12}{9.8} = 2.4 \text{ s}\)

(ii) For particle Q, it is projected at \(t = 1\) with speed 10 m/s from 5 m above the ground. We need to find when both particles are moving in the same direction.

Critical points occur when the velocity of each particle is zero. For P, this occurs at:

\(t = \frac{12}{9.8} = 1.2 \text{ s}\)

For Q, the critical point occurs at:

\(t = 1 + \frac{10}{9.8} = 2 \text{ s}\)

Thus, the particles move in the same direction in the intervals:

• \(1 < t < 1.2\)
• \(2 < t < 2.4\)

For the second particle, using the equation of motion:

\(s_2 = 20t - 0.5gt^2\)

For the first particle, which was projected 2 seconds earlier:

\(s_1 = 12(t + 2) - 0.5g(t + 2)^2\)

Setting \(s_1 = s_2\) to find when they collide:

\(12(t + 2) - 0.5g(t + 2)^2 = 20t - 0.5gt^2\)

Solving for \(t\):

\(t = \frac{1}{7} = 0.143 \text{ s}\)

For the height at which they collide:

Substitute \(t = \frac{1}{7}\) into the equation for \(s_2\):

\(s = 20 \times \frac{1}{7} - 5 \times \left(\frac{1}{7}\right)^2 \approx 2.755\)

Height is approximately 2.76 m.

(i) For ball A, using the equation of motion:

\(s_A = \frac{1}{2} g \times 2.5^2 = 31.25 \text{ m}\)

For ball B, using the equation of motion:

\(s_B = 20 \times 1.5 - \frac{1}{2} g \times 1.5^2 = 18.75 \text{ m}\)

Equating the heights:

\(\frac{1}{2} g \times 2.5^2 + 20 \times 1.5 - \frac{1}{2} g \times 1.5^2 = 50 \text{ m}\)

Thus, the height of the tower is 50 m.

(ii) For ball A, using \(s = \frac{1}{2} g t_A^2\):

\(50 = 0.5 g t_A^2 \Rightarrow t_A = 3.16 \text{ s}\)

For ball B, time in motion:

\(t_B = \sqrt{10} - 1 = 2.16 \text{ s}\)

To the top, using \(v^2 = u^2 - 2gs_B\):

\(0^2 = 20^2 - 2gs_B \Rightarrow s_B = 20 \text{ m}\)

To the top, using \(v = u + at\):

\(0 = 20 - gt_B \Rightarrow t_B = 2 \text{ s}\)

Downwards, using \(s = \frac{1}{2} g (0.16)^2\):

\(s_B = 0.13 \text{ m}\)

Total distance for B:

\(20 + 0.13 = 20.1 \text{ m}\)