Solution :-

(a) Finding the tension in the string and the time taken for B to reach the plane:

Equations of motion:

• \(T – 3g = \pm 3a\) (*B1 for one correct equation)
• \(5g – T = \pm 5a\)
• \(5g – 3g = \pm (3+5)a\) (*B1 for any two correct consistent equations)

Attempt to solve for T: (DM1 – Must get to ‘T=’. Dep on both B marks)

May find a first [a = 2.5]

\(T = 37.5\) N or \(\frac{75}{2}\) N (A1 – Allow without working)

Correct use of suvat with their a and solve for t: (DM1 – Must get to “t=”. Dep on both B marks)

E.g. \(2 = 0.5 \times 2.5 \times t^2\)

Could find \(v = \sqrt{10}\) then use \(2 = \frac{0 + their \sqrt{10}}{2} t\)

\(t = 1.26\) or \(\frac{\sqrt{40}}{5}\) or \(\frac{2\sqrt{10}}{5}\) (A1 – \(t = 1.2649…\))

Alternative answer 1:

Use of energy to find velocity at plane and then suvat to find t:

PE loss = KE gain: \(5g \times 2 – 3g \times 2 = \frac{1}{2} \times 5v^2 + \frac{1}{2} \times 3v^2\) (M1)

Or PE loss – WD by tension = KE gain: \(5g \times 2 – their\ 37.5 \times 2 = \frac{1}{2} \times 5v^2\)

Or WD by tension – PE gain = KE gain: \(their\ 37.5 \times 2 – 3g \times 2 = \frac{1}{2} \times 3v^2\)

\(v = \sqrt{10}\) or \(v = 3.16\), then \(2 = \frac{0 + v}{2} t\)

Dependent on both B marks only if candidate uses tension, but otherwise not dependent on either B mark.

\(t = \sqrt{1.6} = 1.26\) or \(\frac{\sqrt{40}}{5}\) or \(\frac{2\sqrt{10}}{5}\) (A1) \(t = 1.2649…\)

(b) Finding the time for which A is at least 3.25 m above the plane:

Correct use of suvat before B hits the plane or when A has risen 2 m, to attempt to find velocity of A when string becomes slack:

\(v^2 = 0 + 2 \times 2.5 \times 2\) or \(v = 0 + 2.5 \times \sqrt{1.6}\) or \(2 = \frac{1}{2}(0+v)\sqrt{1.6}\)

Must use \(s=2\) and \(u=0\). Must be complete method to find v or \(v^2\) [\(\sqrt{10}\)].

Could have found \(v = \sqrt{10}\) in part (a) and give M1 if used in part (b).

Correct use of suvat for motion of A (between height of 3m and 3.25 m), to form an equation in t, using \(a=-g\):

\(3.25 – 3 = (their \sqrt{10})t + 0.5 \times (-10) \times t^2\)

Solving a 3 term quadratic for t to get at least one (unsimplified) value using their 2.5 (or using any other correct method):

If correct should get \(t = 0.093, 0.540\) or \(\frac{\sqrt{10}-\sqrt{5}}{10}, \frac{\sqrt{10}+\sqrt{5}}{10}\), 0.09262… or 0.53978…

Could use formula and realise that the time for at least 3.25 m, \( = 2\frac{\sqrt{b^2 – 4ac}}{2a} = 2\frac{\sqrt{10 – 4 \times 5 \times 0.25}}{2 \times 5}\) which gets DM1.

Time = 0.447 s or \(\frac{\sqrt{5}}{5}\) s or \(\frac{1}{\sqrt{5}}\):

Time = 0.53978… – 0.09262… = 0.44721… or \(2(\frac{\sqrt{10}}{10} – \frac{\sqrt{10} – \sqrt{5}}{10}) = 0.44721…\)

Alternative answer 1:

For attempt to find max height using correct suvat with \(a=-g\):

\(0^2 = (their \sqrt{10})^2 + 2 \times (-10)s\) Where s is distance above 3 metres. (which leads to a maximum height of 3.5 m).

For attempt to find time from .25 m below top to top:

\(t = \pm 0.224\) or \(\frac{\sqrt{5}}{10}\) \([0.22360…]\) probably from \([3.5 – 3.25 = 0t + 0.5 \times 10t^2]\). Dep on both previous M1s.

Time = 0.447 s or \(\frac{\sqrt{5}}{5}\) s: For doubling.

Alternative answer 2:

Correct use of suvat for motion of A (between height of 3m and 3.25 m), to form an equation to find speed at height of 3.25 m:

\(w^2 = their \sqrt{10}^2 + 2 \times (-g) \times 0.25\). \([w^2 = 5]\) or \([w = \sqrt{5}]\).

For correct use of suvat to find time to max height:

\(0 = \sqrt{5} + (-g) \times t\) \([t = \frac{\sqrt{5}}{10}]\).

Time = 0.447 s or \(\frac{\sqrt{5}}{5}\) s: For doubling.

Alternative answer 3:

Correct use of energy before B