October 2020 Paper 3 Mechanics Q5

EdexcelCurrent spec11 marksProjectiles

Figure 2: ball projected from O at the top of a vertical cliff 25 m above N, with speed U m/s at 45 degrees above the horizontal, landing at A on horizontal ground 100 m from N — Figure 2

A small ball is projected with speed \(U\ \text{m s}^{-1}\) from a point \(O\) at the top of a vertical cliff.

The point \(O\) is 25 m vertically above the point \(N\) which is on horizontal ground.

The ball is projected at an angle of \(45^\circ\) above the horizontal.

The ball hits the ground at a point \(A\), where \(AN = 100\) m, as shown in Figure 2.

The motion of the ball is modelled as that of a particle moving freely under gravity.

Using this initial model,

(a) show that \(U = 28\) (6)

(b) find the greatest height of the ball above the horizontal ground \(NA\). (3)

In a refinement to the model of the motion of the ball from \(O\) to \(A\), the effect of air resistance is included.

This refined model is used to find a new value of \(U\).

(d) State one further refinement to the model that would make the model more realistic. (1)

Mark scheme (a)

Scheme	Marks	AO
Using horizontal motion	M1	3.3
\(U\cos 45^\circ t = 100\)	A1	1.1b
Using vertical motion	M1	3.4
\(U\sin 45^\circ t - \dfrac{1}{2}gt^2 = -25\)	A1	1.1b
Solve problem by eliminating \(t\) and solving for \(U\)	M1	3.1b
\(U = 28\)*	A1*	1.1b
	(6)

Notes

M1: Complete method to give equation in \(U\) and \(t\) only, condone sin/cos confusion and sign errors

A1: Correct equation

M1: Complete method to give equation in \(U\) and \(t\) only, condone sin/cos confusion and sign errors

A1: Correct equation (\(g\) does not need to be substituted)

M1: Must have earned the previous two M marks.
Eliminate \(t\) and solve for \(U\).
N.B. They may solve for \(t\) first \(\left(100 - \dfrac{1}{2}gt^2 = -25\right)\) and then use it to find \(U\).

A1*: Exact given answer correctly obtained with no wrong working (e.g. \(g = 9.81\) used) or approximation seen.

Mark scheme (b)

Scheme	Marks	AO
Using vertical motion	M1	3.4
\(0^2 = (28\sin 45^\circ)^2 - 2gh\)	A1	1.1b
Greatest height = 45 m	A1	1.1b
	(3)

Notes

M1: Complete method to give equation in \(h\) only (allow if \(U\) not substituted), condone sin/cos confusion and sign errors

A1: Correct equation (\(g\) does not need to be substituted) (A0 if \(U\) is used instead of 28)

A1: cao

Scheme	Marks	AO
New value \(\gt 28\)	B1	3.5a
	(1)

Mark scheme (d)

Scheme	Marks	AO
e.g. wind effects, more accurate value of \(g\), spin of ball, include size of the ball, not model as a particle, shape of ball	B1	3.5c
	(1)
	(11 marks)

Notes

B1: Penalise incorrect extras i.e. B0 if there are incorrect extras.
The ground being horizontal, the cliff being vertical, .. are not part of the model so B0
Include weight/mass of the ball B0