October 2021 Paper 3 Q11

OCR ACurrent spec10 marksProjectiles

A ball is projected from point A on level ground with speed 25 m/s at 15 degrees above the horizontal; the ground slopes down beyond A to a lower level, and the ball's path lands at B, which is 4 m below the level of A

A golfer hits a ball from a point \(A\) with a speed of \(25\,\mathrm{m\,s^{-1}}\) at an angle of \(15^\circ\) above the horizontal. While the ball is in the air, it is modelled as a particle moving under the influence of gravity. Take the acceleration due to gravity to be \(10\,\mathrm{m\,s^{-2}}\).

The ball first lands at a point \(B\) which is \(4\,\mathrm{m}\) below the level of \(A\) (see diagram).

(a) Determine the time taken for the ball to travel from \(A\) to \(B\). [3]

(b) Determine the horizontal distance of \(B\) from \(A\). [2]

The horizontal distance from \(A\) to \(B\) is found to be greater than the answer to part (b).

(d) State one factor that could account for this difference. [1]

Mark scheme (a)

Scheme	Marks	AO
\(-4 = (25\sin 15)t - \frac{1}{2}(10)t^2\)	M1 A1	3.3 1.1
\(t = 1.75\ (\mathrm{s})\)	A1	2.2a
	[3]

Notes

M1: Use of \(s = ut + \frac{1}{2}at^2\) with \(a = \pm g\) and \(s = \pm 4\)
Allow sin/cos confusion

A1: BC (1.750981765…) 1.75 only
For reference: 1.779296952… (if \(g = 9.8\) used)
Penalise \(g = 9.8\) only once in the question

Alternative method

Scheme	Marks
\(0 = (25\sin 15)^2 + 2(-10)s_1\) and \(0 = 25\sin 15 + (-10)t_1\)	M1*
\(4 + s_1 = \frac{1}{2}\ 10\ t_2^{\,2}\) and \(t = t_1 + t_2\)	M1dep*
\(t = 0.6470476\ldots + 1.1039341\ldots = 1.75\,(\mathrm{s})\)	A1

Notes

M1*: Finding the maximum height \(s_1\,(= 2.093353\ldots)\) above \(A\) and corresponding time \(t_1\,(= 0.647047\ldots)\)
Using \(v = 0\) and \(a = \pm 10\)

M1dep*: Complete correct method to find \(t\)
Using \(u = 0\) and where \(t_2\,(= 1.1039341\ldots)\) is the time from the maximum height to the ground

Mark scheme (b)

Scheme	Marks	AO
\((25\cos 15)t\)	M1	3.4
42.3 (m)	A1FT	1.1
	[2]

Notes

M1: Use of \(s = ut\) with their \(t\) from (a)
Allow sin/cos confusion

A1FT: 42.2829627… - ft their positive value of \(t\) from (a) but must be using \((25\cos 15)t\)
For reference: 42.96672196… (if \(g = 9.8\) used)

Mark scheme (c)

Scheme	Marks	AO
\(v_h = 25\cos 15\)	B1	1.2
\(v_v = 25\sin 15 - 10(1.5)\)	B1	3.3
\(\tan\theta = \dfrac{v_v}{v_h}\)	M1	3.1b
\(19.5^\circ\) below the horizontal	A1	3.2a
	[4]

Notes

B1: Correct expression for horizontal velocity component (soi)
24.14814…

B1: Correct expression for vertical velocity component at \(t = 1.5\) (condone positive value)
\(-8.529523\ldots\)

M1: Use of tan to find angle (allow reciprocal) – dependent on one B mark earned
M0 if using expressions for displacements

A1: oe (e.g., \(70.5^\circ\) to the downward vertical)
For reference: \(18.8^\circ\) (if \(g = 9.8\) used)

Mark scheme (d)

Scheme	Marks	AO
e.g., a less accurate value of \(g\) was used e.g., no consideration of the wind e.g., no consideration of (back)spin on the ball (but not topspin)	B1	3.5a
	[1]

Notes

B1: Any valid reason (do not accept mention of resistance e.g., air/wind resistance)