June 2025 Paper 1 Q12

OCR MEICurrent spec8 marksProjectiles SUVAT

12 In this question \(x\) and \(y\) are the horizontal and upwards vertical directions respectively.

An astronaut is standing on the surface of the moon exploring the motion of a ball.

(a) The astronaut drops a ball from rest from 1 m above the surface. It takes 1.1 s to hit the surface.

Calculate the value of the acceleration due to the moon’s gravity. Give your answer correct to 3 significant figures. [2]

The astronaut stands in a crater of the moon and hits the ball with a golf club from the moon’s surface. The initial velocity of the ball is \(25\,\mathrm{m\,s^{-1}}\) at an angle of \(40^\circ\) above the horizontal in the \(x\)-direction.

(b) Taking the origin to be the point of projection, determine the equation of the trajectory of the ball. Give your answer in the form \(y = \mathrm{f}(x)\), with each of the coefficients correct to 3 significant figures. [4]

(c) The edge of the crater is 40 m away from the point of projection and 15 m above it.

Determine whether the ball goes over the crater’s edge. [2]

Scheme	Marks	AO
\(s = ut + \frac{1}{2}at^2\) with \(s = -1,\ u = 0,\ t = 1.1\)	M1	1.1a
\(a = -1.65\,\mathrm{m\,s^{-2}}\)	A1	1.1
	[2]

Mark scheme (b)

Scheme	Marks	AO
\(x = 25\cos 40^\circ\, t\)	B1	1.1
\(y = 25\sin 40^\circ\, t + \frac{1}{2}at^2\)	M1	3.3
\(y = 25\sin 40^\circ\left(\dfrac{x}{25\cos 40^\circ}\right) + \dfrac{1}{2}a\left(\dfrac{x}{25\cos 40^\circ}\right)^2\)	M1	2.1
\(y = 0.839x - 0.00225x^2\)	A1	1.1
	[4]

Notes

B1: Oe eg \(t = \dfrac{x}{25\cos 40^\circ}\), \(t = 0.0522x\) or \(\dfrac{x}{19.15}\) or \(\dfrac{x}{19.2}\)

M1: Forms equation of motion in the vertical direction
Soi. Allow sign errors.
Do not allow if \(a = \pm 9.8\) used here or subsequently

M1: Substitutes expression for \(t\) in their \(y\) equation

A1: Cao. Coefficients must be 3sf.

Mark scheme (c)

Scheme	Marks	AO
When \(x = 40,\ y = 30.0\ [\gt 15]\)	M1	3.4
So the ball goes over the crater’s edge	A1	3.2a
	[2]

Notes

M1: Use their model and \(x = 40\)
(29.72 if exact values used)

A1: Conclusion based on correct working.
No FT from wrong (b)
Possible interpretation of the question gives \(x = \sqrt{40^2 - 15^2} = 5\sqrt{55} = 37.1\) giving \(y = 28.0\)

Alternative method

Scheme	Marks
When \(y = 15,\ x = 18.8,\ 354\)	M1
So the ball is above 15m as 40 m is between these values	A1

M1: Use the model with \(y = 15\)

A1: Conclusion based on correct working
No FT for wrong (b)
Possible interpretation \(x = \sqrt{1375} = 37.1\) which is between these values.

When \(x = 40,\ t = 2.09\) when \(y = 15,\ t = 0.983,\ 18.5\)