Mathematics · May / June 2024 · Paper 31 · Question 8
YMA01
As printed in the paper
The question as text
8. h O d x Figure 4 Figure 4 is a graph showing the path of a golf ball after the ball has been hit until it first hits the ground. The vertical height, h metres, of the ball above the ground has been plotted against the horizontal distance travelled, x metres, measured from where the ball was hit. The ball travels a horizontal distance of d metres before it first hits the ground. The ball is modelled as a particle travelling in a vertical plane above horizontal ground. The path of the ball is modelled by the equation h = 1.5x – 0.5x e 0.02x 0 x d Use the model to answer parts (a), (b) and (c).
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
8(a)[3]
0
=
1.5
−
0.5 e
d
e
d
=
3
d
d
0.02
=
ln3
=
54.93
d
d
)
d
(
h
0.02
x
0.02
x
8(b)[4]
=
1.5
−
0.5e
+
0.5
x
×
0.02e
d
x
0.02
0.02
0.02
(
)
Sets
0
=
1.5
−
0.5e
x
−
0.5
×
0.02e
x
e
x
0.5
+
0.5
×
0.02
=
1.5
x
x
1.5
0.02
e
x
=
(
)
0.5
+
0.5
×
0.02
x
1.5
150
150
0.02
x
e
=
=
=
50ln
*
x
(
)
0.5
+
0.5
×
0.02
50
+
x
50
+
x
x
(i) awrt 31.43
8(c)[3]
(ii) awrt 30.88 metres (including units)
(a)
Allow to be solved in either x or d
±
0.02
M1:
Sets h = 0, cancels by d (or takes out a factor of d) and proceeds to a form
e
d
=
a
±
0.2
d
and a may be 1. Condone a misread or miscopy proceeding to the form
e
=
a
b
dM1: Solves the equation of the required form using logs. It is dependent on the previous method mark.
Look for an expression for d (or x) or a multiple of d (or x) before proceeding to a value for d.
ln3
0.02
e.g.
e
d
3
=
d
=
0.02
1
1
e.g.
−
0.02
d
=
e
=
−
0.02 lne
d
=
ln
d
3
3
Condone the slip on the index to be 0.2d
±
It cannot be implied by a numerical value for d.
A1:
54.93 (m) cao (units not required but if given they must be correct) following correct log work seen
and at least one intermediate stage of working. Condone
x =
54.93
. isw if they round after a correct
answer is seen.
ln3
0.02
d
Minimum acceptable e.g.
e
=
3
d
=
=
54.93
(M1dM1A1)
0.02
0.02
d
e.g.
0
=
1.5
−
0.5 e
0.02
=
ln3
=
54.93
(M1dM1A1)
d
d
d
d
0.02
d
0.02
d
e.g.
0
=
1.5
d
−
0.5 e
d
3
=
e
d
=
54.93
(M1dM0A0)
0.02
M1:
Attempts the product rule on
e
x
achieving
±
e
0.02
±
e
0.02
where A can be 1. It is likely to
x
Ax
x
B
x
be part of an expression.
Condone a misread or miscopy proceeding to the form
±
e
0.2
±
e
0.2
Ax
x
B
x
(
)
d
h
0.02
0.02
A1:
1.5
0.5e
x
0.5
0.02e
x
o.e.
=
−
+
x
×
d
x
0.02
x
0.02
x
±
0.02
x
±
0.02
x
dM1: Sets
0
=
1.5 "0.5e
−
−
0.5
x
×
0.02e
"
, attempts to make
e
(or
C
e
) the subject and
D
±
0.02
proceeds to the form
e
x
=
where C can be 1.
C
E
+
Fx
150
0.02
x
0.02
x
0.02
x
May see
0
=
150 "50e
−
−
x
e
"
e
=
50
+
x
±
0.02
x
Condone slips in their rearrangement but they must take out a factor of
e
(or
C
divide by their bracket. If they take logs of both sides first, the rearrangement must be correct.
Condone invisible brackets to be recovered or implied by further work.
0.02
0.02
Condone a misread or miscopy proceeding from
0
1.5 "0.5e
x
0.5
0.02e
=
−
−
x
×
±
0.2
x
±
0.2
x
attempting to make
e
(or
e
) the subject.
C
A1*: Achieves the given answer with no errors seen provided all previous marks have been scored.
Do not allow this mark to be scored for proceeding directly from
150
0.02
x
(
)
e
0.5
+
0.5
×
0.02
=
1.5
to
.
x
x
=
50ln 50
+
x
1.5
0.02
x
We must see either
e
=
o.e. or an unsimplified expression for x
(
)
0.5
+
0.5
×
0.02
x
1.5
e.g.
=
50ln 0.5
before achieving the given answer.
x
+
0.01
x
(c)(i) Check by the question. If there is a contradiction between answers, the answer in the main
body of the work takes precedence.
M1:
Attempts the iteration formula ''correctly'' seen once.
150
E.g. Award for
=
50ln
or awrt 31.4
x
2
80
A1:
awrt 31.43
(c)(ii)
A1:
awrt 30.88 m (including units)
π
(
)
=
4sin
2
−=
1
2
*
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