Mathematics · May / June 2024 · Paper 51 · Question 5
YMA01
As printed in the paper
The question as text
5. A parachute is used to deliver a box of supplies. The parachute is attached to the box. • the parachute and box are dropped from rest from a helicopter that is hovering at a height of 520 m above the ground • the parachute and box fall vertically and freely under gravity for 5 seconds, then the parachute opens • from the instant the parachute opens, it provides a resistance to motion of magnitude 3200 N • the parachute and box continue to fall vertically downwards after the parachute opens • the parachute and box are modelled throughout the motion as a particle P of mass 250 kg
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[2]
s =
×
9.8 5
×
2
2
= 123 or 120 (m)
v =
9.8 5
×=
49
OR
v =
2 9.8 122.5
×
×
= 49
5(b)[7]
OR
122.5
=
5
−×
1
9.8 5
×
2
=> v = 49
v
2
250
−
3200
= ±
250
g
a
Correct value for their a ( 3 or 3
−)
v =
2
49
2
−××
2 3 (520 122.5)
−
v = 4 (
m s−)
1
N.B. They may do (c) first and then use their t value to obtain v:
v =
49
+ −×
( 3 15)
OR
(520 122.5)
15
( 3) 15
2
M1A1ft
−
=
v
−×−
1
×
2
v = 4 (
m s−) A1
1
5(c)[3]
4
=
49
−
3t
(49
+
4)
OR
(520 122.5)
−
=
t
2
1
OR
(520 122.5)
−
=
49
t
−
×
3
t
2
2
1
OR
(520 122.5)
−
=
4
t
+
×
3
t
2
2
t = 15 (other root of quadratic is 53
3 which leads to v < 0)
Total time = 5 + 15 = 20 (s)
5(d)[2]
49
4
20
Notes
5(a)
M1: Complete method to find the distance
A1: cao
(b)
B1: 49 or 49. Allow 5g or 5g or 492 (2401) seen.
M1: Equation of motion, correct terms, condone sign errors
A1: Correct equation (allow + or −)
A1: cao
M1: Complete method to find speed at ground (must have found a new a)
M0 if they use u = 0 either explicitly in (b) or implicitly, by using it in (c) to get the
time, which is then used in (b).
A1ft: Correct equation, ft on their s, v and a .
N.B. This mark can be awarded even if it leads to a negative value for v2
A1: cao
(c)
M1: Complete method to find time from when the parachute opens to when P lands
on the ground (must have found and use a new a)
M0 if they use s = 520 and/or u = 0
A1:cao.
A1ft: Their t + 5
N.B. The final answer should be rounded to 2 or 3 sf, if they haven’t already been
penalised following use of g= 9.8 earlier in the question.
B1: Correct shape (B0 if continuous vertical line at the end or graph ends on the t
(d)
axis)
B1ft:Correct figs, ft on their 49, 4 and 20, but B0ft if they just assume it stops as it
reaches the ground.
The ft is only available if the graph has just 2 straight lines, one starting at the origin
with positive gradient and the second line has negative gradient with second line not
meeting the taxis.
This B1ft is available if the graph has a vertical line at the end but is otherwise
correct.
+
sin
θ
=
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