Question 6:
6 | (i) | T cos30(cid:32)2g
PQ
T (cid:32)22.6N
PQ | M1
A1
[2] | 3.3
1.1 | Resolving vertically for Q | 22.632130…
6 | (ii) | r(cid:32)1.5sin60(cid:14)0.75sin30
T sin30(cid:32)2r(cid:90)2
PQ
(cid:90)(cid:32)1.84rads-1 | B1
M1
A1
[3] | 3.1b
3.3
1.1 | May be implied in N2L for Q
N2L for Q with their r
n | 1.8384425…
6 | (iii) | v (cid:32)2.39ms-1
P | B1FT
[1] | 1.1 | e
(1.5sin60)(their (cid:90))
6 | (iv) | T cos60(cid:32)2.5g(cid:14)T cos30
AP PQ
T (cid:32)88.2N
AP | M1
M1
A1
[3] | 3.1b
1.1
i
c
1.1 | m
Resolving vertically for P
Substituting cv for T from (i) and
PQ
solving for T
AP
6 | (v) | p
T (cid:14)T sin60(cid:16)T sin30(cid:32)2.5(1.5sin60)(cid:90)2
BP AP PQ
S
T (cid:32)(cid:16)54.1N so T (cid:32)54.1
BP BP
T (cid:31)0(cid:63) rod BP is under compression
BP | e
M1
A1
A1
A1FT
[4] | 3.1b
1.1
1.1
3.2a | N2L for P (4 terms)
FT the sign of T
BP | –54.0909226…
PPMMTT
Y533 Mark Scheme June 20XX
Assessment Objectives (AO) Grid
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M = Modelling
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Y533 Mark Scheme June 20XX
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Y533 Mark Scheme June 20XX
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PPMMTT
AS Level Further Mathematics A
Y533 Mechanics
Printed Answer Booklet
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