Mathematics · Oct / Nov 2021 · Paper 51 · Question 1
YMA01
As printed in the paper
The question as text
1. A 2.5 m C D 2 m B 9 m Figure 1 A non‑uniform rod AB has length 9 m and mass M kg. The rod rests in equilibrium in a horizontal position on two supports, one at C where AC = 2.5 m and the other at D where DB = 2 m, as shown in Figure 1. The magnitude of the force acting on the rod at D is twice the magnitude of the force acting on the rod at C. The centre of mass of the rod is d metres from A. Find the value of d. (6)
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Show the answer and mark scheme
2.5 m 4.5 m 2 m
A C G D B
d m Mg
1st Equation in d (or another defined unknown), R and M (as appropriate) only M1 A1
2nd Equation in d (or the same defined unknown, R and M (as appropriate)
only
Possible equations:
( ),
↑
R
+
2
R
=
Mg
M( ),2
C
R
×
4.5
=
Mg d
(
−
2.5)
M(
D R
),
×
4.5
=
Mg
(7
−
d
)
M( ),2.5
A
R
+
(7
×
2 )
R
=
Mgd
M( ),6.5
B
R
+
(2 2 )
×
R
=
Mg
(9
−
d
)
SC: M( ), (
G R d
−
2.5)
=
2 (7
R
−
d
)
Solve for d, must be a numerical value
d = 5.5
Notes for question 1
N.B. Allow M marks for equations if they use Rc and RD
M1 Correct number of terms, dimensionally correct, condone sign errors and
missing g
A1 Correct equation
M1 Correct number of terms, dimensionally correct, condone sign errors and
missing g
A1 Correct equation
DM1 Dependent on previous two M marks, for solving for d
A1 d = 5.5 oe Ignore an extra m (but not M)
N.B.
If g is omitted consistently in both equations, all three A marks are available.
If they use Rg consistently in both equations, all three A marks are available.
If they have 3 equations, mark the ones that are used to obtain d.
If R and 2R are consistently the wrong way round, apply the scheme, unless an
MR gives a better total.
3u 0
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