Mathematics · Oct / Nov 2021 · Paper 51 · Question 5
YMA01
As printed in the paper
The question as text
5. P α Q Figure 2 A small bead of mass 0.2 kg is attached to the end P of a light rod PQ. The bead is threaded onto a fixed vertical rough wire. The bead is held in equilibrium with the rod PQ inclined to the wire at an angle α, where tan α = 4 3 , as shown in Figure 2. The thrust in the rod is T newtons. The bead is modelled as a particle.
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5(a)[3]
( )
↑
±
F
=
0.2
g
−
2.5cos
α
Allow use of (
µ
R
) for F
F = 0.46 (N) oe including fractions , upwards
5(b)[7]
( )
↑
F
+
0.2
g
=
6.125cos
α
(
→
)
R
=
6.125sin
α
(4.9)
F
=
µ
R
Solve for µ
µ = 0.35 oe including fractions.
N.B. If F and R are interchanged in their equations, max B1 can be scored.
Notes for question 5
M1 Correct no. of terms, condone sin/cos confusion and sign errors, allow if
5(a)
they have T instead of 2.5
A1 Correct equation . Allow + F or −F
A1 Need both magnitude (must be positive) and direction
M1 Correct terms, condone sin/cos confusion and sign errors errors
5(b)
allow if they have T instead of 6.125 (but M0 if using T = 2.5)
A1 Correct equation
M1 Correct terms, condone sin/cos confusion and sign error
allow if they have T instead of 6.125 (but M0 if using T = 2.5)
A1 Correct equation
B1 F
=
µ
R
seen but B0 if they use a value for R found in (a)
DM1 Dependent on both M’s
A1 cao
0
=
u
−
9.8 2.5
×
oe using gradient of graph.
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