Mathematics · Oct / Nov 2022 · Paper 51 · Question 3
YMA01
As printed in the paper
The question as text
3. P X N α Figure 2 A rough plane is inclined to the horizontal at an angle α, where tan α = 3 4 A particle P of mass 2 kg is held in equilibrium on the plane by a horizontal force of magnitude X newtons, as shown in Figure 2. The force acts in a vertical plane which contains a line of greatest slope of the inclined plane.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[3]
2g
(
),14.7cos
α
=
2 sin
α
+
(could be – F)
g
F
(
→
),14.7
+
cos
α
=
sin
α
F
R
OR:
AND eliminate R to give an
( ),
R
cos
α
+
F
sin
α
=
2
g
equation in F only.
Verificaton methods
14.7cos
α
=
(11.76)
=
2 sin
α
g
(i.e. verification that X = 14.7 => F = 0)
OR:
cos
α
=
2 sin
α
=> X = 14.7
X
g
(i.e. verification that F = 0 => X =14.7)
so F = 0* oe
S
F1
X
2g
F1 = 0.5S
3(b)[8]
Two equations taken from:
(
),
cos
2 sin
X
α
+
F
=
g
α
1
(
),
S
=
X
sin
α
+
2 cos
g
α
(
→
),
X
+
F
cos
α
=
S
sin
α
1
( ),
cos
α
+
sin
α
=
2
S
F
g
1
N.B. M0 for both equations if they put X = 14.7 anywhere
X = 4g/11, 3.6 or 3.56 or 3.57
N.B. Enter marks for the equations on ePen in the order in which
they appear above.
Notes for question 3
M1 Equation in F only, correct no of terms, condone sign errors and
sin/cos confusion (M0 if they use F = 0.5R)
3(a)
N.B. Allow the equation without F
Allow use of m instead of 2 for the Mmark
A1 Correct equation
A1* cao Must state a conclusion or ,
if verifying, must state clearly
X = 14.7 => F = 0 OR F = 0 => X = 14.7
3(b)
B1 F1 = 0.5S seen e.g. on a diagram (even if wrong direction)
M1 A resolution, correct no of terms, condone sign errors and sin/cos
confusion
Allow use of m instead of 2 for the A mark
A2 Correct equation, 1 each error
M1 A resolution, correct no of terms, condone sign errors and sin/cos
confusion
Allow use of m instead of 2 for the A mark
A2 Correct equation, 1 each error
A1 cao
3616
−
250
−
565
−
226
=
250
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