Mathematics · May / June 2022 · Paper 51 · Question 4
YMA01
As printed in the paper
The question as text
4. P N 30° Figure 1 A small block of mass 5 kg lies at rest on a rough horizontal plane. The coefficient of friction between the block and the plane is 3 7 A force of magnitude P newtons is applied to the block in a direction which makes an angle of 30° with the plane, as shown in Figure 1. The block is modelled as a particle. Given that P = 14
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Show the answer and mark scheme
4(a)[6]
R
g
R = 42 (N)
(Max Friction =) 3
7 ×
42
= 18 (N) (18 only, with no working can
score this M mark)
Horiz cpt of P = 14cos30o = 12.124… and 12 < 18 (their max friction)
They must be comparing with a maximum friction i.e. the word
‘maximum’ oe must have been clearly stated somewhere.
N.B. M0 if they state or imply that the friction acting on the block is 18
N.
Friction = 12 or better (N) and block doesn’t move
4(b)[6]
( ) sin
P
30
o
+
S
=
5
g
3
(
→
) cos
3
0
o
=
(Allow M1A0 if they use the max friction from (a)
P
S
7
or 3
×
wrong value for
) (allow M1A0 for
cos 0o
3
=
)
S
P
F
7
Solve for P
P = 19 or 19.4 (N)
Notes for question 4
4(a)
M1 Correct no. of terms, condone sin/cos confusion and sign errors
A1 Correct equation in R only.
A1 Correct value (seen or implied)
3
M1 Use of
with their R substituted.
F
=
R
7
M1 Condone sin/cos confusion
A1 cao and any equivalent correct statement and justification
4(b)
M1 Correct no. of terms, condone sin/cos confusion and sign errors
A1 Correct equation
M1 Correct no. of terms, condone sin/cos confusion and sign errors
A1 Correct equation
DM1 Dependent on both M marks; must be solving two equations in P
and one other unknown
A1 cao
RC RD
C D B
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