Mathematics · Oct / Nov 2022 · Paper 51 · Question 6
YMA01
As printed in the paper
The question as text
6. [In this question, i and j are horizontal unit vectors.] A particle A of mass 0.5 kg is at rest on a smooth horizontal plane. At time t = 0, two forces, F1 = (–3i + 2j) N and F2 = (pi + qj) N, where p and q are constants, are applied to A. Given that A moves in the direction of the vector (i – 2j),
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Show the answer and mark scheme
6(a)[4]
i
p
i
q
p
i
q
( 3
−+
)
1
p
−
=
(2
+
q
)
2
2
+
−
4
=
0
* Allow 0
=
2
+
−
4
but nothing else
p
q
p
q
p = 5 => q = 6
− => Resultant force = (2
i
−
4 )
j
6(b)[5]
(2
−
4 )
i
j = 0.5a
v = (4
−
8 ) 4
×
i
j
Speed =
16
2
( 32)
2
= 1280
= 36 (
m s−) or better
1
+ −
=
16 5
Notes for question 6
M1 For adding and collecting i’s and j’s.
6(a)
N.B. Could be implied by p = 4 and q = 4
M1 Using ratios oe to set up an equation in p and q only, allow the
ratio the wrong way round.
M0 if they write down: 3
−+
p
=
1 and 2
+
q
= − and NEVER use
2
ratios, but ignore these equations if they go on to use ratios
A1 Correct equation
A1* Correct answer correctly obtained
6(b)
B1 Correct resultant force seen
M1 Use of F = ma OR F = ma where F (F) is their resultant (must
have attempted to add the two forces) (M0 if they include g)
M1 Use of v = at OR v = at with t = 4 where a or a is their
acceleration. (M0 if or u
u
is nonzero)
M1 Use of Pythagoras to find magnitude of v OR a OR F, including
square root
N.B. The above 3 steps may appear in any order but must be entered on
ePen in the order as above.
A1 Any equivalent surd or correct to at least 2 SF
=
μ
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