Mathematics · Oct / Nov 2023 · Paper 51 · Question 4
YMA01
As printed in the paper
The question as text
4. [In this question i and j are horizontal unit vectors directed due east and due north respectively.] A particle P moves with constant acceleration (–λi + 2λj) m s –2, where λ is a positive constant. At time t = 0, the velocity of P is (5i – 8j) m s –1
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[2]
i
j
i
4(b)[3]
13
=
(5
−
5 )
λ
2
+ −+
( 8 10 )
λ
2
2
2
169
=
25 50
−
λ
+
25
λ
+
64 160
−
λ
+
100
λ
25
λ
2
−
42
λ
−
16
=
0
*
( 2
−
+
4 )
j seen or implied
4(c)[5]
i
(5
−
8 )
+ −
( 2
+
4 )4
i
j
i
j
8
3
8
−
−
−
e.g.
tan
1
±
,
tan
1
±
,
sin
1
±
,…
3
8
73
339o
Notes for question 4
(a)
Use of v=u+at to form a vector expression in λ and t
M1
Correct unsimplified expression with t = 5
A1
N.B. Allow use of column vectors for the M mark but not for
the A mark.
(b)
M1
Collect i’s and j’s and correct use of Pythagoras to form an
equation in λ
Correct equation
A1
A1*
cso. Expand brackets and correctly reach the GIVEN answer.
N.B. Allow
0
=
25
λ
2
−
42
λ
−
16
(c)
B1
Or column vector
Complete method to find the velocity when t = 4.
M1
Correct unsimplified expression. Note the correct velocity is
A1
j
v
= −
3
i
+
8
M1
Use their velocity vector at t = 4 with trig to find a relevant
angle.
A1
Cao. Degrees sign not required.
N.B. if they work with both values of λ, can score max all
the marks except the last one.
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