Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9709/41 Cambridge International AS & A Level – Mark Scheme May/June 2023
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2023 Page 5 of 13
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | R0.1gcos60[0.5] | B1 | Correctly resolving perpendicular to the plane for Q.
F 0.70.1gcos60[0.35] | M1 | Use of F R for Q where R is a component of
weight but not mass; allow sin/cos mix.
For whole system: LHS of Newton’s second law:
0.2gsin600.1gsin60F [0.866...F]
Or separately for P and Q:
0.2gsin60T(0.2a) and T 0.1gsin60F(0.1a), and eliminate T to
get 0.2gsin600.1gsin60F(0.3a) | M1 | Complete method to determine the resultant force for
the whole system. Allow sign errors and sin/cos mix,
but must include all required terms and be
dimensionally correct. If considering either the while
system or P and Q separately then ignore the RHS of
their Newton’s second law equations.
3
As 0.350the particles do move.
2 | A1 | Correct indication (with no incorrect working) that the
resultant force is positive (e.g. 0.8660...0.350or
0.516… (to at least 1 sf) which is positive) together
with a correct conclusion. Candidates may calculate
10 37
the acceleration which is 1.72008... and
6
then say that the particles are moving.
4
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | Attempt to use Newton’s second law for P:
0.2gsin60( 31)0.2a | M1 | Allow sign errors, sin/cos mix. but must be
dimensionally consistent.
a5(ms-2) | A1
Newton’s second law for system:0.2gsin600.1gsin0.3(5)
or Newton’s second law for Q: ( 31)0.1gsin0.1(5) | M1 | Attempt Newton’s second law for Q, or for the whole
system. Allow sign errors, sin/cos mix, but must be
dimensionally consistent.
13.4 | A1 | 13.41784…
4
Question | Answer | Marks | Guidance