Mathematics · May / June 2024 · Paper 51 · Question 3
YMA01
As printed in the paper
The question as text
3. Q (2 m) P (4 m) Figure 1 Two particles, P and Q, have masses 4m and 2m respectively. The particles are connected by a light inextensible string. A second light inextensible string has one end attached to Q. Both strings are taut and vertical, as shown in Figure 1. The particles are accelerating vertically downwards. Given that the tension in the string connecting the two particles is 3mg, find, in terms of m and g, the tension in the upper string. (6) 6 7
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Show the answer and mark scheme
For P: 4
3
4
mg
−
mg
=
ma
For both: 4
mg
+
2
mg
−
T
=
(4
+
2)
ma
=
6
ma
Any two of these
For Q: 2
+
3
−
=
2
mg
mg
T
ma
Solve for T
mg , 4.5mg oe
9
2
Notes
N.B. Use the mass in the ‘ma’ term in each equation of motion to
determine to which part of the system the equation refers.
Allow a replaced by a
− in both equations.
Enter marks on ePEN in the order in which equations appear.
M1: Equation of motion with correct terms, condone sign errors
A1: Correct equation
M1: Equation of motion with correct terms, condone sign errors
A1: Correct equation
DM1: Dependent on both M’s, for solving for T (must be in terms of
mg)
A1: Any equivalent expression of the form kmg.
N.B. For the P and Q equations, allow M1 if they have T instead of 3mg
and it’s VERY clear that T is the tension in the connecting string and
not the tension they are trying to find.
( ),
×
4.5 1.2
+
×
2
=
4
×
1.5
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