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Mathematics · May / June 2024 · Paper 51 · Question 8

YMA01

As printed in the paper

The question as text

8. P B (3m) A (2m) h α Figure 4 One end of a light inextensible string is attached to a particle A of mass 2m. The other end of the string is attached to a particle B of mass 3m. Particle A is held at rest on a rough plane which is inclined to horizontal ground at an angle α, where tan 5 12 The string passes over a small smooth pulley P which is fixed at the top of the plane. Particle B hangs vertically below P with the string taut, at a height h above the ground, as shown in Figure 4. The part of the string between A and P lies along a line of greatest slope of the plane. The two particles, the string and the pulley all lie in the same vertical plane. The coefficient of friction between A and the plane is 11 36 The particle A is released from rest and begins to move up the plane.

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme
  • 8(a)[3]

    R

    mg

    11

    12

    22

    mg

    =

    ×

    2

    ×

    =

    *

    F

    mg

    36

    13

    39

  • 8(b)[2]

    3

    −

    =

    3

    mg

    T

    ma

  • 8(c)[4]

    22

    mg

    4

    mg

    −

    −

    2

    sin

    α

    =

    2

    (

    −

    =

    2

    )

    T

    mg

    ma

    T

    ma

    39

    3

    22

    mg

    OR:

    3

    −

    −

    2

    sin

    α

    =

    5

    mg

    mg

    ma

    39

    Solve for a in terms of g

    N.B. Must reach a = kg from their equations

    a = 1

    3 g *

    2

    gh

  • 8(d)[6]

    v =

    2

    3

    22

    mg

    2

    sin

    2

    OR PE Gain = 2

    sin

    α

    −

    −

    mg

    α

    = ±

    ma

    mgd

    39

    2

    = 10

    g

    mgd

    =

    ±

    a

    3

    13

    2

    gh

    2

    g

    0

    2

    =

    −

    ×

    ×

    d

    3

    3

    22

    1

    2

    10

    mgd

    gh

    mgd

    =

    ×

    2

    m

    ×

    −

    39

    2

    3

    13

    1

    1

    =

    =

    d

    h

    d

    h

    2

    2

    Total distance = 1

    3

    +

    =

    h

    h

    h

    2

    2

    Notes

    N.B. For (a) and (c), if fractions for the trig ratios are not seen or

    implied, only penalise once.

    M1: Resolve perp to the plane, correct terms, condone cos/sin confusion

    8(a)

    and sign errors. Allow if they use

    cos(

    12

    ) or similar.

    13

    A1: Correct equation

    A1*: Correct given answer correctly obtained, must see 12

    13

    8(b)

    M1:Equation of motion for B, correct terms, condone sign errors

    A1: Correct equation

    8(c)

    M1:Equation of motion for A OR whole system, correct terms but

    allow F, condone sign errors and sin/cos confusion.

    Allow if they

    use

    sin(

    13)

    5

    or similar.

    A1: Correct equation

    M1: Solve for a in terms of g, need to see trig substituted and must be

    solving 2 equations in T and a OR using a whole system equation with

    correct terms.

    A1*: Correct given answer correctly obtained, must see 5

    13

    N.B. Allow a full verification, using the equations of motion for A and

    B OR the whole system equation.

    8(d)

    B1: seen or implied

    M1:Equation of motion for A, correct terms, condone sign errors and

    sin/cos confusion

    A1: Correct acceleration or deceleration of A

    M1: Complete method to find an equation in d, g and h only, using a

    new calculated a

    A1: cao

    A1ft: Their d (which must be a multiple of h) + h.

    N.B. This mark is only dependent on the previous M.

    OR: Using Workenergy

    B1: cao

    M1: PE gain of A, condone sign errors and sin/cos confusion

    A1: Correct PE gain

    M1: Use of workenergy principle to obtain an equation in m, d, g and h

    only, using their PE expression

    A1: cao

    A1ft: Their d (which must be a multiple of h) + h, with the final answer

    of the form kh.

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