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Mathematics · Oct / Nov 2021 · Paper 51 · Question 6

YMA01

As printed in the paper

The question as text

6. v m s–1 O 2.5 7 t seconds Figure 3 A small ball is thrown vertically upwards at time t = 0 from a point A which is above horizontal ground. The ball hits the ground 7 s later. The ball is modelled as a particle moving freely under gravity. The velocity‑time graph shown in Figure 3 represents the motion of the ball for 0  t  7

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

Show the answer and mark scheme

  • 6(a)[2]

    Allow g or 9.81 instead of 9.8

    u =

    24.5

    or 25 (

    m s−) Allow 2.5g

    1

    Many other methods

  • 6(b)[3]

    1

    s =

    24.5 2

    × +

    ×

    9.8 2

    ×

    2

    2

    1

    24.5 7

    9.8 7

    2

    OR

    s =

    × −

    ×

    ×

    2

    1

    1

    OR

    s =

    ×

    9.8 4.5

    ×

    2

    −

    (24.5 2.5

    ×

    +

    × −

    ( 9.8) 2.5 )

    ×

    2

    2

    2

    1

    1

    OR s =

    ×

    9.8 4.5

    ×

    2

    −

    ×

    9.8 2.5

    ×

    2

    2

    2

    Many other methods, using suvat and/or the graph (e.g. similar triangles and

    area under graph)

    Allow g or 9.81 instead of 9.8 in all equations.

    68.6 or 69 (m)

    Notes for question 6

    For use of g = 9.81, which will only affect the final A mark in each part,

    penalise once for whole question

    M1 for complete method using suvat or the graph to produce an equation in u

    6(a)

    only, with correct number of terms, condone sign errors.

    A1 cao (must be positive)

    M1 Complete method to give a final displacement,

    6(b)

    condone sign errors within a suvat equation.

    A1ft Correct equation ft on their u

    A1 cao

    T

    −

    2

    mg

    sin

    α

    −

    F

    =

    2

    ma

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