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Mathematics · May / June 2023 · Paper 51 · Question 1

YMA01

As printed in the paper

The question as text

1. A particle A has mass 4 kg and a particle B has mass 2 kg. The particles move towards each other in opposite directions along the same straight line on a smooth horizontal table and collide directly. Immediately before the collision, the speed of A is 2u m s –1 and the speed of B is 3u m s –1 Immediately after the collision, the speed of B is 2u m s –1 The direction of motion of B is reversed by the collision.

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

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

    2u 3u

    A (4 kg) B (2 kg)

    v 2u

    (

    ) (

    )

    (

    )

    CLM:

    4 2

    ×

    u

    + −

    3

    u

    ×

    2

    =

    4

    v

    +

    2 2

    ×

    u

    OR

    Equating impulses: (

    )

    2 2 3

    u

    −−

    u

    =

    4(

    −−−

    v

    2 )

    u

    1

    (

    )

    2 u

    m s

    −

    1

  • 1(b)[1]

    The direction of motion is reversed.

    (

    )

    I

    = ±

    2 2 3

    u

    −−

    u

    For B:

  • 1(c)[3]

    u

    OR For A:

    I

    = ±

    4

    2

    −−

    u

    2

    I

    =

    10 Ns

    u

    or 10u

    kgms−

    1

    Notes

    (a)

    M1

    Dimensionally correct CLM equation or equating of impulses equation.

    Allow consistent extra g’s. Ignore sign errors. May be +v or v

    −

    A1

    Correct unsimplified equation

    A1

    Cao. Must be positive.

    (b)

    B1

    Accept opposite direction. Do not accept changed or to the left or

    backwards, away from B

    N.B. This mark is dependent on correctly obtaining 1

    1

    2 u or

    2 u

    −

    in (a)

    (c)

    M1

    Dimensionally correct impulsemomentum equation using A or B.

    Condone sign errors with appropriate velocities. M0 if g is included

    A1

    Correct unsimplified equation

    A1

    Cao with units. Accept kg m/s

    F

    +

    (3

    c

    i

    +

    4 )

    c

    j

    + −

    ( 14

    i

    +

    7 )

    j

    =

    0 oe

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