Exam-style Questions: Equations of Motion

  1. Fig. 1.1 shows a long rope that is tied at one end to a high support. A woman swings forwards and backwards across a pool using the other end of the rope.

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    Fig. 1.2 shows the variation with time t of the displacement x, of the woman from A to B and back to A.

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    a) State what the gradient of the graph represents and explain why the graph shows both negative and positive gradients.

    (2 Marks)

    b) Mark on Fig. 1.2 with a corss

    • a position where the speed of the woman is zero (label this cross Z)
    • a position where the speed of the woman is a maximum (label this cross M)

    (2 Marks)

    c) Use Fig. 1.2 to calculate the maximum positive spee of the woman. Show on the Fig. 1.2 how you determined your answer.

    Maximum speed = ................... m s1-

    (3 Marks)

    (Marks available: 7)

  2. Fig. 3.1 shows the path of a golf ball from the time it ends contact with a golf club, point C, until it hits the ground at G. Assume that there is no air resistance.

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    The ball leaves the club with a velcotiy 42 m s-1 at an angle of 36° to the horizontal.

    a) Show that the horizontal component of the velocity is 34 m s-1

    (1 Mark)

    b) The distance C to G is 170m. Sow that the time taken for the ball to travel from C to G is 5.0s.

    (1 Mark)

    c) Calculate

    i) the initial vertical component of the velocity

    vertical velocity component = .................... m s-1

    (2 Marks)

    ii) the maximum height reached.

    maximum height = .................... m

    (3 Marks)

    d) The ball has a mass of 50g. Calculate the kinetic energy of the ball at maximum height.

    kinetic energy = .................... J

    (3 Marks)

    e) ON Fig. 3.1 sketch the path of the golf ball if air resistance is assumed not to be negligible.

    (2 Marks)

    f) Explain the shape of your sketch

    (1 Mark)

    (Marks available: 13)

  3. An aircraft of total mass 1.5 x 105 kg accelerates, at maximum thrust from the engines, from rest along a runwau for 25s before reaching the required speed to take-off of 65m s-1.

    Assume that the acceleration of the aircraft is constant.

    Calculate

    a) the acceleration of the aircraft

    acceleration = .......................... m s-2

    (3 Marks)

    b) the force acting on the aircraft to produce this acceleration

    force = .......................... N

    (2 Marks)

    c) the distance travelled by the aircraft in this time.

    distance = .......................... m

    (2 Marks)

    (Marks available: 7 )

Answer

Answer outline and marking scheme for question:

  1. a) velocity

    travels in two oposite directions or equivalent words / increasing and decreasing displacement

    (2 Marks)

    b) Z any peak or trough / A / B / 0 / 3.0 / 6.0s

    M any point where gradient is a maximum (1.0 - 1.6 or 4.4 - 5.0 s)

    If M and Z are given on Fig. 1.1 then max 1

    (2 Marks)

    c) tangent to curve drawn

    values given correct from graph

    answers correct for maximum in range of 1.3 to 1.5

    (3 Marks)

    (Marks available: 7)

  2. a) Vx = 42 cos36

    = 34 (m s-1)

    (1 Mark)

    b) time = 170 / 34

    = 5 (s)

    (1 Mark)

    c) i) Vy = V sin36 / cos54

    = 24.7 (25 allowed)

    (2 Marks)

    ii) v2 = u2 + 2as / s = ut + ½ at2 / S = [(u + v)/2]t

    0 = (24.7)2 - 2 x 9.81 x s ecf from c) (i)

    s = 3.(1) (m) allow 31 to 32 as answer depends on sig figs and equation used.

    (3 Marks)

    d) k.e = ½ m v2

    = 0.5 x 50 x 10-3 x (34)2

    = 28.9 (J)

    (3 Marks)

    e) i) height less

    range less

    (2 Marks)

    ii) force acting against the motion and this effect on the acceleration / ball does not work against air resistance / k.e. reduced due to air resistance / velocity is reduced by air resistance

    (1 Mark)

    (Marks available: 13)

  3. a) acceleration = (v-u) / t

    = (65 - 0) / 25

    = 2.6 (m s-2)

    (3 Marks)

    b) force = ma

    = 150000 x 2.6

    = 390000 (N) / 3.9 x 105

    (2 Marks)

    c) distance = (u+v) t / 2

    = (0 + 65) x 25 / 2

    = 813 (m) (allow 810)

    (2 Marks)

    (Marks available: 7)