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CHAPTER 2 : FORCE & MOTION
ANSWER                       (d) (i) the first tape

2.1 Linear motion
Question 1
(a) Scalar quantity            (ii) the last tape
(b) Arrow:

                             (e) acceleration



                             (f)
(c) 700
(d) Zero
                             2.2 Motion Graph
Question 2                   Question 4
(a) A.c current              (a)(i) velocity
(b) (i) 0.02 s
    (ii) constant velocity         (ii)
    (iii)

                             (b)(i) acceleration
                             (b)(ii)
Question 3
(a) Ac current               (b)(iii)
(b) Tape chart


                             Question 5
                             (a) Displacement
                                   Time

                             (b) R : from AB


                             (c)

                                    (ii)




(c) Constant acceleration


           FIZIKMOZAC 2010
Question 6
(a) 15 m                                        (e) mPuP = (mP + mQ)v
(b) Velocity                                        (2)(1) = (2 + m) (0.4)
(c) (i) constant velocity                           m = (2 – 0.8)/ 0.4
    (ii) at rest                                       = 3 kg

Question 7                                   Question 10
(a)(i) acceleration                          (a) quantity of matter in object
(a)(ii) displacement                         (b) (i) equal but opposite direction
(b)                                              (ii) 0 = MV + mv
                                                     MV = - mv
       Section of         Type of motion
                                                 (iii) principle of conservation of
       the graph            of the car
                                                       momentum
                                             (c) (1200)v = (4)(60) = 240
                         Constant
           OA                                            v = 240/1200 = 0.2 m s-1
                         acceleration
                         Constant velocity
           AB                                Question 11
                                             (a) Principle of conservation of
                                                 momentum
Question 8
                                             (b) Momentum of the man (forward)
(a)(i) constant velocity
                                                 is equal with the momentum of
(a)(ii) constant acceleration and
                                                 the boat (backward) but in
         constant velocity after 2.01 pm
                                                 opposite direction
(b)(i) zero
                                             (c) 0 = (50 x 2) + (20)v
(b)(ii)the car moves with constant
                                                    v = - (100)/20 = - 5 m s-1
         velocity
                                             (d) Rocket
(c)

                                             Question 12
                                             (a) Mass x velocity
                                             (b) Momentum = 0.08 x 100
                                                     = 8.0 kg ms-1
                                             (c) Decrease
                                             (d) To lengthen the time of impact /
                                                 to reduce the impulsive force

                                             Question 13
                                               (a) Inelastic collision
2.4 Momentum                                   (b) (1200)(30 ) + (1000)( – 20)


            =
(a) Inelastic collision                             = (1200 + 1000)v
                                                    v = 7.27 m s-1

       .
(b)                                            (c) Inertia
                                               (d) The larger the mass and
                                                   velocity,    the    higher  the
c) mv = 2 x 100 = 2 x 1 = 2 kg ms-1

                =        = .
                                                   momentum.
                                                   Momentum = mass x velocity
            .
      (d)

                FIZIKMOZAC 2010
Truck with higher momentum          Question 17
      is difficult to stop or higher      (a) The position between the two
      impact if accident occurs               balls are equal.
                                              The distance between the two
2.6 Impulsive force                           balls increase.
Question 14                               (b) Gravitational force
   (a) Change of momentum                 (c) (i) gravitational acceleration, g
   (b) (i) Force on surface B is larger       (ii) 10 m/s2
       than surface A                         (iii) Mass does not affect g
       (ii) Time of impact on surface     (d) (i) Velocity decreases
            A is longer than on surface       (ii) moves against gravitational
            B                                      force.
       (iii) Constant/unchanged
       (iv)   The shorter the time of     Question 18
              impact, the larger the        (a) Gravitational force
              force //force Inversely       (b) Surface area of the feather is
              proportional to the time          larger
              of impact                     (c) Velocity increases constantly /
       (v)    Sponge/mattress                   constant acceleration
   (c) Higher mass high velocity,               Final velocity is constant.
       high impact / high momentum /            Final velocity for water droplet
       high impulsive force                     is higher than final velocity for
                                                feather.
Question 15                                 (d) Final velocity is inversely
(a)(i) gravitational potential energy           proportional to surface area
     (ii) kinetic energy                    (e) Graph
(b)(i) mgh = ½ mv2
          v2 = 2gh = 2 x 10 x 2.5
          v = 7.07 m s-1
(ii)



(c)(i) soft/spongy
        Lengthen the time of impact
        Reduce the impulsive force
(ii) use parachute
     To reduce velocity/momentum

Question 16
   (a) Impulsive force
   (b) High impulsive force/short
       impact time
   (c) 0.05 x 5 = 0.25 kg m/s
   (d) Use sponge, mattress / soft
       material                           2.9 Equilibrium of forces
2.8 Gravitational Force

          FIZIKMOZAC 2010
Question 19                                   (ii) R = mg – ma / R < mg
  (a) Diagram
                                            Question 22
                                               (a) Force is anything that can
                                                   move a stationary object //
                                                   stop a moving object //
                                                   change direction / shape /
                                                   speed of an object.
                                               (b) Label force


   (b) F = ma
       a = 5 / 2 = 2.5 m s-2
   (c) 5 N but in opposite direction

Question 19(spm 08)                             (c) (i) Fx = 120 cos 30⁰ = 104.4 N
(a)(i) magnitude: equal                             (ii) FY = 120 sin 30⁰ = 60 N
        Direction: opposite
  (ii) zero
                                            Question 23
  (iii) equilibrium of forces
                                              (a) Force is anything that can
(b)(i) Acceleration
                                                  move a stationary object //
  (ii) there is resultant force, force is
                                                  stop a moving object // change
        directly proportional to
                                                  direction / shape / speed of an
        acceleration
                                                  object.
Question 20
                                               (b) Diagram
  (a) 650 = 300 + P
      P = 350 N
  (b) The bicycle moves with
      constant velocity. The
      resultant force equals zero.
  (c) Velocity increases / accelerate
  (d) Inertia
  (e) The time of impact is short.             (c) Fx = 1500 x cos 60⁰ = 750 N
      High impulsive force.                    (d) (i) Fx : to make car move
                                                            forward // overcome
Question 21                                                 frictional force
    (a) 500 N
    (b) 1. At rest                                (ii) Fy : to lift the car off the
        2. moving down or up with                          ground // to move the car
        constant velocity                                  upwards// to overcome
     (c) F = R – mg                                        the weight of the car
     (d) There is a resultant force acts
         upward.
         R = mg + ma
(e)(i) decreases                            Question 24

           FIZIKMOZAC 2010
(a) 500 N                                     (ii) energy lost to heat/frictional
   (b) (i) increased                             force
       (ii) decreased
       (iii) Unchanged                     Question 28
   (c) Resultant force = 0                   (a) Gravitational potential energy
   (d) R = (46)(10) = 460 N                  (b) (i) E = mgh = (30)(10)(2.5)
        F = mg - R = ma                                 = 750 J
        50a = 500 – 460                          (ii) E = ½ mv2 = ½ (30)(4)(4)
            a = 0.8 ms-2                                 = 240 J
                                             (c) (i) Energy consumes in (b)(i) is
Question 25                                            larger than in (b)(ii)
  (a) Attraction force                            (ii) energy lost to heat /
  (b) (i) same magnitude, opposite                      frictional force
      direction
      (ii) equal in magnitude,             Question 29
      opposite direction                     (a) (i) Diagram 29.2 is further than
  (c) (i) zero                                        in diagram 29.1
      (ii) equal                                 (ii) decreases
  (d) The net force is zero, at rest         (b) Streamline
      The net force is zero, moves           (c) W = 70 x 10 = 700 J
      with constant velocity.                (d) (i) kinetic energy to
  (e) Equilibrium of forces                           gravitational potential
                                                      energy to kinetic energy
2.10 Energy                                      (ii) sound / heat
Question 26
   (a) To gain maximum kinetic             Question 30
       energy before he begins to            (a) (i) Trolley Q has less mass
       jump.                                     (ii) Newton’s second law
       Kinetic energy increases with             (b)
       velocity.
   (b) To gain elastic potential energy
       from the pole.
       Change to gravitational
       potential energy.
   (c) mgh = 60 x 10 x 52 = 3 120 J
   (d) 10 m s-2
   (e) To lengthen the time of impact
       between the athlete and the            (c) Velocity of the trolley
       mattress.                                  decreases
       So reduce the impulsive force          (d) Trolley Q moves faster along
                                                  frictionless slope
Question 27                                       (e)(i) mgh = 0.5 x 10 x 0.7
  (a) Work = force x displacement                              = 3.5 J
  (b) (i) 220 x 0.5 = 110 J
      (ii) 20 x 10 x 0.5 = 100 J                   (ii) Total energy at the top of
  (c) (i) Work done in 4(b)(i) is larger           the slope:

          FIZIKMOZAC 2010
Kinetic energy +
        gravitational potential energy
        = 3.5 + 3.0 = 6.5 J

2.11 Elasticity
Question 31
   (a) (i) Newton, N
       (ii) e is directly proportional to
            W
       (iii) Hooke’s Law
   (b) (i) 12 N
   (c) Gradient = 12 = 240 N/m
                     0.05
   (d) E = ½ Fx = ½ x 10 x 0.04 = 0.2 J
   (e)




Question 32
  (a) (i) Hooke’s Law
  (b) (i) elastic potential energy
       (ii) E = ½ x 60 x 0.08 = 2.4 J
  (c) (i) 80 N
      (ii) Spring Q
          The elastic limit for spring Q
          is 100 N which more than 80
          N.




           FIZIKMOZAC 2010

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  • 1. CHAPTER 2 : FORCE & MOTION ANSWER (d) (i) the first tape 2.1 Linear motion Question 1 (a) Scalar quantity (ii) the last tape (b) Arrow: (e) acceleration (f) (c) 700 (d) Zero 2.2 Motion Graph Question 2 Question 4 (a) A.c current (a)(i) velocity (b) (i) 0.02 s (ii) constant velocity (ii) (iii) (b)(i) acceleration (b)(ii) Question 3 (a) Ac current (b)(iii) (b) Tape chart Question 5 (a) Displacement Time (b) R : from AB (c) (ii) (c) Constant acceleration FIZIKMOZAC 2010
  • 2. Question 6 (a) 15 m (e) mPuP = (mP + mQ)v (b) Velocity (2)(1) = (2 + m) (0.4) (c) (i) constant velocity m = (2 – 0.8)/ 0.4 (ii) at rest = 3 kg Question 7 Question 10 (a)(i) acceleration (a) quantity of matter in object (a)(ii) displacement (b) (i) equal but opposite direction (b) (ii) 0 = MV + mv MV = - mv Section of Type of motion (iii) principle of conservation of the graph of the car momentum (c) (1200)v = (4)(60) = 240 Constant OA v = 240/1200 = 0.2 m s-1 acceleration Constant velocity AB Question 11 (a) Principle of conservation of momentum Question 8 (b) Momentum of the man (forward) (a)(i) constant velocity is equal with the momentum of (a)(ii) constant acceleration and the boat (backward) but in constant velocity after 2.01 pm opposite direction (b)(i) zero (c) 0 = (50 x 2) + (20)v (b)(ii)the car moves with constant v = - (100)/20 = - 5 m s-1 velocity (d) Rocket (c) Question 12 (a) Mass x velocity (b) Momentum = 0.08 x 100 = 8.0 kg ms-1 (c) Decrease (d) To lengthen the time of impact / to reduce the impulsive force Question 13 (a) Inelastic collision 2.4 Momentum (b) (1200)(30 ) + (1000)( – 20) = (a) Inelastic collision = (1200 + 1000)v v = 7.27 m s-1 . (b) (c) Inertia (d) The larger the mass and velocity, the higher the c) mv = 2 x 100 = 2 x 1 = 2 kg ms-1 = = . momentum. Momentum = mass x velocity . (d) FIZIKMOZAC 2010
  • 3. Truck with higher momentum Question 17 is difficult to stop or higher (a) The position between the two impact if accident occurs balls are equal. The distance between the two 2.6 Impulsive force balls increase. Question 14 (b) Gravitational force (a) Change of momentum (c) (i) gravitational acceleration, g (b) (i) Force on surface B is larger (ii) 10 m/s2 than surface A (iii) Mass does not affect g (ii) Time of impact on surface (d) (i) Velocity decreases A is longer than on surface (ii) moves against gravitational B force. (iii) Constant/unchanged (iv) The shorter the time of Question 18 impact, the larger the (a) Gravitational force force //force Inversely (b) Surface area of the feather is proportional to the time larger of impact (c) Velocity increases constantly / (v) Sponge/mattress constant acceleration (c) Higher mass high velocity, Final velocity is constant. high impact / high momentum / Final velocity for water droplet high impulsive force is higher than final velocity for feather. Question 15 (d) Final velocity is inversely (a)(i) gravitational potential energy proportional to surface area (ii) kinetic energy (e) Graph (b)(i) mgh = ½ mv2 v2 = 2gh = 2 x 10 x 2.5 v = 7.07 m s-1 (ii) (c)(i) soft/spongy Lengthen the time of impact Reduce the impulsive force (ii) use parachute To reduce velocity/momentum Question 16 (a) Impulsive force (b) High impulsive force/short impact time (c) 0.05 x 5 = 0.25 kg m/s (d) Use sponge, mattress / soft material 2.9 Equilibrium of forces 2.8 Gravitational Force FIZIKMOZAC 2010
  • 4. Question 19 (ii) R = mg – ma / R < mg (a) Diagram Question 22 (a) Force is anything that can move a stationary object // stop a moving object // change direction / shape / speed of an object. (b) Label force (b) F = ma a = 5 / 2 = 2.5 m s-2 (c) 5 N but in opposite direction Question 19(spm 08) (c) (i) Fx = 120 cos 30⁰ = 104.4 N (a)(i) magnitude: equal (ii) FY = 120 sin 30⁰ = 60 N Direction: opposite (ii) zero Question 23 (iii) equilibrium of forces (a) Force is anything that can (b)(i) Acceleration move a stationary object // (ii) there is resultant force, force is stop a moving object // change directly proportional to direction / shape / speed of an acceleration object. Question 20 (b) Diagram (a) 650 = 300 + P P = 350 N (b) The bicycle moves with constant velocity. The resultant force equals zero. (c) Velocity increases / accelerate (d) Inertia (e) The time of impact is short. (c) Fx = 1500 x cos 60⁰ = 750 N High impulsive force. (d) (i) Fx : to make car move forward // overcome Question 21 frictional force (a) 500 N (b) 1. At rest (ii) Fy : to lift the car off the 2. moving down or up with ground // to move the car constant velocity upwards// to overcome (c) F = R – mg the weight of the car (d) There is a resultant force acts upward. R = mg + ma (e)(i) decreases Question 24 FIZIKMOZAC 2010
  • 5. (a) 500 N (ii) energy lost to heat/frictional (b) (i) increased force (ii) decreased (iii) Unchanged Question 28 (c) Resultant force = 0 (a) Gravitational potential energy (d) R = (46)(10) = 460 N (b) (i) E = mgh = (30)(10)(2.5) F = mg - R = ma = 750 J 50a = 500 – 460 (ii) E = ½ mv2 = ½ (30)(4)(4) a = 0.8 ms-2 = 240 J (c) (i) Energy consumes in (b)(i) is Question 25 larger than in (b)(ii) (a) Attraction force (ii) energy lost to heat / (b) (i) same magnitude, opposite frictional force direction (ii) equal in magnitude, Question 29 opposite direction (a) (i) Diagram 29.2 is further than (c) (i) zero in diagram 29.1 (ii) equal (ii) decreases (d) The net force is zero, at rest (b) Streamline The net force is zero, moves (c) W = 70 x 10 = 700 J with constant velocity. (d) (i) kinetic energy to (e) Equilibrium of forces gravitational potential energy to kinetic energy 2.10 Energy (ii) sound / heat Question 26 (a) To gain maximum kinetic Question 30 energy before he begins to (a) (i) Trolley Q has less mass jump. (ii) Newton’s second law Kinetic energy increases with (b) velocity. (b) To gain elastic potential energy from the pole. Change to gravitational potential energy. (c) mgh = 60 x 10 x 52 = 3 120 J (d) 10 m s-2 (e) To lengthen the time of impact between the athlete and the (c) Velocity of the trolley mattress. decreases So reduce the impulsive force (d) Trolley Q moves faster along frictionless slope Question 27 (e)(i) mgh = 0.5 x 10 x 0.7 (a) Work = force x displacement = 3.5 J (b) (i) 220 x 0.5 = 110 J (ii) 20 x 10 x 0.5 = 100 J (ii) Total energy at the top of (c) (i) Work done in 4(b)(i) is larger the slope: FIZIKMOZAC 2010
  • 6. Kinetic energy + gravitational potential energy = 3.5 + 3.0 = 6.5 J 2.11 Elasticity Question 31 (a) (i) Newton, N (ii) e is directly proportional to W (iii) Hooke’s Law (b) (i) 12 N (c) Gradient = 12 = 240 N/m 0.05 (d) E = ½ Fx = ½ x 10 x 0.04 = 0.2 J (e) Question 32 (a) (i) Hooke’s Law (b) (i) elastic potential energy (ii) E = ½ x 60 x 0.08 = 2.4 J (c) (i) 80 N (ii) Spring Q The elastic limit for spring Q is 100 N which more than 80 N. FIZIKMOZAC 2010