PHYA2 answers to booklet vectors moments etc

PHYA2 answers vectors, moments, graphing motion

1.         (a)        (i)         resultant force acting on tray is zero [or P + W = Q] (1)resultant torque is zero
[or correct moments equation
or anticlockwise moments = clockwise moments] (1)

(ii)        W= 0.12 × 9.81 = 1.2N (1) (1.18N)

(iii)       (taking moments about P gives)
Q × 0.1 = 0.12 × 9.81 × 0.25 (1)Q = 2.9 N (2.94 N) (1)P = 2.9 – 1.2 = 1.7 N (1) (or 2.94 – 1.18 = 1.76 N)
                (allow C.E. for values of W and Q)                                                                       6

(b)        placed at Q (1)no additional turning moment about Q (1)                                                                                      2

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2.         (a)        (i)         acceleration  (1)

(ii)        both represent acceleration of free fall
[or same acceleration]  (1)

(iii)       height/distance ball is dropped from above the ground
[or displacement]  (1)

(iv)      moving in the opposite direction  (1)

(v)       kinetic energy is lost in the collision
[or inelastic collision]  (1)                                                                                                        5

(b)        (i)         v2 = 2 × 9.81 × 1.2  (1)
v = 4.9 m s–1  (1)    (4.85 m s–1)

(ii)        u2 = 2 × 9.81 ×0.75  (1)
u = 3.8 m s–1  (1)    (3.84 m s–1)

(iii)       change in momentum = 0.15 × 3.84 – 0.15 × 4.85  (1)
= –1.3 kg m s–1  (1)            (1.25 kg m s–1)

             (allow C.E. from (b) (i) and (b)(ii))

(iv)      F =   (1)

            = 13 N  (1)

             (allow C.E. from (b)(iii))                                                                                              8

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3.         (a)        (i)         component velocity North = 20 cos68° (1)
= 7.5 m s–1
which is supplied by wind (1)
by triangle of velocities [or by components] (aircraft must
point East) (1)

            alternative (a)(i)
triangle or parallelogram of velocities (1)
find angle between aircraft component and wind using sine and cosine
formulae – prove 90° (1) (1)

(ii)        work done = Fs cosq [or force × distance moved in direction
of force or 2.0 × 103 × 10 × 103 cos22°] (1)
= 1.8(5) × 107 J (1)

(iii)       power =  = 1.8(5) × 107 ÷ (1)
= 3.6 × 104 W (1)

            alternative (iii)
power = force × vel. component East = 2.0 × 103 × 20 cos22° (1)
              = 3.6 × 104 W (1)                                                                                               max 6

(b)        return time =  = 714 s \ total time = 1214 s (1)
average speed =  = 16[16.5]m s–1 (1)                                                                                 2

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4.         (a)        (i)         a force multiplied by a distance
perpendicular distance from line of action of the force to the
point P (1)
                         (stated or from diagram)

(ii)        N m (1)                                                                                                                                       3

(b)        (i)         force up at pivot (1)

two downward forces at correct points (1)

(ii)        weight of tube ( = mg) = 12.0 × 9.81 = 118 N (1)

(iii)       moments about pivot equated (1)118 × 1.6 = W × 0.3 gives W = 629 (N) (1)(allow e.c.f. for weight in (ii))
mass =  = 64.1 kg (1)               (allow e c f for W)                                                      5

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5.         (a)        (i)        

two forces opposing (1)forces parallel (1)s correct (1)

(ii)        N m (1)                                                                                                                                       4

(b)        (i)         anticlockwise moments = clockwise moments (1)

(ii)        weight of beam acts at centre (1)this is through the pivot (1)                                                                                                     3

(c)        (equating moments gives) 400 × 1.0 = 200 × 0.50 + 250 × d (1)\400 – 100 = 250 × d and d = 1.2 m (1)                                                                                         2

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6.         (a)        (i)         rate of change of velocity
[or a = ] (1)

(ii)        (acceleration) has (magnitude and) direction (1)                                                               2

(b)        (i)         (acceleration) is the gradient (or slope) of the graph (1)

(ii)        (displacement) is the area (under the graph)                                                                       2

(c)       
    4

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7.         (a)        (i)        

n.b. B must make an appreciable angle with wall and bar

(ii)        A          weight of sign and bar (accept gravity) (1)
B          reaction of wall (1)
C          tension in wire (1)                                                                                               max 5

(b)       
use of mg (1)
clockwise moments 118 × 0.375 (1)

             = anticlockwise moments (Tcos40° (1)) × 0.750 (1)
T = 77 N (1)                                                                                                                                   max 4

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8.         (a)        (i)         gradient =  = 3.0 ms–2 (1)

(ii)        distance is area under graph (to t = 0.1 s)
or  × 0.7 × 2.1 0.3  (1) = 1.4(2) m (1)                                                        3

(b)        (i)         T – mg = ma [or T = 1500(9.8+3.0)] (1)
= 1.9 × 104 N (1)

T = mg = l.5 × 104 N (1)

(ii)        EF (1)                                                                                                                                          4

(c)        power = Fu or l.5 × 104 × 2.5 (1)
= 3.7[3.8] × 104 W (1)                                                                                                                          2

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9.         (i)         a =  = 11 ms–2 (1)

F = ma =1.1 × 105 N (1)

(ii)        D  = 236 m s–1

a =  = 29.5 ms–2 (1)

(iii)       sone =  × t = × 4.0 = 88m (1)

stwo =  × t =  × 8.0 (1) = 1296(m) (1)

total distance = 1384 m (1)

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10.       (a)        (i)         region A: uniform acceleration

                                                                     (or (free-fall) acceleration = g( = 9.8(i) m s–2))

force acting on parachutist is entirely his weight

             (or other forces are very small) (1)

(ii)        region B: speed is still increasing

acceleration is decreasing (2)                  (any two)

because frictional (drag) forces become significant
(at higher speeds)

(iii)       region C: uniform speed (50 m s–1)

because resultant force on parachutist is zero (2) (any two)

weight balanced exactly by resistive force upwards                                                          6
                                                                                                                                 QWC

(b)        deceleration is gradient of the graph (at t = 13s) (1)

(e.g. 20/1 or 40/2) = 20 m s–2 (1)                                                                                                        2

(c)        distance = area under graph (1)suitable method used to determine area (e.g. counting squares) (1)with a suitable scaling factor (e.g. area of each square = 5 m2) (1)distance=335m (±15m) (1)                                                                                                                 4

(d)        (i)         speed = Ö(5.02 + 3.02) = 5.8 m s–1 (1)

(ii)        tan q =     gives q = 31°(1)                                                                                                  2

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