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)
[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
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
[8]
2. (a) (i) acceleration (1)
(ii) both represent acceleration of free fall
[or same acceleration] (1)
[or same acceleration] (1)
(iii) height/distance ball is dropped from above the ground
[or displacement] (1)
[or displacement] (1)
(iv) moving in the opposite direction (1)
(v) kinetic energy is lost in the collision
[or inelastic collision] (1) 5
[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)
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)
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)
= –1.3 kg m s–1 (1) (1.25 kg m s–1)
(allow C.E. from (b) (i) and (b)(ii))
(iv) F =
= 13 N (1)
(allow C.E. from (b)(iii)) 8
[13]
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)
= 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)
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)
of force or 2.0 × 103 × 10 × 103 cos22°] (1)
= 1.8(5) × 107 J (1)
(iii) power =
= 3.6 × 104 W (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
power = force × vel. component East = 2.0 × 103 × 20 cos22° (1)
= 3.6 × 104 W (1) max 6
(b) return time =
average speed =
average speed =
[8]
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)
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 =
mass =
[8]
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
[9]
6. (a) (i) rate of change of velocity
[or a = ] (1)
[or a =
(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)
[8]
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 reaction of wall (1)
C tension in wire (1) max 5
(b)
use of mg (1)
clockwise moments 118 × 0.375 (1)
use of mg (1)
clockwise moments 118 × 0.375 (1)
= anticlockwise moments (Tcos40° (1)) × 0.750 (1)
T = 77 N (1) max 4
T = 77 N (1) max 4
[9]
8. (a) (i) gradient =
(ii) distance is area under graph (to t = 0.1 s)
or
or
(b) (i) T – mg = ma [or T = 1500(9.8+3.0)] (1)
= 1.9 × 104 N (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
= 3.7[3.8] × 104 W (1) 2
[9]
9. (i) a =
F = ma =1.1 × 105 N (1)
(ii) D = 236 m s–1
a =
(iii) sone =
stwo =
total distance = 1384 m (1)
[6]
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)
(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
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 =
[14]
