theonlinephysicstutor.com
Mark schemes
B
1 [1]
A
2 [1]
B
3 [1]
C
4 [1]
D
5 [1]
A
6 [1]
B
7 [1]
C
8 [1]
(a) km h–1 → ms–1 (27.8 m s–1) or 100000/(5.8 × 3600)
9
C1
acceleration equation or correctly substituted values
C1
4.79 cao
A1
3
(b) equation of motion or correctly substituted values
(s = ut + at2; s = (v + u)t/2; v2 = u2 + 2as)
C1
80.6 m e.c.f. from (a)
A1
2
[5]
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(a) correct substitution in (v2 = u2 + 2as)
10
or correct rearrangement g = or
= 9.6 (9.61 m s–1)
2
(b) g = W/m or W = mg (= ma) and weight is proportional
to mass/doubling the mass doubles the weight/’masses
cancel’/the factor of two cancels (so g remains the same)
1
(c) ball’s acceleration will decrease/be less than card’s or card’s
acceleration will be unaffected/nearly constant
air resistance affects cards less or card is more streamlined
or card does less work against air resistance
alternative timing/(velocity/speed/acceleration)
uncertain/(inaccurate /imprecise/less reliable)
indication that full width of ball may not pass through gate/difficulty
in determining ‘length’ of ball passing through gate
2
[5]
(a) states area under graph = distance or clear evidence of
11 graph use
B1
½ × 30 × 25 seen
B1
2
(b) accel = grad of graph or uses a = Δv/Δt
M1
30/20 = 1.5 m s–2
A1
2
(c) 300 + 375 = 675 m
B1
1
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(d) 675/680 m (ecf) at 30m/s takes 22.5/22.7 s
C1
but actually took 90 s
C1
so loss of time = 67.5/67.3 s
A1
3
[8]
(a) AB: (uniform) acceleration (1)
12 BC: constant velocity / speed or zero acceleration (1)
CD: negative acceleration or deceleration or decreasing speed / velocity (1)
DE: stationary or zero velocity (1)
EF ; (uniform) acceleration in opposite direction (1)
5
(b) area under the graph (1)
1
(c) distance is a scalar and thus is the total area under the graph
[or the idea that the train travels in the opposite direction] (1)
displacement is a vector and therefore the areas cancel (1)
2
[8]
clear attempt to use area under graph/statement that distance is
13 equivalent to area under graph
C1
38 to 40 squares/1 square is equivalent to 0.05 m
C1
1.9 to 2.0 m
A1
[3]
@TOPhysicsTutor facebook.com/TheOnlinePhysicsTuto
Page 33 of 50
Mark schemes
B
1 [1]
A
2 [1]
B
3 [1]
C
4 [1]
D
5 [1]
A
6 [1]
B
7 [1]
C
8 [1]
(a) km h–1 → ms–1 (27.8 m s–1) or 100000/(5.8 × 3600)
9
C1
acceleration equation or correctly substituted values
C1
4.79 cao
A1
3
(b) equation of motion or correctly substituted values
(s = ut + at2; s = (v + u)t/2; v2 = u2 + 2as)
C1
80.6 m e.c.f. from (a)
A1
2
[5]
@TOPhysicsTutor facebook.com/TheOnlinePhysicsTuto
Page 31 of 50
, theonlinephysicstutor.com
(a) correct substitution in (v2 = u2 + 2as)
10
or correct rearrangement g = or
= 9.6 (9.61 m s–1)
2
(b) g = W/m or W = mg (= ma) and weight is proportional
to mass/doubling the mass doubles the weight/’masses
cancel’/the factor of two cancels (so g remains the same)
1
(c) ball’s acceleration will decrease/be less than card’s or card’s
acceleration will be unaffected/nearly constant
air resistance affects cards less or card is more streamlined
or card does less work against air resistance
alternative timing/(velocity/speed/acceleration)
uncertain/(inaccurate /imprecise/less reliable)
indication that full width of ball may not pass through gate/difficulty
in determining ‘length’ of ball passing through gate
2
[5]
(a) states area under graph = distance or clear evidence of
11 graph use
B1
½ × 30 × 25 seen
B1
2
(b) accel = grad of graph or uses a = Δv/Δt
M1
30/20 = 1.5 m s–2
A1
2
(c) 300 + 375 = 675 m
B1
1
@TOPhysicsTutor facebook.com/TheOnlinePhysicsTuto
Page 32 of 50
, theonlinephysicstutor.com
(d) 675/680 m (ecf) at 30m/s takes 22.5/22.7 s
C1
but actually took 90 s
C1
so loss of time = 67.5/67.3 s
A1
3
[8]
(a) AB: (uniform) acceleration (1)
12 BC: constant velocity / speed or zero acceleration (1)
CD: negative acceleration or deceleration or decreasing speed / velocity (1)
DE: stationary or zero velocity (1)
EF ; (uniform) acceleration in opposite direction (1)
5
(b) area under the graph (1)
1
(c) distance is a scalar and thus is the total area under the graph
[or the idea that the train travels in the opposite direction] (1)
displacement is a vector and therefore the areas cancel (1)
2
[8]
clear attempt to use area under graph/statement that distance is
13 equivalent to area under graph
C1
38 to 40 squares/1 square is equivalent to 0.05 m
C1
1.9 to 2.0 m
A1
[3]
@TOPhysicsTutor facebook.com/TheOnlinePhysicsTuto
Page 33 of 50
