1.
(a)
line of best fit is not straight / line of best fit does not go through origin;
1
(b)
smooth curve;
that does not go outside the error bars;
Ignore extrapolations below n = 1.
2
(c)
(d)
we can re-write the suggested relation as log D = log c + p log n;
now we can plot a graph of log D versus log n;
the slope of the (straight line) graph is equal to p;
Accept logs in any base.
(i)
(ii)
absolute uncertainty in diameter D is ±0.08cm;
0.08
giving a relative uncertainty in D2 of 2 ×
= 0.13 or 13%;
1.26
Award [2] if uncertainty is calculated for a different ring number.
3
2
it is possible to draw a straight line that passes through the
origin (and lies within the error bars);
or
the ratio of
(iii)
(iv)
D2
is constant for all data points;
n
gradient = k;
calculation of gradient to give 0.23 (accept answers in
range 0.21 to 0.25);
evidence for drawing or working with lines of maximum
and minimum slope;
answers in the form k = 0.23± 0.03;
Accept an uncertainty in k in range 0.02 to 0.04.
First marking point does not need to be explicit.
cm2;
1
4
1
[14]
2.
(a)
(b)
two error bars in the correct position;
two error bars of the correct length; (allow a square-length each
side of the data points judge by eye and allow for the thickness
of the line drawn)
2
suitable curve that goes through the two error bars;
and through (0,0);
2
IB Questionbank Physics
1
(c)
(d)
a straight line cannot be drawn through the error bars and through
the origin;
so height is not directly proportional to the diameter;
(and) height is proportional to energy;
3
the gradient of the line gives the value of the power coefficient/exponent/
1
1
lg d = lg h + constant and lg d = lg h + constant;
3
4
gradient of data line = 0.33 (±0.02) ;
gradient of max and gradient of min = 0.37 (±0.02) and 0.29(±0.02);
some conclusion as to why this supports theory 1 e.g. therefore
the uncertainties do not allow for n = 0.25 so data supports theory 1;
4
[11]
3.
(a)
line going through each error bar;
1
(b)
(i)
line does not go through (0,0)/origin so no;
Watch for ECF from (a).
1
(ii)
line is curved / gradient not constant so no;
1
(c)
(d)
line drawn to find intercept on y-axis;
4.2(±0.1)m;
2
large (at least half of line) triangle from straight line portion of graph;
slope = 0.012(±0.001)(m s–1);
volume per second=area × slope;
(0.022 m3 s–1)
3
Alternatively for [2 max].
determines height difference over time range within 0 and 120 s;
1.8  [differencein heights]
volume per second =
;
time between heights
(0.022 m3 s–1)
(e)
(850 × 0.022) = 19 m3 or (850 × 0.02) = 17 m3;
IB Questionbank Physics
1
2
(f)
graph starts at same point but half initial gradient by eye;
1
way down y-axis;
4
Original line need not be shown. Allow ECF from (c) if the curve
begins at (0, 3.5).
line always lower than original by eye and ending about
2
[11]
4.
B
[1]
5.
D
[1]
6.
D
[1]
7.
B
[1]
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3
8.
B
[1]
9.
B
[1]
IB Questionbank Physics
4