IB Physics Fall Refresh Work
Odell says, "Physics is phun!"
Fall 2014
38 min
25 marks
Complete this packet and turn in to Mr. Odell the first day after Thanksgiving Break
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1.
Linear motion
At a sports event, a skier descends a slope AB. At B there is a dip BC of width 12 m. The slope
and dip are shown in the diagram below. The vertical height of the slope is 41 m.
A
(not to scale)
slope
41m
C D
B
1.8m
dip
12m
The graph below shows the variation with time t of the speed v down the slope of the skier.
25.0
20.0
15.0
v / ms –1
10.0
5.0
0.0
0.0
1.0
2.0
3.0
4.0 5.0
t/s
6.0
7.0
8.0
The skier, of mass 72 kg, takes 8.0 s to ski, from rest, down the length AB of the slope.
(a)
Use the graph to
(i)
calculate the kinetic energy EK of the skier at point B.
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(2)
(ii)
determine the length of the slope.
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2
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(4)
(b)
(i)
Calculate the magnitude of the change EP in the gravitational potential energy of the
skier between point A and point B.
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(2)
(ii)
Use your anwers to (a)(i) and (b)(i) to determine the ratio
EP  E K  .
E P
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(2)
(iii)
Suggest what this ration represents.
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(1)
(c)
At point B of the slope, the skier leaves the ground. He “flies” across the dip and lands on
the lower side at point D. The lower side C of the dip is 1.8 m below the upper side B.
(i)
Calculate the time taken for an object to fall, from rest, through a vertical distance of
1.8 m. Assume negligible air resistance.
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(2)
(ii)
The time calculated in (c)(i) is the time of flight of the skier across the dip. Determine
the horizontal distance travelled by the skier during this time, assuming that the skier
has the constant speed at which he leaves the slope at B.
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(2)
(Total 15 marks)
2.
Momentum
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(a)
State the law of conservation of linear momentum.
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(2)
(b)
A toy rocket of mass 0.12 kg contains 0.59 kg of water as shown in the diagram below.
high-pressure air
water
nozzle, radius 1.4mm
The space above the water contains high-pressure air. The nozzle of the rocket has a circular
cross-section of radius 1.4 mm. When the nozzle is opened, water emerges from the nozzle
at a constant speed of 18 m s–1. The density of water is 1000 kg m–3.
(i)
Deduce that the volume of water ejected per second through the nozzle is 1.1  10–4
m3.
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(2)
(ii)
Deduce that the upward force that the ejected water exerts on the rocket is
approximately 2.0 N. Explain your working by reference to Newton’s laws of motion.
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(4)
(iii)
Calculate the time delay between opening the nozzle and the rocket achieving lift-off.
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(2)
(Total 10 marks)
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IB Physics Fall Refresh Work
Markscheme
1.
Linear motion
(a)
(i)
E K = 12 × 72 × 23 2 ;
= 1.9  104 J;
(ii)
(b)
(i)
uses area between the t-axis and the line;
correctly converts area  distance (one 1cm  1cm square  5.0m);
distance between 90 m and 105 m;
improved accuracy, distance between 95 m and 100 m;
Do not accept kinematic formulas. Distance can only be found
from area.
2
4
EP = 72  9.8  41;
= 2.9  104 J;
2
Accept 3.0  104J for responses using g = 10ms2.
(ii)
ratio =
(2.9 ×10 4 1.9 ×10 4 )
(2.9 ×10 4 ) ;
= 0.34;
2
Accept 0.37 for responses using g = 10ms2 .
(iii)
fraction of energy lost due to air resistance / friction between skis and
slope / work to push snow away from skis;
1
5
(c)
(i)
(ii)
1.8 = 12 × 9.8× t 2 ;
t = 0.61s;
2
distance = 23  0.61;
= 14m;
2
[15]
2.
Momentum
(a)
(b)
the momentum of a system (of interacting particles) is constant;
if no external force acts on system / net force on system is zero /
isolated system;
A statement of “momentum before = momentum after” achieves first
mark only.
(i)
2
use of volume = r2  v;
=   (1.4  103)2  18;
= 1.1  104 m3
(ii)
(iii)
mass ejected per second = 1.1  104  1000 = 0.11kg;
change in momentum per second = 0.11  18;
by Newton’s 2nd, this is force on (ejected) water;
by Newton’s 3rd, equal force acts upwards on rocket;
so force is 2.0N
Do not accept references to momentum conservation.
weight of water to be ejected = 5.0N / mass of water to be
ejected = 0.51kg;
time delay = 4.6s;
2
4
2
[10]
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IB Physics Fall Refresh Work
Examiner Report
1.
[Part 1 HL and SL (a)] – Linear motion
This question was a very popular choice at both levels and was, generally, done well.
Part (b) asked for the distance travelled by the skier down the slope. Since the path was not a
straight line and the acceleration was not constant the usual formulae of kinematics did not apply
and students had to use the area under the curve. Unfortunately very few candidates chose this
route. In (b) (iii) candidates had to identify two causes of the retarding force on the skier. Friction
was a common answer without, however, identifying where the friction force was coming from
(for example friction between the skis and the snow).
[HL only part (c)] This was very well done by most candidates.
[SL only part (c)] This was very well done by most candidates.
[HL and SL part (d)] Most realized that the skier would land further to the right and many
correctly identified a softer landing but very few could give a coherent argument as to why.
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2.
[Part 1 HL and SL] Momentum
Most candidates managed to score many marks in this part of the question.
Part (a) was completed by almost all candidates. Many used the term “closed system” without
defining it. In (b) (ii) the question explicitly asked for reference to Newton‟ s laws. In such
situations it is not sufficient to simply state the laws – one must state the laws in the concrete
context of the particular question. Thus in this case, the examiners expected statements such as
“the rate of change of the momentum of the ejected water gives the force on the water” and “the
force exerted on the water downwards by the rocket is equal and opposite to the force exerted by
the water on the rocket upwards” etc.
Part (a) was completed by almost all candidates. Many used the term “closed system” without
defining it. In (b) (ii) the question explicitly asked for reference to Newton‟ s laws. In such
situations it is not sufficient to simply state the laws – one must state the laws in the concrete
context of the particular question. Thus in this case, the examiners expected statements such as
“the rate of change of the momentum of the ejected water gives the force on the water” and “the
force exerted on the water downwards by the rocket is equal and opposite to the force exerted by
the water on the rocket upwards” etc.
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