# IB Physics 2 winter

```IB Physics Winter Break Refresh Work
Odell Says, &quot;My gift to you is Physics&quot;
Winter 2013/14 &amp; Winter 2014/15
39 min
26 marks
Complete this packet and turn in to Mr. Odell the first day of Spring Semester.
1
1.
Specific heat and a domestic shower
(a)
Define specific heat capacity.
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(1)
(b)
Equal masses of two different solid substances A and B are at the same temperature. The
specific heat capacity of substance A is greater than the specific heat capacity of substance
B. The two substances now have their temperatures raised by the same amount.
Explain which substance will have the greater increase in internal energy assuming both
remain in the solid phase.
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(2)
(c)
The diagram below shows part of the heating circuit of a domestic shower.
Cold water enters the shower unit and flows over an insulated heating element. The heating
element is rated at 7.2 kW, 240 V. The water enters at a temperature of 14C and leaves at a
temperature of 40C. The specific heat capacity of water is 4.2  103 J kg–1 K–1.
(i)
Estimate the flow rate of the water.
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2
(4)
(ii)
Suggest one reason why your answer to (c)(i) is only an estimate.
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(1)
(Total 8 marks)
2.
Mechanical power
(a)
Define power.
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(1)
(b)
A car is travelling with constant speed v along a horizontal straight road. There is a total
resistive force F acting on the car.
Deduce that the power P to overcome the force F is P = Fv.
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(2)
(c)
A car drives up a straight incline that is 4.8 km long. The total height of the incline is 0.30
km.
The car moves up the incline at a steady speed of 16 m s–1. During the climb, the average
friction force acting on the car is 5.0  102 N. The total weight of the car and the driver is
1.2  104 N.
(i)
Determine the time it takes the car to travel from the bottom to the top of the incline.
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(2)
(ii)
Determine the work done against the gravitational force in travelling from the bottom
to the top of the incline.
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(1)
3
(iii)
Using your answers to (c)(i) and (c)(ii), calculate a value for the minimum power
output of the car engine needed to move the car from the bottom to the top of the
incline.
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(4)
(d)
From the top of the incline, the road continues downwards in a straight line. At the point
where the road starts to go downwards, the driver of the car in (c), stops the car to look at the
view. In continuing his journey, the driver decides to save fuel. He switches off the engine
and allows the car to move freely down the hill. The car descends a height of 0.30 km in a
distance of 6.4 km before levelling out.
The average resistive force acting on the car is 5.0  102 N.
Estimate
(i)
the acceleration of the car down the incline.
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(5)
(ii)
the speed of the car at the bottom of the incline.
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(2)
(e)
In fact, for the last few hundred metres of its journey down the hill, the car travels at
constant speed. State the value of the frictional force acting on the car whilst it is moving at
constant speed.
4
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(1)
(Total 18 marks)
5
IB Physics Winter Break Refresh Work
Markscheme
1.
Specific heat and a domestic shower
(a)
(b)
(c)
the amount of energy / heat required to raise the temperature of 1 kg
of a substance through 1K / 1C;
1
the internal energy is the total energy of the molecules of a substance;
the greater the specific heat (the more energy required to raise unit mass
through 1 K) this means that to increase the temperature by the same amount,
more energy most be given to substance A than to substance B (so internal
energy is greater) / OWTTE;
2
(i)
(ii)
energy supplied by heater in 1s = 7.2  103 J;
energy per second = mass per second  sp ht  rise in temperature;
7.2  103 = mass per second  4.2  103  26;
to give mass per second = 0.066kg;
energy is lost to the surroundings;
flow rate is not uniform;
Do not allow “the heating element is not in contact with all the water
flowing in the unit”.
4
1
[8]
2.
Mechanical power
(a)
the rate of working / work  time;
If equation is given, then symbols must be defined.
1
6
(b)
P
v
(c)
(i)
W F d

;
t
t
d
therefore, P  Fv ;
t
d
t ;
v
=
4800
 300 s ;
16
(ii)
W = mgh = 1.2  104  300 = 3.6  106 J;
(iii)
work done against friction = 4.8  103  5.0  102;
total work done = 2.4  106 + 3.6  106;
total work done = P  t = 6.0  106;
to give P 
(d)
(i)
2
sin =
6.0 10 6
 20 kW ;
300
2
1
4
0.30
 0.047 ;
6 .4
weight down the plane = W sin = 1.2  104  0.047 = 5.6  102N;
net force on car F = 5.6  102 5.0  102 = 60N;
a
F
;
m
60
 5.0 10  2 ms  2 ;
3
1.2 10
(ii)
v2 = 2as = 2  5.0  102  6.4  103;
to give v = 25ms1;
(e)
5
5.6  102 N;
2
1
[18]
7
IB Physics Winter Break Refresh Work
Examiner Report
1.
Specific heat and a domestic shower
(a)
(b)
(c)
the amount of energy / heat required to raise the temperature of 1 kg
of a substance through 1K / 1C;
1
the internal energy is the total energy of the molecules of a substance;
the greater the specific heat (the more energy required to raise unit mass
through 1 K) this means that to increase the temperature by the same amount,
more energy most be given to substance A than to substance B (so internal
energy is greater) / OWTTE;
2
(i)
(ii)
energy supplied by heater in 1s = 7.2  103 J;
energy per second = mass per second  sp ht  rise in temperature;
7.2  103 = mass per second  4.2  103  26;
to give mass per second = 0.066kg;
energy is lost to the surroundings;
flow rate is not uniform;
Do not allow “the heating element is not in contact with all the water
flowing in the unit”.
4
1
[8]
2.
Mechanical power
(a)
the rate of working / work  time;
If equation is given, then symbols must be defined.
1
8
(b)
P
v
(c)
(i)
W F d

;
t
t
d
therefore, P  Fv ;
t
d
t ;
v
=
4800
 300 s ;
16
(ii)
W = mgh = 1.2  104  300 = 3.6  106 J;
(iii)
work done against friction = 4.8  103  5.0  102;
total work done = 2.4  106 + 3.6  106;
total work done = P  t = 6.0  106;
to give P 
(d)
(i)
2
sin =
6.0 10 6
 20 kW ;
300
2
1
4
0.30
 0.047 ;
6 .4
weight down the plane = W sin = 1.2  104  0.047 = 5.6  102N;
net force on car F = 5.6  102 5.0  102 = 60N;
a
F
;
m
60
 5.0 10  2 ms  2 ;
3
1.2 10
(ii)
v2 = 2as = 2  5.0  102  6.4  103;
to give v = 25ms1;
(e)
5
5.6  102 N;
2
1
[18]
9
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