Physics 200
Homework #5
Due: October 26, 2005
1. We observe two objects. Object A has a mass of 10 kilograms and moves at 2 m/s.
Object B has a mass of 5 kilograms and moves at 4 m/s.
(a) Which object has more kinetic energy? (show all work)
Object A: KE = ½ mv2 = ½ (10 kg) ( 2m/s)2 = 20 kg m2 / s2 = 20 J
Object B: KE = ½ mv2 = ½ (5 kg) ( 4m/s)2 = 40 kg m2 / s2 = 40 J
Object B has more kinetic energy
(b) Which object has more linear momentum? (show all work)
Object A: momentum = mv = (10 kg) ( 2m/s) = 20 kg m/s
Object B: momentum = mv = (5 kg) ( 4m/s) = 20 kg m/s
Object A and B have the same amount of momentum
2. The bright star Sirius emits a continuous spectrum with a maximum at 302.564 nm.
Estimate the surface temperature of Sirius.
peakT = constant = 2.9 x 10-3 meter Kelvin or 2.9 x 106 nanometer Kelvin (nm K)
T = (2.9x106 nm K) / (302.564 nm) = 9600 K
3. (a) Calculate the time (T) for one complete oscillation of the electric field of yellow
light that has a wavelength of:   5.8 x107 m
[Hint: find the frequency and then the period ( T  1 f )]
f = v and v = c , so f = c
f = c/ = (3 x 108 m/s) / (5.8 x 10 -7 m) = 5.17 x 10 14 /s
T = 1/f = 1/ (5.17 x 10 14 /s) = 1.9 x 10 -15 seconds
(b) About how many wavelengths N are included in the light wave produced by a single
atom when it emits a burst of light lasting t  1 x109 sec onds ? ( N  t / T )
Since this is 3(b), use the period we found in part 3(a)
N = (1 x 10 -9 s) / (1.9 x 10 -15 s) = 530,.000
(c) How long in space is the light wave? ( L  N )
L = N = (530,000) (5.8 x 10 -7 m) = 0.3 meters
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4. A standing wave experiment is performed with microwaves in air. The distance
between nodes is found to be 1 centimeter.
(a) What is the wavelength of the microwaves?
The distance between two successive nodes is half a wavelength, so half a wavelength is
1 centimeter.
That means a whole wavelength is 2 centimeters.
(b) What is the frequency of the microwaves?
f = v and v = c , so f = c
f = c/ = (3 x 108 m/s) / (2 x 10 -2 m) = 1.5 x 10 10 /s which you can write 1.5 x 10 10 Hz
5. Suppose that we see two stars in the sky (A and B) that have the same surface
temperature and the same surface area. We note that star A is 4 times brighter than star
B. If there is no absorbing material between us and the stars, what might the cause of the
brightness difference?
A happens to be closer to our eyes than B. (Since the intensity drops off as 1/r2, this tells
us that star B is 2 times farther away than star A.)
6. Suppose that we concentrate on one of the Planck oscillators having a frequency of
f  2 x1013 hertz . What is the difference between two successive energy values for this
oscillator? Express your answer in units of joules and then in units of electron volts.
For a Planck oscillator (also called a quantum oscillator) the energy of the different states
is En = nhf.
The energy of the first state (n=1) is E1 = 1 h f
The energy of the second state (n=2) is E2 = 2 h f
The energy of the third state (n=3) is E3 = 3 h f
etc…
We can find the energy between successive states by subtracting:
For example:
E2 – E1 = 2 h f – 1 h f = 1 h f = (6.626 x 10-34 J s) (2 x 1013 /s) = 1.3 x 10 – 20 J
Using Planck’s constant: h = 6.626 x10-34 joule•seconds
7. A 10 kg mass at rest in the laboratory suddenly explodes into two masses of 4 kg and 6
kg. If the velocity of the 4 kg mass is 36 m/s what is the velocity of the 6 kg mass?
Momentum is conserved, so the initial momentum (which is 0) is equal to the final
momentum (that is, the final momentum must be 0). We’ll call the unknown velocity X
Initial mv = (10 kg)(0 m/s) = 0 kg m/s
Final mv = (4 kg)(36 m/s) + (6 kg)( X m/s) = 0 kg m/s
Subtract (4 kg)( 36 m/s) from both sides
(6 kg)( X m/s) = - (4 kg)(36 m/s)
divide by 6 kg
(X m/s ) = - (4 kg)(36 m/s) / (6 kg) = - (4/6) ( 36 m/s) = - 24 m/s
The minus sign tells its direction. The 6 kg mass is moving in the opposite direction than
the 4 kg mass.
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