Day 5.5 Momentum and Relativity
Instructions : The next two classes deal with momentum and energy. Momentum was introduced in the quantum unit. These first three questions are here to review what is already known and deepen the understanding of energy, kinetic energy and momentum.
1) Momentum is easily confused with kinetic energy. Compare them by considering; a) their defining equations
Momentum goes as speed (p = mv) and kinetic energy as speed squared (E = ½ mv
2
). b) the equations that show how force changes them
Kinetic energy comes from work – a force acting across a distance (W =

E k
= F
 d ). This is a dot product of two vectors and produces a scalar. Momentum comes from impulse – a force acting over time
( p = F
 t). This is a vector times a scalar and produces a vector.
In any interaction, energy is always conserved but kinetic energy usually is not. Kinetic energy is only conserved in a perfectly elastic collision. Momentum may be exchanged between the objects, but the total momentum will be conserved if there are no outside forces.
Example #1: An elastic collision
2)
Two balls of a Newton’s cradle are released and hit the three stationary ones. What happens?
A) The three balls move away at 2/3 of the speed to conserve momentum.
B) The furthest ball moves at 2 times the speed to conserve momentum
C) The furthest ball moves away at 1.4 times the speed to conserve energy.
D) The furthest two balls move at the same speed to conserve energy and momentum.
Explain.
The result must conserve both momentum and kinetic energy. (The result must conserve kinetic energy minus a small loss into sound, vibration and heat.) This is only possible if the farthest two balls swing out with the same momentum and kinetic energy. This collision is largely elastic but not perfectly so. Examples of perfectly elastic collisions come from astronomy (slingshot effect with a spacecraft and planet) and particle physics (alpha particle deflected by gold nucleus).
Example #2: An inelastic collision
3) A cart moving at speed v, hits and sticks to a stationary cart. They move together with a speed of
A)
½ v to conserve momentum
B) square root(½ v) to conserve momentum
C)
½ v to conserve energy
D) square root(½ v) to conserve energy.
Explain.
Kinetic energy is not evenly roughly conserved in a collision like this however, momentum is. The missing kinetic energy went into sound, heat and deformation. Momentum is conserved so mv = 2m (½v).
Example #3: An explosion
4) A gun with a mass of 1 kg is hanging from a string and fires a bullet of 10 g horizontally at 100 m/s.
What is conserved?
A) momentum, total energy
C) kinetic energy, momentum
B) kinetic energy, total energy
D) all three
Explain.
The gun recoils at 1 m/s to conserve momentum (Mv = mV). The huge increase in kinetic energy (1/2 mV
2
+ ½ Mv
2
) came from chemical energy.
What kills - momentum or kinetic energy?
The bullet and gun have equal momenta and getting hit by either would push you back by a similar amount. The bullet has 10 times the kinetic energy. If you attached a bullet onto the butt of the gun, it would hurt but it still wouldn`t kill you. If each end was directed at a block of wood, the bullet stuck to the gun would make a hole that was 1/10 th
as deep. Momentum bumps you and kinetic energy rips you.
Watch the first 9½ minutes of Approaching the Speed of Light . http://www.triumf.ca/home/multimedia/videos/speed-light
5) The video mentions a number of particles.

a) What particles can be found in the nucleus? Protons, neutrons and pions.
b) What particles are produced when the proton hits the carbon atom? Pions.
c) What particles are produced when these decay? muons, pions and electrons d) The electromagnets select particles with equal charge and momentum
6) You will be graphing momentum against velocity for subatomic particles. What will a graph of momentum (vertical) against sped look like?
A) . B) p p
C) p
D) p v v v v
Explain: The equation p = mv, suggests B. However, we already have a sense that there is a speed limit of c, which suggests that the equation is wrong and that it will look like C.
7) Each histogram shows the time of flight for the particles of equal momentum to travel 4.4 m. Each graph has two and possibly three peaks. The peaks are due to the three different particles present; electrons, muons and pions. The electrons move fastest and the pions move the slowest. a) Which particle produces the peak that is furthest to the left? electrons b) Which particle is heaviest? pion
Instructions : There are twelve different histograms available. The proportions of the three particles changes dramatically with momentum. At high speeds there are lots of pions. At low speeds there are very few, because by the time they get to the detector, most of them have decayed.
8) Use your histogram to get the speed as a fraction of c for each particle. Show your steps below. a) Electron b) Muon c) Pion v = 4.4/ct
9) Look at the class`s data. We can’t use the electrons’ velocities. Why not?
The velocities are all almost c. Some may even be over.
10) Graph p (units of MeV/c) vs. v (in units of c) for the muons and pions. Guess what you need to graph p against to get a straight line and test it. a) What must you graph on the horizontal axis instead of v to get a straight line?
Graph p vs.
 mv b) What is the value of the slopes of your two straight-line graphs? c) What is the physical meaning of the slopes on your graph?
Mass of the pion and muon in units of MeV. d) Calculate the % difference of your slopes and the accepted values; 105.7 MeV/c 2 , 139.6 MeV/c 2 e) What do you think the major source of error was?
The time could only be measured to two digits. It was +/- 0.2 ns, which for small times like the electrons there was an error of close to 15%. The distance was given as just two digits - 4.4 m - but they could have easily measured it to the nearest mm or 4 digits. f) Why do the histogram peaks have a spread to them?
The particles were selected with a range of momenta that would go down the beam line.
7) Your friend thinks that it is just a matter of time before we have rockets that can go faster than the speed of light. How would you try to convince them that this will not happen?
As we can see from the graph and the equation, as the speed increases it gets harder and harder to increase it further. There is an asymptote at v = c, so an object with mass cannot go at the speed of light let alone faster. This is not a technical limit like the speed of sound was – but a theoretical limit like absolute zero.
Textbook : 11.2 (576 -578) p. 578 #10-12, p. 579 #10-12

8) Many popularizations of physics say that mass increases as you go faster. Most physicists say that mass does not change with speed. What do physicists think is the correct interpretation and why?
Physicists interpret p =
 mv as meaning p =
 mv) whereas the popularizations are interpreting it as p =
(
 m) v. Both interpretations give the same calculated values. Einstein is quoted on p. 576, giving an example of the physicists’ view.