Contemporary Physics I – HW 2

HW 2 - SolutionsDue Friday, October 13, 2017

1. What is the charge number (Q), Baryon number, (B – how many baryons), and lepton number (L) ofthe following [5 pts]:

Particle Q B Lp +1 1 0e+ +1 0 -1γ 0 0 0ν 0 0 -1

42He 2 4 0

Remember, we are only considering the nucleus of He (also known as an α particle)

2. Draw a Feynman diagram illustrating the following:

(a) An electron scattering off a photon [5 pts].

(b) νe + n→ p+ e− [5 pts]

3. What is the γ factor of a particle moving at: [5pts each]

(a) 3× 106 m/s

γ =1√

1− v2

c2

(1)

=1√

1− (3×106 ms )2

(3×108 ms )2

=1√

1− (1× 10−2)2

γ = 1.00005

(b) 0.5c

γ =1√

1− v2

c2

(2)

=1√

1− (0.5�c)2

(�c)2

=1√

1− (0.5)2

γ = 1.15

(c) 0.999c

γ =1√

1− v2

c2

(3)

=1√

1− (0.999�c)2

(�c)2

=1√

1− (0.999)2

γ = 22.4

For which of these can you assume non-relativistic physics?

For (a), γ is very near to 1 so we can assume any relativistic effects are negligible and can be ignored.For (b) and (c), the velocities are a significant fraction of the speed of light, c, so we should takerelativistic effects into account.

4. Consider a 2kg block. What is the momentum if the speed is [10 pts]:

(a) 10 m/s ?

p = m× v

= (2 kg)× (10m

s)

p = 20kg m

s

(b) 400 m/s?

p = m× v

= (2 kg)× (400m

s)

p = 800kg m

s

(c) 2× 108 m/s

p = mvγ

=mv√1− v2

c2

=(2 kg)× (2× 108m

s )√1− 22

32

p = 5.37× 108kg m

s

(d) 2.9× 108 m/s

p = mvγ

=mv√1− v2

c2

=(2 kg)× (2.9× 108m

s )√1− 2.92

32

p = 2.27× 109kg m

s

5. A particle with a mass of 10−26kg has a momentum of p = 4 · 10−18kg ·m/s. How fast is it traveling?

If you get an answer that’s faster than the speed of light, you should at least consider the possibilitythat you’ve made a mistake. [10 pts]

Naively, we can try |~p| = m|~v|

|~v| = |~p|/m = 4 · 10−18(kg ·m/s)/10−26(kg) = 4 · 108m/s

We are incorrect as we are above the speed of light. The fact that you have a momentum many ordersof magnitude larger than the mass could have suggested this. We need to try the relativistic approach.

|~v| =

(|~p|2m2

(1 + |~p|2m2 /(c)2)

).5

|~v| =

4·10−18(kg·m/s)2

10−26(kg)2

(1 + 4·10−18(kg·m/s)2

10−26(kg)2 /(c)2)

.5

|~v| ≈ 2.4 · 10−8 m/s

6. P2.14: A .7 kg block of ice is sliding by you on a very slippery floor at 2.5 m/s. As it goes by, yougive it a kick perpendicular to its path. your foot is in contact with the ice block for .003 s. The blockeventually slides at an angle of 22 degrees from its original direction. The overhead view shown inFigure 2.54 is approximately to scale. The arrow represents the average force your tow applies brieflyto the block of ice. [25 pts total]

(a) Which of the possible paths shown in the diagram corresponds to the correct overhead view ofthe block’s path? [5pts]Path A reflects a path that stops and restarts motion in the x direction, clearly indicating morethan one direction of force applied. Path C is a path that is potentially correct, however theproblem says that the problem is drawn to scale. A block of ice .7 kg is approximately the samesize as your foot, so the idea that you can begin the .003 seconds where the red arrow is and thenapply the force with so much control so that you create a visibly curved path for the ice is ludicrous.

Path B

is the correct path because for .003s of contact time, drawn to scale, the ice will simply look as ifit has instantaneously changed direction from above.

(b) Which components of the block’s momentum are changed by the impulse applied by your foot?(Check all that apply. The diagram shows a top view, looking down on the xz plane.) [2pts]Since you kicked perpendicular to its path, which is shown to be in the i direction in the figure,only the component of momentum in the i direction changes

(c) What is the unit vector p in the direction of the block’s momentum after the kick? [3pts]We are asked for the unit vector, which must have a magnitude of one and is unitless. The unitvector in the p direction simply is a vector of magnitude one in the same direction as ~p, which weknow from the problem statement is 22 degrees above the horizontal x direction. Therefore,

p = cos(22◦)i+ sin(22◦)i

p = .927i+ .375j

(d) What is the x component of the block’s momentum after the kick? [5pts]Since the force was applied perpendicular to the x direction, we know that the velocity in thatdirection did not change, so that the final x component of the velocity is

px = .7kg ∗ 2.5m/s = 1.8kg ·m/s

(e) Remember that ~p = |~p|p. What is the magnitude of the block’s momentum after the kick? [3pts]The reminder part e is a vector equation. Using our answer to part c and writing only the x-

component, we get that

px = |~p| · cos(22◦)

|~p| = pxcos(22◦)

|~p| = 1.8kg ·m/scos(22◦)

= 1.9kg ·m/s

(f) Use your answers to the preceding questions to find the z component of the block’s momentumafter the kick (drawing a diagram is helpful). [2pts]

sin(22◦) =pz| ~py|

pz = |~p| · sin(22◦) = 1.9 · sin(22◦) = .71kg ·m/s

(g) What was the magnitude of the average force you applied to the block? [5pts]Since you only applied force in the z direction, we only have to consider one direction of impulseto get the average force.

∆Pz/∆t = Fzavg

∆Pz = Pzf − Pzi = .71kg ·m/s

Fzavg = (.71kg ·m/s)/(.003s) = 2.4 · 102N

7. You are standing at the top of a 10m cliff (as shown), throwing stones over the edge [25 pts]

You throw a 0.2 kg stone at an angle of 30 degrees (π/6) above the horizontal with a speed of 10m/s.

(a) What is the x-component of the stone’s velocity immediately after it leaves your hand? [3pts]

vix = vi cos(π

6

)= 10m/s(0.866)

= 8.66m/s

(b) What is the y-component of the stone’s velocity immediately after it leaves your hand? [2pts]

viy = vi sin(π

6

)= 10m/s(0.5)

= 5m/s

(c) How long after you throw it does it reach it’s maximum height? [5pts]

vfy = viy − gt

→ th =vfy − viy−g

=0− 5m/s

−9.81m/s2

= 0.51s

(d) How long after you throw the stone does it take to hit the ground? [10pts]First calculate the maximum height the ball reaches

hmax =yi + viy th −1

2gt2h

=(10m) + (5m/s)(0.51s)− 1

2(9.81m/s2)(0.51s)2

=11.3m

Then how long the ball falls from that height

yground =hmax + viy tground −1

2gt2ground

0 =11.3m+ (0m/s)tg −1

2(9.81m/s2)t2g

→ tg =

√2(11.3m)

9.81m/s2

=1.52s

So the total time is this sum of this time to that height, and the time from that height to theground

ttotal =th + tg = 0.51s+ 1.51s

= 2.02s

(e) How far from the edge of the cliff does the rock hit? [5pts]

xf =xi + vixtg +1

2axt

2g

=(0m) + (8.66m/s)(2.02s) +1

2(0m/s2)(2.02s)2

= 17.5m