Name___________________________________
Physics 120 Quiz #6, March 2, 2012
Please show all work, thoughts and/or reasoning in order to receive partial credit. The
quiz is worth 10 points total.
1. Substance A has a large specific heat capacity (on a per gram basis), while substance
B has a smaller specific heat capacity. If the same amount of energy is put into a
100-gram block of each substance, and if both blocks were initially at the same
temperature, which one will now have the higher temperature?
a. Substance A
b. Substance B
c. They will have equal temperatures.
d. There is not enough information to answer.
2. A chain of metal links with total mass m = 3 kg is coiled up in a tight ball on a lowfriction table. You pull on a link at one end of the chain with a constant force F = 68
N. Eventually the chain straightens out to its full length L = 1.0 m, and you keep
pulling until you have pulled your end of the chain a total distance d = 3.0 m
(diagram is not to scale).
a. What is the speed of the center of mass of the chain at this instant?

L
WTrans = F  d −  = 68N × ( 3m − 0.5m) = 170J = 12 mv 2f ,com → v f ,com

2
v f ,com
€

L
2F  d − 

2
=
m

L
2F  d − 
2 × 68N ( 3m − 0.5m)

2
=
=
= 10.7 ms
m
3kg
b. What is the change in energy of the chain?
ΔE = WTotal = Fd = 68N × 3m = 204J
€
c. In straightening out, the links of the chain bang against each other, and their
temperature rises. Assume that the process is so fast that there is insufficient time
for significant thermal transfer of energy from the chain to the table, and ignore
the small amount of energy radiated away as sound produced in the collisions
among the links. Calculate the increase in thermal energy of the chain.
ΔE = ΔWTrans + ΔQ → ΔQ = ΔE − ΔWTrans = 204J −170J = 34J
€
Useful formulas:


p = γmv
γ=
keff , parallel = nparallel kindividual ;
1
keff , series =
2
stress = Ystrain →
F
ΔL
=Y
A
L
kindividual
nseries
v
c2
 

v +v
  


 
vavg = i f ⇒ vavg = Rω avg ; special case : constant acceleration rf = ri + vi t + 12 at 2 ; vf = vi + at
2


v2
Fg = mg;
FC = m
Constants
R
g = 9.8 sm

GM1M 2
ˆ
Fgravity =
r
12
G = 6.67 "10#11 Nm
r122
kg

 
1
m
1e =1.6 "10#19 C
Fspring = − ks ; s = ( L − Lo ) sˆ; T = = 2π
f
keff
1
C
k=
= 9 "10 9 Nm
 
4$%o
W = ∫ F ⋅ dr = ΔKE = − ΔU
%o = 8.85 "10#12 NmC
GM1M 2
Ug = −
1eV =1.6 "10#19 J
r
µo = 4$ "10#7 TmA
Ug = mgy
1−
2
2
2
2
2
2
2
c = 3"10 8 ms
1 2
ks
2
1
KET = mv 2
2
1 2
KER = Iω
2
KE = (γ − 1) mc 2
Us =
h = 6.63"10#34 Js
0.511MeV
c2
937.1MeV
m p =1.67 "10#27 kg =
c2
948.3MeV
mn =1.69 "10#27 kg =
c2
931.5MeV
1amu =1.66 "10#27 kg =
c2
23
N A = 6.02 "10
me = 9.11"10#31 kg =
Erest = mc 2
€

 
p f = pi + Fnet Δt
Momentum Principle:  
dp
= Fnet
dt

 
  
L f = Li + τ net Δt; τ = r × F
Angular Momentum Principle: 
  

dL 
= τ net ; L = r × p = Iω;
€
dt
Position-update: 
 

rf = ri + v avg Δt = ri +
€
Energy Principle:

p
m 1+
p2
m 2c 2
Δt
ΔE = W = ΔU g + ΔU s + ΔKE T + ΔKE R
 
 
W = ∫ F ⋅ dr + ∫ τ ⋅ dθ
Vectors €

magnitude of a vector : a = ax2 + ay2 + az2
€


writing a vector : a = ax ,ay ,az = a aˆ = ax iˆ + ay ˆj + az kˆ
 
dot product : A ⋅ B = AB cosθ
cross product : C = A × B = ( Ay Bz − Az By ), ( Az Bx − Ax Bz ), ( Ax By − Ay Bx )
 
magnitude of the cross product : A × B = ABsin θ
€
Geometry
Circles : C = 2πr = πD
Triangles : A = 12 bh
Spheres : A = 4 πr 2
€
A = πr 2
V = 43 πr 3
Ax 2 + Bx + C = 0 & x =
v = Rω; a =
!
#B ± B 2 # 4AC
2A
dv
dω
=R
= Rα
dt
dt