 # Model Solutions to Sample Exam 2

```Oberlin College Physics 110, Fall 2011
Model Solutions to Sample Exam 2
Key idea: Energy is conserved because poles aren’t used, friction is negligible, and the normal force does no
work.
Initial energy:
Final energy:
1
2
2 mvi
1
2
2 mvf
+ mghi = 12 m(0)2 + mghi .
+ mghf .
Energy conservation:
1
2
2 mvf
+ mghf = mghi =⇒ vf2 = 2g(hi − hf ).
Change in height: hf − hi = −12.3 m + 6.5 m − 3.5 m + 4.4 m − 10.7 m = −15.6 m.
p
Final velocity: vf = 2g(hi − hf ) = 17.5 m/s.
See model solutions to assignment 4.
Additional problem 80: Ice mound, part III.
2
v2
R+ 0
5
3g
v2
2
R+ 0
3
3g
2
v0
R+
3
3g
2
v2
R− 0
3
3g
s
2v02 R
3g
Wrong value when v0 = 0.
OK.
Dimensionally incorrect.
Height of departure will increase with v0 .
Wrong value when v0 = 0.
Additional problem 66: Simple harmonic motion graph, II.
a. We have
x(t)
= A sin(ωt)
v(t)
= Aω cos(ωt)
a(t)
= −Aω 2 sin(ωt).
Because sin θ varies between +1 and −1, the maximum acceleration will come when the sine is equal to −1,
namely
2
2π
2
maximum acceleration = Aω = A
= 12.0 cm/s2 .
T
1
b. Comparing a(t) and x(t) above, a(t) = −ω 2 x(t). Alternatively,
a=
F
m
but
2
3
F = −kx
The graph is thus
a (cm/s2)
14
12
10
8
6
4
2
−4
−3
−2
x (cm)
−1
1
4
−2
−4
−6
−8
−10
−12
−14
2
so
a=−
k
x.
m
Additional problem 70: Work with force preconceptions.
P~
In truth
F = m~a, from which we proved in class that (where i stands for “initial” and f stands for
“final”)
~
rf
XZ
F~ · d~r
=
~a · d~r
m
~
ri
X
~
rf
Z
~
ri
tf
Z
W
=
=
=
=
d~v d~r
·
dt
dt dt
t
Z itf
d~v
· ~v dt
m
ti dt
Z tf 2
d~v
1
dt
2m
dt
ti
2 tf
1
v t .
2m ~
m
i
[[The above is not needed to solve the problem. I put it in only to facilitate comparison between what
happens in truth and what would happen if the preconception were true.]]
P~
But if you held the pseudo-Aristotelian preconception that
F = m~v , then you would have to hold that
XZ
~
rf
F~ · d~r
Z
~
ri
X
~
rf
~v · d~r
= m
~
ri
tf
Z
W
~v ·
= m
ti
tf
Z
d~r
dt
dt
~v 2 dt,
= m
ti
and the right-hand side is always positive or zero.
Additional problem 89: Impulse with force preconceptions.
P~
In truth
F = m~a, from which we proved in class that (where i stands for “initial” and f stands for
“final”)
Z tf
X Z tf
F~ dt = m
~a dt
ti
ti
Z tf
X
d~v
dt
J~ = m
ti dt
t
= m [~v ]tfi .
[[The above is not needed to solve the problem. I put it in only to facilitate comparison between what
happens in truth and what would happen if the preconception were true.]]
3
P~
But if you held the pseudo-Aristotelian preconception that
F = m~v , then you would have to hold that
Z
Z
tf
X tf
F~ dt = m
~v dt
ti
ti
Z
tf
X
d~r
J~ = m
dt
ti dt
=
t
m [~r]tfi .
7. Cracked up cantaloupe.
at first:
later:
V1
v
V2
m
M
M/2
M/2 + m
M
M
V1 +
+ m V2
momentum: mv =
2
2
M
M
kinetic energy: 21 mv 2 = 12 V12 + 12
+ m V22
2
2
Because M/2 is about 500 times m, approximate M/2 + m by M/2.
M
(V1 + V2 )
from momentum: v =
2m
M
from kinetic energy: v 2 =
(V 2 + V 2 )
2m 1 2 2
M
square of “from momentum”: v 2 =
(V12 + 2V1 V2 + V22 )
2m
2
equating the two expressions
2for v gives:
M
M
(V 2 + V22 ) =
(V12 + 2V1 V2 + V22 )
2m 1
2m
M
V12 + V22 =
(V 2 + 2V1 V2 + V22 )
2m 1
Remembering that M/2m is about 500, this is
M
0≈
(V 2 + 2V1 V2 + V22 )
2m 1 2
2V1 V2 ≈ −(V1 + V22 )
The right hand side is negative, so V1 and V2 are of opposite sign: they fly away in different
directions.
4
8. Atomic collision.
initial
final
alpha
64.0 degrees
alpha
O
51.0 degrees
O
Conservation of momentum:
Horizontal:
Vertical:
mα vα,i = mα vα,f cos(64.0◦ ) + mO vO,f cos(51.0◦ )
◦
◦
0 = mα vα,f sin(64.0 ) − mO vO,f sin(51.0 )
(1)
(2)
We know the speed of the outgoing nucleus, vO,f = 1.20 × 105 m/s, and the two masses, so these are two
equations in two unknowns. Equation (2) tells us that
vα,f = vO,f
mO sin(51.0◦ )
= 4.15 × 105 m/s,
mα sin(64.0◦ )
while equation (1) tells us that
vα,i = vα,f cos(64.0◦ ) +
mO
vO,f cos(51.0◦ ) = 4.84 × 105 m/s.
mα
5
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