Download PHYSICS 101 MIDTERM

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PHYSICS 101 MIDTERM
October 23, 2007
1
3
5
2 hours
Please circle your section
9 am
Galbiati
2 10 am McDonald
11 am
McDonald
4 11 am Staggs
12:30 pm Sondhi
Problem
1
2
3
4
5
6
7
Total
Score
/10
/10
/15
/15
/15
/20
/15
/100
Instructions: When you are told to begin, check that this examination booklet
contains all the numbered pages from 2 through 18. The exam contains 7 problems.
Read each problem carefully. You must show your work. The grade you get depends
on your solution even when you write down the correct answer. BOX your final
answer. Do not panic or be discouraged if you cannot do every problem; there are
both easy and hard parts in this exam. If a part of a problem depends on
a previous answer you have not obtained, assume it and proceed. Keep
moving and finish as much as you can!
Possibly useful constants and equations are on the last page, which you may
want to tear off and keep handy
Rewrite and sign the pledge: I pledge my honor that I have not violated the Honor Code
during this examination.
Signature
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Problem 1: Grab Bag
(a) This problems deals with collisions in a 1-dimensional system. Given the masses of the
two colliding bodies (m1 , m2 ), their initial velocities (v1,i , v2,i ), and the kind of collision
specified in each line, you are requested to calculate the final velocities: v1,f and v2,f .
m1
v2,i
v1,i
m2
a.1 [2 pts] Perfectly elastic collision; m1 =m2 =10 kg; v1,i =12 m/s; v2,i =−5 m/s.
The velocity of the center of mass vCM is:
vCM =
10 kg · 12 m/s + 10 kg · (−5 m/s)
= 3.5 m/s
2 kg
The initial velocities in the frame of the center of mass frame of reference are obtained by
subtracting the velocity of the center of mass:
u1,i = 8.5 m/s, u2,i = −8.5 m/s.
Since the collision is elastic, the final velocites in the center of mass frame of reference are
obtained from the initial velocities (in the same frame) by flipping the sign:
u1,f = −8.5 m/s, u2,f = 8.5 m/s.
We finally find the final velocities in the original frame of reference by adding the velocity of
the center of mass:
v1,f = -5 m/s
v2,f = 12 m/s
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a.2 [1 pts] Perfectly elastic collision; m1 =20 kg; m2 =10 kg; v1,i =12 m/s; v2,i =−5 m/s.
The velocity of the center of mass vCM is:
vCM =
20 kg · 12 m/s + 10 kg · (−5 m/s)
= 6.3 m/s
2 kg
The initial velocities in the frame of the center of mass frame of reference are obtained by
subtracting the velocity of the center of mass:
u1,i = 5.7 m/s, u2,i = −11.3 m/s.
Since the collision is elastic, the final velocites in the center of mass frame of reference are
obtained from the initial velocities (in the same frame) by flipping the sign:
u1,f = −5.7 m/s, u2,f = 11.3 m/s.
We finally find the final velocities in the original frame of reference by adding the velocity of
the center of mass:
v1,f = 0.6 m/s
v2,f = 17.6 m/s
a.3 [2 pts] Perfectly inelastic collision; m1 =20 kg; m2 =10 kg; v1,i =12 m/s; v2,i =−5 m/s.
The velocity of the center of mass vCM is:
vCM =
20 kg · 12 m/s + 10 kg · (−5 m/s)
= 6.3 m/s
2 kg
Both balls move with the velocity of the center of mass after the collision.
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(b) [5 pts] A PHY101 student on winter break snowboards down the backside of Spherical
Half Dome, starting form rest on the top of the Dome. At what angle does she become
airborne?
R
R
R
The take off condition requires the normal force to be null. At takeoff, the only force providing
the centripetal acceleration is the component of the weight force in the radial direction:
mg cos θ = m
v2
R
The kinetic energy can be obtained by applying the energy conservation principle between the
start of the slide and the point at of takeoff at the angle θ. Note the height at the start of the
slide is R, and at the takeoff point is R cos θ.
1
mgR = mgR cos θ + mv 2 ,
2
Combining the two equations:
1
mgR = mgR cos θ + mgR cos θ,
2
cos θ = 2/3
θ = 48◦ .
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Problem 2: Ladder [10 pts]
A ladder of length L = 2.3 m and mass M = 8.7 kg leans against a wall at angle θ = 27◦
to the vertical. What is the minimum coefficient µs of static sufficient friction between the
foot of the ladder and the floor to keep the ladder from slipping, supposing that there is no
friction between the ladder and the vertical wall?
Static equilibrium ⇔
P
F=0=
P
τ.
The force diagram is:
Since the wall exerts no vertical force on the ladder, the (upward) normal force NF of the floor
on the ladder obeys:
NF = M g (= 85.3 N).
The normal force NF is perpendicular to the floor, not to the ladder. Similarly, the normal
force NW of the wall on ladder is perpnedicular to the wall, not to the ladder. Because there
is no friction between the ladder and the wall there is no vertical force between the ladder and
the wall.
The force of friction at the floor is Ffriction ≤ µNF = µM g, which must balance the (horizontal)
normal force NW of the wall on the ladder. If the coefficient of friction is minimal, we have
that:
NW = Ffriction = µmin M g.
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For static equilibrium, the sum of the torques on the ladder must be zero, so if we calculate
the torques about the foot of the ladder, we have:
0 = M g(L/2) sin θ − NW L cos θ,
noting that the force of gravity acts on the center of mass (CM) of the ladder, at distance L/2
from its ends. Thus,
NW = (M g/2) tan θ = Ffriction = µmin M g (= 21.7 N),
and so,
µmin = 0.5 tan θ = 0.255 (= NW /M g = 21.7/85.3).
Taking torques about the CM of the ladder:
0 = NF (L/2) sin θ − (NW + Ffriction )(L/2) cos θ,
or:
M g sin θ = 2µmin M g cos θ,
which also leads to µmin = 0.5 tan θ.
We could also calculate the torques about the upper end of the ladder:
0 = NF L sin θ − Ffriction L cos θ − M g(L/2) sin θ,
or:
which also leads to µmin
Mg
sin θ = µmin M g cos θ,
2
= 0.5 tan θ.
We could even calculate the torques about the axis where the floor meets the wall:
0 = NF L sin θ − NW L cos θ − M g(L/2) sin θ,
or
which also leads to µmin
Mg
sin θ = µmin M g cos θ,
2
= 0.5 tan θ.
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Problem 3. Pulley.
A mass M = 7.5 kg is suspended from a pulley of mass m = 1.3 kg, radius r = 0.053 m,
and moment of inertia I = 2mr2 /3. A (massless) rope is attached to a roof beam and run
around the pulley, as shown in the figure. The tension T on the free end of the rope is 10%
more than that needed to keep the system at rest.
a. [5 pts] What is the tension T ?
Two segments of the rope hold the masses at rest, so Trest = (M + m)/2. Then,
T = 1.1(M + m)g/2 = 1.1(7.5 + 1.3)9.8/2 = 47.4 N.
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b. [5 pts] What is the relation between the angular acceleration α of the pulley and the
vertical acceleration a of mass M ?
α = a/r.
This is straightforward. However, if tension T is applied by your hand, then the vertical
acceleration of your hand must be 2a not a.
c. [5 pts] What is the acceleration a of mass M ?
The (upward) acceleration a of mass M is the same as the acceleration a of the
center of mass of the pulley.
Let T 0 be the tension in the portion of the rope that is connected to the roof
beam. Then, the vertical component of f = ma on the system of pulley plus mass
M is:
T + T 0 − (M + m)g = (M + m)a.
If we only analyze the acceleration of the pulley, we have:
T + T 0 − T 00 − mg = ma,
while f = ma for mass M alone gives:
T 00 − M g = M a,
so if we add these two equations we get the first equation, which is for masses m
and M combined.
The torque equation for the motion of the pulley is:
τ = (T − T 0 )r = Iα =
2mr2 a
,
3 r
so that:
2
ma.
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Adding this equation to the 1st equation to eliminate T 0 , we find:
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2T − (M + m)g = M + m a.
3
T − T0 =
Inserting the relation T = 1.1(M + m)g/2, the acceleration of mass M is:
a=
0.1(M + m)
0.1(7.5 + 1.3)
g=
9.8 = 0.89 m/s2 .
M + 5m/3
7.5 + 1.67 · 1.3
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Problem 4. Sledding.
Two girls, each of mass M = 20 kg, are sledding on identical plastic sleds each of mass
m = 2.0 kg. At the bottom of a hill, they find themselves sliding next to each other without
friction on a horizontal area at the same velocity v0 = 5.0 x̂ m/s. Suddenly, child A leaps
off her sled and lands on the sled of child B. While she is in the air, the x-component of her
velocity is v1x = v0 = 5.0 m/s and the y-component is v1y = v0 /2 = 2.5 m/s.
Note: in this problem, the x and y axis cover the horizontal area where the two sleds slide
after coming down the hill. The y axis is not vertical!
a. [8 pts] What is the final velocity v2 of A’s sled after she leaps off it?
The initial momentum of A and her sled is P0 = (M + m)v0 x̂ = 11mv0 . The total horizontal
momentum of A and her sled just after she leaps is the sum of her momentum, P1 = M v1 =
10m(v0 x̂ + (v0 /2) ŷ) and the sled’s final momentum which we may write as P2 = mv2 . Since
there are no horizontal forces acting on the isolated system of A and her sled, momentum is
conserved so that P0 = P1 + P2 . In other words,
P2x = P0x − P1x = 11mv0 − 10mv0 = mv0 ,
so that v2x = v0 since the mass of the sled alone is just m. For the y-component we find
P2y = −P1y = −5mv0 ,
so that v2y = −5v0 .
Thus, v2 = (5.0 m/s)x̂ − (25.0 m/s)ŷ.
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b. [7 pts] What is the final velocity vF of A, B and B’s sled after A lands on it? If you did
not get part a, leave your answer in terms of v2x and v2y as needed.
Here we can either consider the initial state to be that of A flying through the air with v1 while
B and her sled slide in the x direction, or we can consider the original initial state, in which
we found above for one sled-plus-child P0 = 11mv0 . We’ll do the latter, noting that there are
no external horizontal forces acting on the isolated system comprising the two children and the
two sleds. We then consider the total initial momentum (for two children and two sleds) to be
P0,tot = 2P0 = 22mv0 x̂.
Using the notation from above for A’s sled’s final momentum, we can write: P0,tot = P2 + Pf ,
where the final momentum of the two children plus sled is Pf = (2M + m)vf = 21mvf .
Therefore:
Pf x = 22mv0 − P2x = 22mv0 − mv0 = 21mv0 = (21m)vf x
so that vf x = v0 . And for the y-component,
Pf y = −P2y = +5mv0 = (21m)vf y,
so that vf y = (5/21)v0 .
Thus, v2 = (5.0 m/s)x̂ + (1.2 m/s)ŷ.
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Problem 5. Kayaking.
A man stands on a cliff, ready to toss a water bottle down to his wife. His wife paddles past
in a kayak. As she draws even with him, he tosses the bottle up and forward (by “forward”
we mean here the direction of the motion of the kayak), from a height h = 2.6 m above her.
At the same instant, the kayak speed is v0k = 5.0 m/s and beginning to slow, so that its
constant acceleration is a = −2.0 m/s2 . After the kayak has gone forward d = 4.0 m, the
bottle lands in the kayak, as shown in the figure.
h
d
a. [5 pts] How long is the bottle in the air?
The time is entirely determined by how long it takes the kayak to get to d to receive the bottle!
Solve the equation d = v0k t + 12 at2 . Then,
t=
v0k ±
p
2 − 2|a|d
v0k
5.0 m/s ±
=
|a|
p
25 m2 /s2 − 2(2.0 m/s2 )(4.0 m)
= (2.5 ± 1.5) s.
2.0 m/s2
We identify the appropriate time as t = 1.0 s. (The longer time solution would only result if
the kayak decelerated until it was going backwards, and in that way returned to the spot a
distance d from the starting place.)
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b. [5 pt] At what angle θ above the horizontal was the bottle launched?
We can find the angle from tan θ = v0y /v0x , if we can find the inital x- and y− components of the velocity of the bottle. We solve for v0x from d = v0x t, so that v0x = d/t =
(4.0 m)/(1.0 s) = 4.0 m/s = v0x . Next we solve for v0y from the equation
−h = v0y t − 1/2gt2 ,
which can be rearranged to give
v0y =
(9.81 m/s2 )(1.0 s) 2.6 m
gt h
− =
−
= 2.3 m/s .
2
t
2
1.0 s
Then we write
θ = tan
−1
v0y
v0x
= tan
−1
2.3
4.0
= 30◦ .
Thus, θ = 30◦ .
c. [5 pts] What was the maximum height the bottle attained above the river?
Note that the maximum height does not occur at t/2, where t is the time for the bottle to
make its complete trip. Instead, we use the fact that at the ball’s maximum height, vy = 0.
The kinematic variables in y that we know are then a = −g, v0y = 2.3 m/s, and vy = 0, and
the unknown variables are t0 and y 0 . We find t0 from
vy = v0y + at0 = v0y − gt0 = 0,
2
so that t0 = v0y /g = (2.3 m/s)/(9.81 m/s ) = 0.23 s. The height above the initial position then
follows from
1
2
y 0 = v0y t − gt2 = (2.3 m/s)(0.23 s) − (0.5)(9.81 m/s )(0.23 s)2 = 0.27 m.
2
Finally then, the maximum height above the water is y 0 + h = 2.6 m + 0.27 m = 2.9 m .
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Problem 6. Sliding Mass.
A mass m = 1.00 kg slides in a circle of radius r1 = 1.00 m on a frictionless table as shown.
The mass is attached to a massless string which passes through a hole in the table and is
held by by an experimenter. The mass has an angular velocity ω1 = 10.0 rad/s.
r1
m
a. [5 pt] What is the moment of inertia of the mass-string system?
I = mr12 = 1.00 kg · m2 .
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b. [5 pts] What is the tension in the string?
The tension supplies all of the centripetal acceleration and thus equals:
T1 = mω12 r1 = 100 N
c. [5 pt] The experimenter now slowly pulls the string until the mass is at radius r2 =
0.500 m. How much work is done on the mass-string system in this process?
As the force is always radial, it produces no torque about the center of the (instantaneous)
orbit which is always at the hole in the table. Thus angular momentum is conserved. Hence,
L1 = I1 ω1 = mr12 ω1 = L2 = I2 ω2 = mr22 ω2
whence:
ω2 = ω1
r12
= 40 rad/s.
r22
The increase in kinetic energy is:
∆KE =
2
1
1
1
1
r4
1
r1
I2 ω22 − I1 ω12 = m r22 ω22 − r12 ω12 = m r22 ω12 14 − r12 ω12 = mr12 ω12
−
1
.
2
2
2
2
r2
2
r22
The increase in kinetic energy comes from the work done on the mass-string system by the
experimenter pulling on the string. This can be rewritten as:
"
#
2
1 2 2 r12
1
1m
2
2
− 1 = 150 J .
W = ∆KE = mr1 ω1 2 − 1 = 1 kg (1 m) (10 rad/s)
2
r2
2
0.5 m
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d. [5 pts] The string will break if its tension exceeds 9.00 × 102 N. What is the smallest
radius rmin to which the mass can be safely pulled by the experimenter?
By our earlier reasoning the angular velocity at radius r is:
ω(r) =
r12
ω1 .
r2
Hence the tension required to supply the centripetal acceleration is:
T = mω 2 (r)r = mω12
r3
r14
r4
r3
r3
r = mω12 13 = mω12 r1 13 = T1 13 = 100 N 13 ,
4
r
r
r
r
r
We can now solve for r by requiring that T = 900 N. We obtain:
r = r1
T1
T
1/3
= r1
100 N
900 N
1/3
= 0.481 m ,
which is then the farthest in that the mass can be pulled without endangering those nearby.
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Problem 7. Sliding Block.
A mass m=1.00 kg slides down the edge of the block of mass M =8.00 kg shown in the figure.
The block of mass m slides down from its initial position, located at an height h=0.100 m
above the ground. There is no friction between the two masses m and M . The two masses
are initially at rest on a frictionless table.
m
M
h
a. [5 pt] What are the velocities of m and M right after m reaches the table?
Let v be the speed of m to the right and V be the speed of M to the left after m slides off M .
Conservation of energy and momentum lead to the equations:
mv + M (−V ) = 0 and mgh =
which imply:
V =−
1
1
mv 2 + M V 2 ,
2
2
m
v.
M
Substituting into the energy derived from the energy conservation principle, we obtain:
s
s
2gR
2 · 9.8 ms−2 · 0.1 m
=
= 1.32 m/s .
v=
1 + (m/M )
1 + 1 kg/8 kg
Then:
V =−
m
1 kg
v=−
1.32 m/s = - 0.165 m/s,
M
8 kg
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b. [5 pts] Subsequently, m slides further on the table and makes an elastic collision with a
wall, which is of course at rest and unmovable. What is the final velocity of m?
The mass m collides elastically with an “infinite” mass. The velocity of the center of mass
frame of reference is zero, before and after the collision: i.e., the system of the observer is
already the center of mass frame of reference!
Therefore, the velocity of mass m is simply reversed:
v2 = −v = - 1.32 m/s .
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c. [5 pt] Show that m catches up with its old friend M. What is the highest point on M ’s
back that m climbs up to now?
As m starts back to the left with a speed v2 = −v = −1.32 m/s it will overtake M which is
m
ahead but is trudging along at a mere speed −v M
= −0.165 m/s. After reaching M it will
start to climb up its back again. At the highest point, whose height above the table we will
denote by h0 , m will be at rest relative to M , i.e. they will both be moving at the same velocity
u towards the left. Our favorite conservation laws applied before m reaches M and at this
point of maximum height imply that:
m
− mv = (M + m)u,
M −v
M
and therefore:
u = −v
2m
.
m+M
Applying the principle of energy conservation we obtain:
1 m 2 1
1
2
M −v
+ m (−v) = (M + m) u2 + mgh0 ,
2
M
2
2
which becomes:
2
m 1
2mv
1
2
mv 1 +
+ mgh0 ,
= (M + m) −
2
M
2
m+M
and:
mgh0 =
1
m 1
4m
mv 2 1 +
− mv 2
,
2
M
2
m+M
Therefore:
v2
h =
2g
0
m
4m
1+
−
M
m+M
v2
=
2g
1 4
1+ −
8 9
2
=
49 v 2
49 (1.32 m/s)
=
= 0.061 m .
72 2g
72 2 · 9.8 ms−2
19
POSSIBLY USEFUL CONSTANTS AND EQUATIONS
You may want to tear this out to keep at your side
L = Iω
PE = mgh
ω = ω0 + αt
v = v0 + at
F = µN
Στ = Iα
ac = v 2 /r
Wnc = ∆KE + ∆PE
I = 25 mr2 [sphere]
REarth = 6400 km
I = Σmi ri2
KE = 12 Iω 2
ω 2 = ω02 + 2α∆θ
F~ ∆t = ∆~p
s = Rθ
v = Rω
W = F s cos θ
a = Rα
I = 31 M w2 [thin sheet]
MEarth = 6.0 × 1024 kg
x = x0 + v0 t + at2 /2
KE = 12 mv 2
∆θ = ω0 t + 12 αt2
F = −GM m/r2
τ = F ` sin θ
p~ = m~v
v 2 = v02 + 2a∆x
I = 12 mr2 [disk]
L = mvr
G = 6.67 × 10−11 Nm2 /kg2