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Physics 221.
Exam II
Summer 2003
The following situation refers to the next two problems:
A bobsled slides down a frictionless icy slope and is observed to have a speed of 35 m/s when it
crosses the final line (point B). The ground behind the finish line is horizontal and has sand spread
on it to slow the bobsled. The coefficient of kinetic friction between the horizontal, rough section
and the bobsled is µk = 0.72.
A×
h
B (finish line)
×
C
×
D
51. What was the speed of the bobsled when it was at point A, at height h = 60 m above the finish
line?
a.
b.
c.
d.
e.
vA = 2.1 m/s
vA = 7.0 m/s
vA = 9.2 m/s
vA = 17 m/s
vA = 31 m/s
Use conservation of energy:
1 2
1
mv A + mgh = mvB2
2
2
v A = vB2 − 2 gh = 7.0 m/s
52. The bobsled comes to a stop at point C. Find the distance D between points B and C.
a.
b.
c.
d.
e.
D = 20 m
D = 42 m
D = 61 m
D = 87 m
D = 110 m
Use the work-kinetic energy theorem. The only force doing work on the bobsled between B and C is
friction, so
1
0 − mvB2 = − µ k mgD
2
v2
D = B = 87 m
2µ k g
Page 1 of 13
Physics 221.
Exam II
Summer 2003
The following situation refers to the next two problems:
A uniform flat plank of mass M = 1000 kg and length L = 10 m floats at rest on a perfectly still lake.
A man of mass m = 100 kg stands on one end of the plank. The man begins running to the left.
x=0
L
x
L
53. When the man reaches the other end of the plank, how far has the plank moved relative to the
ground?
a.
b.
c.
d.
e.
0.34 m
0.56 m
0.91 m
1.32 m
4.44 m
There are no external forces on the system, so the position of the CM must remain the same before
and after the man moves along the plank. Take x = 0 at the left end of the plank before the man
starts moving:
L
M + mL
Before : xCM = 2
m+M
L

M  + x  + mx
2

After, the plank has moved a distance x: xCM = 
m+M
L
L

+ mL =M  + x  + mx
2
2

These two quantities must be the same:
m
x=
L = 0.91 m
m+M
M
Page 2 of 13
Physics 221.
Exam II
Summer 2003
54. After he gets to the leftt end of the plank, he turns around and walks back to where he started.
His walking speed is less than his running speed. At the end, the position of the plank will be:
a. Where it was at the beginning.
b. A little to the left of where it was at the beginning because the linear momentum of the plank is
now smaller.
c. A little to the left of where it was at the beginning because the force exerted by the man on the
plank is now smaller.
d. A little to the right of where it was at the beginning because the plank has now more time to
move.
e. A little to the right of where it was at the beginning because the impulse on the plank is now
larger.
Using the same argument as before, the position of the plank must be the same. The motion will take
longer, but the final result is the same as if the man runs back.
Page 3 of 13
Physics 221.
Exam II
Summer 2003
The following situation refers to the next two problems:
A compressed spring of constant k = 2000 N/m is used to shoot a dart of mass m = 100 g at a
wooden block of mass M = 1.0 kg as shown in the figure. When the dart hits the block, it stays
pinned on it. Right after the dart hits the block, the speed of the block is measured to be vf = 0.4 m/s.
M
m
55. How much had the spring been compressed to shoot the dart?
a.
b.
c.
d.
e.
∆x = 0.9 cm
∆x = 2.3 cm
∆x = 3.1 cm
∆x = 6.7 cm
∆x = 8.3 cm
For the spring-dart system, energy is conserved:
1 2 1 2
m
kx = mvdart ,
so x = vdart
2
2
k
Then there’s a collision, where linear momentum is conserved:
M +m
mvdart = ( M + m)v f ,
so vdart =
vf
m
Putting everything together:
m M +m
M +m
x=
vf =
v f = 0.031 m=3.1 cm
k m
km
Page 4 of 13
Physics 221.
Exam II
Summer 2003
56. The dart is now replaced with a rubber ball of mass m that bounces back after hitting the block.
After being hit by the ball, the block is observed to move at the same speed vf as before. What can be
said about the initial compression ∆x of the spring?
a. It was smaller than before.
b. It was the same as before.
c. It was larger than before.
d. It is not possible to answer this question without knowing the speed of the ball after hitting the
block.
e. It is not possible to answer this question without knowing whether the collision was elastic or
inelastic
When the projectile bounces back instead of sticking to the block, the momentum transfer is larger:
p
p
pball
pblock
|pblock| = |pball| + |p|
If the final speed of the block is the same, the ball was slower than the dart. Thus the spring had
been compressed less.
Page 5 of 13
Physics 221.
Exam II
Summer 2003
57. A 12-kg cannon ball is fired from a 675-kg brass cannon on a frictionless, horizontal surface at
an angle of 23° with respect to the horizontal. The cannon is constrained to move only in the
horizontal direction. The initial velocity of the cannon ball is vi = 52 m/s. What is the recoil speed of
the cannon?
a.
b.
c.
d.
e.
0.85 m/s
0.92 m/s
1.2 m/s
1.8 m/s
2.1 m/s
θ = 23°
Total linear momentum for the cannon+ball system is conserved only along the x-direction. (There
are external forces along the y-direction: the normal by the ground will increase appropriately to
absorb the y-component of the force exerted by the cannon ball on the cannon).
0 = M c vc + mb vb , x
mb
m
vb , x = − b vb cos 23D = −0.85 m/s
Mc
Mc
(The minus sign is dropped because the question asks for the speed.)
vc = −
58. A pear-shaped flat object with uniform mass density can rotate about different axis of rotation.
Rank the moments of inertia about each axis:
Axis 1
a.
b.
c.
d.
e.
Axis 2
Axis 3
I1 = 0; I2 < I3
I1 = 0; I2 > I3
I1 < I2 < I3
I1 > I2 > I3
I1 > I3 > I2
Moment of inertia is larger as more mass is far away from the axis. All the mass is far away from
axis 1, so it’s the largest moment of inertia. For axis 3, the thick side of the object is far from the
axis, whereas for axis 2 it’s only the thin part (with less mass since the object is uniform). So I3 > I2.
Page 6 of 13
Physics 221.
Exam II
Summer 2003
The following situation refers to the next three problems:
A 5-kg solid bowling ball rolls without slipping down a rough, inclined plane. The acceleration of
the ball down the ramp is 4 m/s2.
N m = 5 kg
a = 4 m/s2
fs
h
mg
59. The forces that exert a non-zero torque on the ball with respect to its center of mass are:
a.
b.
c.
d.
e.
Gravity and friction only.
Friction only.
Gravity, friction and the normal force of the plane.
Gravity and the normal force of the plane only.
Friction and the normal force of the plane only.
Gravity is applied at the CM, so it cannot produce any torque about this point.
The normal force is anti-parallel to the lever arm that goes from the CM to the point of application
of the normal, so the cross-product r×N is also zero.
Friction, instead, produces a clockwise torque.
60. Find the magnitude of the frictional force acting on the bowling ball as it rolls down the ramp.
a.
b.
c.
d.
e.
f=0N
f=4N
f=8N
f = 10 N
f = 12 N
Newton’s second law for the rotation of the cylinder reads:
2
f s R = Iα
with I = mR 2
5
If the ball rolls without slipping, aCM = Rα . Then we have:
2
a
f s R = mR 2
R
5
2
f s = ma = 8.0 N
5
Page 7 of 13
Physics 221.
Exam II
Summer 2003
61. If the bowling ball is replaced by a kid’s bowling ball, with half the mass and half the radius of
the original one. The acceleration changes by a factor of:
a.
b.
c.
d.
e.
4
2
1 (Does not change)
½
¼
Changing the dimensions or mass of the ball will not make a difference as long as it is still a solid
sphere. (We saw it in a demo with cylinders). Write 2nd Newton’s law for the translation of the CM:
mg sin θ − f s = ma
And the equation for rotation was:
2
a
f s R = mR 2
R
5
Thus,
2
mg sin θ − ma = ma
5
2
g sin θ − a = a
5
Even without bothering to solve this equation, we see that the answer does not depend on the mass
or the radius of the sphere. It only depends on the fact that it is a sphere (the 2/5 factor).
Page 8 of 13
Physics 221.
Exam II
Summer 2003
The following situation refers to the next three problems:
A block of mass m1 = 1.0 kg sits atop a frictionless plane and is connected to a mass m2 = 2.0 kg
through a massless, ideal string that goes over a pulley of mass M = 4.0 kg and radius R = 0.20 m.
The pulley rotates about its axis without friction and the string moves over the pulley without
slipping. The system starts at rest and mass m2 falls through a height h = 2 m. The pulley can be
considered a uniform cylinder of radius R.
M
m1
y
m2
z
(out of the page)
x
m1 = 1.0 kg
m2 = 2.0 kg
M = 4.0 kg
R = 0.20 m
h=2m
h
62. What is the speed of mass m2 right before it hits the ground?
a.
b.
c.
d.
e.
v = 2.4 m/s
v = 2.8 m/s
v = 3.1 m/s
v = 4.0 m/s
v = 5.1 m/s
Use conservation of energy:
1
1
1
m2 gh = m2 v 2 + m1v 2 + Iω 2
2
2
2
where I =
1
MR 2
2
If the string does not slip on the pulley, v = Rω . Thus,
1
1
11
v
m2 gh = m2 v 2 + m1v 2 +
MR 2  
2
2
22
R
v=
2m2 gh
M
m1 + m2 +
2
2
= 4.0 m/s
Page 9 of 13
Physics 221.
Exam II
Summer 2003
63. The direction of the net torque on the pulley is:
a.
b.
c.
d.
e.
−y
−y
+z
−z
+x
The angular acceleration of the pulley is clockwise, so into the page. The net torque must point in
the same direction. In terms of the axis, this is the −z-direction.
64. Mass m1 is replaced with a cylinder of the same mass and the plane is coated with a rough paint
to produce friction. The cylinder rolls without slipping on the plane (the string is attached to an axle
going through the center of the cylinder). The speed of mass m2 right before it hits the ground will
be:
a.
b.
c.
d.
e.
Larger than before.
The same as before.
Smaller than before.
It depends on the radius of the cylinder.
It depends on the coefficient of static friction between the cylinder and the plane.
In this case, there would be an additional term in the equation for conservation of energy,
corresponding to the rotational kinetic energy of the cylinder of mass m1 and radius r1:
m2 gh =
1
1
1
1
m2 v 2 + m1v 2 + I1ω12 + Iω 2
2
2
2
2
where I =
1
1
2
MR 2 and I1 = m1r1
2
2
The same potential energy needs to be now divided in four parts instead of three, which means less
energy for every part −mass m2 in particular.
Page 10 of 13
Physics 221.
Exam II
Summer 2003
65. Two blocks are attached to two massive wheels by ideal cables as shown in the figure. The two
wheels are solidly attached to one another, so they rotate together on a frictionless axle. The radii of
the wheels are R1 = 1.5 m and R2 = 0.5 m. The system is in equilibrium (everything is at rest).
N
R1
R2
T2
g
T1
T1
T2
m2
m1
m2g
m1g
If m1 = 30 kg, what is m2?
a.
b.
c.
d.
e.
m2 = 20 kg
m2 = 30 kg
m2 = 40 kg
m2 = 65 kg
m2 = 90 kg
The net torque and the net force on each part of the system must be zero. For the hanging masses,
we have:
m2 g − T2 = 0 ⇒ T2 = m2 g
m1 g − T1 = 0
⇒
T1 = m1 g
For the wheels, we have:
T1 + T2 − N = 0 (not interested in this part)
R1T1 − R2T2 = 0
Substitute the values for the tensions:
R1m1 g − R2 m2 g = 0
m2 = m1
R1
= 90 kg
R2
Page 11 of 13
Physics 221.
Exam II
Summer 2003
66. A wooden beam sitting on a frictionless, horizontal surface is pivoted at its center. A dart
traveling with speed v hits and sticks to one end of the beam. Let KE, L and P be the kinetic energy,
angular momentum about the pivot point and linear momentum of the dart-beam system. Which of
the following statements is true?
a.
b.
c.
d.
e.
Only KE is conserved through the collision.
Only L is conserved through the collision.
Only P is conserved through the collision.
Both KE and L are conserved through the collision.
Both KE and P are conserved through the collision.
v
L is conserved because the external torque about the pivot point is zero.
P is not conserved because there is an important external force acting on the pivot (that prevents the
beam from sliding across the table when it is hit by the dart and forces it to rotate instead).
KE is not conserved because this is an inelastic collision. Some energy is used to stick the dart on
the beam (friction, rearrangement of atoms).
Page 12 of 13
Physics 221.
Exam II
Summer 2003
67. Consider the Earth as a uniform, solid sphere that makes one complete revolution in 1 day. As
part of a carefully crafted attack, aliens from outer space drop a small neutron star on the equator of
the Earth. The mass of the neutron star is exactly the same as the mass of the Earth. (Neutron stars
are extremely dense, so you can assume it is a point mass).
North
ω
R
M
Equator
M
After the star is dropped, how long does it take the Earth to make one complete revolution?
a.
b.
c.
d.
e.
0.93 days
1.00 days
2.00 days
2.67 days
3.50 days
There is no external torque about the axis of rotation of the Earth on the Earth+star system, so
angular momentum about this axis is conserved.
2
Lbefore = Iω 0 = MR 2ω 0 (use the moment of inertia of a solid sphere for the Earth)
5
7
2

Lafter = ( I + MR 2 ) ω f =  MR 2 + MR 2  ω f = MR 2ω f
5
5

Lbefore = Lafter
Since T ∝
⇒
2
7
ω0 = ω f
5
5
⇒
2
7
ω f = ω0
1
7
, this means T f = T0 = 3.5 days
2
ω
Page 13 of 13