Physics 120
2013 March 15
— Midterm Exam 2 Solutions —
Surname: Solutions
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Question #
Mark
Maximum Mark
Multiple Choice
20
11
10
12
10
Total =
/40
Your SPECIAL CODE is 1111. Bubble in your Name, Student Number and Special Code on the Bubble
Sheet and check it. Enter your answers to the multiple choice questions on the bubble sheet by blackening in
the circle corresponding to the best answer. There is only one correct answer per question.
There are 10 multiple choice questions. Select the correct answer for each one and mark it on the
bubble form on the cover sheet. Each question has only one correct answer. (2 marks each)
1. An archer pulls the bowstring back a distance of
0.2 m before releasing the arrow. Consider that
the bow and string act like a spring with a spring
constant of 230 N/m. The arrow has a mass of
0.03 kg.
What is the magnitude of the spring potential energy of the drawn bow?
(d) 7.4 J
(a) 2.3 J
(e) 9.2 J
(b) 4.6 J Correct
(c) 6.2 J
Model the bow as an ideal spring. Energy
stored is Uspring = 21 kx2 . x is the extension
of the spring from equilibrium. Energy stored
for an extension of 0.2 m is U(0.2 m)spring =
0.5 × 230 × 0.22 = 4.6 J.
2. What is speed of the arrow when it leaves the
bow?
1
(a) 8.5 m/s
(b) 10.2 m/s
(c) 17.5 m/s Correct
(d) 19.3 m/s
(e) 23.4 m/s
The energy stored in the previous problem is
turned into kinetic energy: 21 mv2 = 4.6 J. v =
q
2×4.6
0.03 .
3. A pendulum is released from rest, as shown in
Figure A. When the bob is directly below the
pivot point, the ideal string encounters an ideal
peg, about which it begins to wrap, as shown in
Figures B and C. The peg is higher than the initial
height of the pendulum.
In Figure C, the bob rises up to a height
Physics 120
2013 March 15
— Midterm Exam 2 Solutions —
(a) equal to the original height of the bob in
Figure A. Correct
At which of the x values shown by a dashed line
is the magnitude of the force the greatest?
(b) greater than the original height of the bob
in Figure A.
F(x) = −dU(x)/dx. The slope is greatest at
point A. It’s magnitude is least where the slope
is zero, point B.
(c) less than the original height of the bob in
Figure A.
In order for to use up all the original potential
energy that the ball had when it started swinging is must return to the same height, no matter
the path it takes.
7. A force F~ acts on mass M during the time period
t = 0 s to t = 1 s. at t = 1s the mass moves with
momentum ~p1 as shown.
4. Two carts collide. During the collision the centre
of mass of the carts travels at a constant velocity
and is unchanged throughout. From this we can
conclude that before and after the collision
Which of the following vectors best represents
~p0 , the momentum of M at t = 0.
(a) total momentum of the carts is conserved
(b) total kinetic energy is conserved
(c) the relative velocities of the carts is conserved
A
(d) all of the above
(e) none of the above
Total momentum is conserved as long as no
external forces are acting on the system. Kinetic energy is only conserved in elastic collisions. Relative velocities of the carts has the
same magnitude, but opposite sign, so it is not
conserved.
5. The Work Kinetic-Energy Theorem is invalid if
work is done by a non-conservative force.
B
C
D
E
~ is to the right can only add to the x
The force F
component of the initial velocity. Therefore the
initial velocity must have a component in the
−x direction in order for it to end up with no x
component.
8. In case A a block is pushed by force F~ for 1 second. In case B a block of one-half the mass is
pushed by the same force F~ for 1 second. In both
cases the track is frictionless and the blocks are
initially at rest.
(a) True
(b) False
Non-conservative forces are those which dissipate energy to heat or forms other than kinetic.
6. The potential energy U(x) of an object as a function of x looks like the plot shown below
Compare the magnitudes of the final momenta of
the two blocks:
(a) pA < pB
(b) pA > pB
U(x)
(c) pA = pB
The force times time is the impulse and equals
the change in momentum. Thus the momentum change is the same for both block. Since
both blocks are initially at rest, their final momenta are the same.
x
A
B
C
2
Physics 120
— Midterm Exam
Solutions —
2013 March
ball, 2neglecting
any effects due to air resistance
and 15
making
reasonable assumptions for bat speeds and incoming pitch speeds?
Is this a “juiced” ball, a “normal” ball, or a “dead” ball?
83
•• the
C ONCEPTUAL
make
puck
easy, hockey
9. The centres of three spheres having masses 1 kg,
end of the To
plank
where
thehandling
man is standing.
pucks are kept frozen until they are used in the game. (a) Explain
2 kg, and 3 kg are placed at the corners of an
why room-temperature pucks would be more difficult to handle on
equilateral triangle whose sides are each 1 meter
the end of a stick than a frozen puck. (Hint: Hockey pucks are made of
long, as shown below. How far from the centre
rubber.) (b) A room-temperature puck rebounds 15 cm when
of the 1 kg sphere is the centre of mass of the
dropped onto a wooden surface from 100 cm. If a frozen puck has
only half the coefficient of restitution of a room-temperature one,
system?
predict how
frozen
would
underand
the same
Thehigh
manthe
walks
to puck
the other
endrebound
of the plank
SSM
conditions.
sits down on the rock. How far did the plank
move along the ice?
COLLISIONS
(d) 2.9 m Correct
(a) 0 m IN MORE
THAN (b)
ONE
DIMENSION
(e) 3.3 m
1.3 m
(c) 1.7 m
•• In Section 8-3, it was proved by using geometry that when
a particle elastically collides with another particle of equal mass that
thethe
c.m.
the boat-man-rock
beis initiallyFind
at rest,
twoofpostcollision
velocities system
are perpendicular.
fore the another
man moves:
Here we examine
way of proving this
result
that
illustrates
S
S
S
the power of vector notation. (a) Given that A ! B " C , square both
(d) 0.73 m Correct sides of this equation (obtain
L
(a) 0.33 m
scalar
product
of each side with itxcm = (mboat 2 + mthe
S S
rock L)/(m
man + mboat + mrock )
(e) 1.00 m
B " C2 S
" 2B # C . (b) Let the momentum of
self) to show that A2 ! 2
(b) 0.50 m
the initially moving particle be P and the momenta of the particles
S
S
(c) 0.67 m
Now find
relative
to the
boat, for
after
p1 and pit2 .is,
after the collision
be where
Write
the vector
equation
the conservation the
of linear
and square both sides (obtain the dot
guy momentum
moves:
Find the y-coordinate of the c.m. which product
is
of each side with itself). Compare it to the equation gotten
half=way between the top and bottom, ycm from
= the elastic
collisionLcondition (kinetic energy is conserved) and
0
S # +m
S
++mman L+mrock L)/(mman
xcm
= (m
1
boat +mrock )
boat two
finally show
that
these
cos 60◦ = 0.43 m. Then find the x compo2 equations imply that p1 p2 ! 0.
84
2
nent: xcm = (0.5√× 3 + 2 × 2)/(1 + 2 + 3) = 0.58
85
•• In a pool game, the cue ball, which has an initial speed of
. Subtract
to get
the change
the ball,
c.m. which
posi- is ini5.0 m>s, makes
an elastic
collision
with theineight
m. Distance = 0.432 + 0.582 = 0.722 m.
tionAfter
of the
tially at rest.
thesystem:
collision, the eight ball moves at an angle of
••
two objec
mass after
collision i
88
m
3
FIGURE
•• A
sion with a
2.0-kg ball i
of motion a
of the 2.0-k
sion. Hint: s
90
•• A
particle wit
is deflected
second part
initial direc
3(v0 # v cos
v ! v0 cos f
89
* CENTE
REFER
30° to the right of the original direction of the cue ball. Assume that
10. A man is standing at one end of a plank of length
the balls have
mass.
(a) Find
theboat
direction
of the cue
∆xcmequal
= mman
L/((m
+mrock )of=motion
100×10/340
= 2.9 m91
man +m
L = 10 m. The man has mass Mman = 100
•• In
ball
immediately
after
the
collision.
(b)
Find
the
speed
of
each
ball
mass m1 an
kg and the plank has mass Mplank = 40 kg
immediately
. after the collision.
a second pa
and the plank is atop a frictionless sheet of
n, collides
collision the
v
86
••
Object
A,
which
has
a
mass
m
and
a
velocity
i
The boat is on ice so the c.m. relative to0 the
ice. At the other end of the plank sits a large
initial energ
with object B, which has a mass 2m and a velocity 12 v0 jn. Following the
ice must remain the same. Therefore, the boat
rock of mass Mrock = 200 kg. The centre of
terms of m1
collision, object B has a velocity of 14 v0 in. (a) Determine the velocity of
moves
the distance
the c.m.
shifts.
mass of the man+plank+rock is 6.5 m from
particle is m
object A after
the by
collision.
(b) Is thethat
collision
elastic?
If not, express
the change in the kinetic energy in terms of m and v0 .
87
•• A puck of mass 5.0 kg moving at 2.0 m>s approaches
There are two written problems. Show all your
andpuck
explain
reasoning
to get fullice.
credit.
anwork
identical
that your
is stationary
on frictionless
After the
collision, the first puck leaves with a speed v1 at 30° to the orig-
11. A puck of mass 5.0 kg moving at 2.0 m/s approaches
anofidentical
puck
that is
stationary
on frictionless
inal line
motion; the
second
puck
leaves with
speed v2 at 60°,
v2 . (b) Was
as in
the speeds
to the original
linev1ofand
motion;
the the
ice. After the collision, the first puck leaves with
a Figure
speed v8-50.
30◦Calculate
1 at (a)
SSM
collision
elastic?
◦
second puck leaves with speed v2 at 60 , as in the figure.
(a) Calculate the speeds v1 and v2 .
[5 pts]
v1
5.0 kg
v = 2.0 m/s
30°
5.0 kg
FIGURE 8-50
Problem 87
(b) How much kinetic energy was lost or gained in the collision, if any? [2 pts]
3
the speed i
solution?
92
•• M
tion at 5.0 m
3.0 m>s. (a)
vcm from the
in the cente
elastic collis
of-mass fra
mass frame
frame by ad
sult by find
the original
93
•• R
of 5.0 kg an
60°
v2
* SYSTE
CONTI
MASS:
94
• E
of 200 kg>s
the rocket. F
Physics 120
2013 March 15
— Midterm Exam 2 Solutions —
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4
Physics 120
— Midterm Exam 2 Solutions —
2013 March 15
12. A bullet with mass m = 20 grams and velocity v = 100 m/s collides with a wooden block of mass M = 2
kg and stays embedded in it. The wooden block is initially at rest, and is connected to a spring with
k = 800 N/m. The other end of the spring is attached to an immovable wall. Note: You may assume
that the spring is massless and that the collision between the bullet and the wooden block is completely
inelastic.
(a) Calculate the momentum of the bullet before the collision. [2]
p = mv = 0.02 × 100 = 2 kg-m/s.
For version b: p = mv = 0.025 × 90 = 2.25 kg-m/s.
(b) What is the kinetic energy of the bullet before collision? [2]
1
2
2
2 mv = .5 × 0.020 × 100 = 100. J
Version b: 21 mv2 = .5 × 0.025 × 902 = 101.25 J.
(c) What is the kinetic energy of the block after the collision? [3]
The momentum is conserved during the impact of the bullet.
mv = (m + M)V
V=
m
v ≈ 1 m/s
m+M
(We neglected the bullet in the approximation. Including the bullet gives V = 0.99 m/s. The
difference is insignificant.)
1
MV 2 = 0.5 × 2 × 12 = 1.0 J
2
Version b:
V=
m
v = 1.111 m/s
m+M
1
MV 2 = 0.5 × 2 × 12 = 1.23 J
2
(d) What is the maximum compression of the spring? [3]
Let x be the maximum compression.
1 2
kx = 1 J
2
p
p
x = 2/k = 1/400 = 1/20 m
Version b:
x=
p
2 × 1.23/k =
5
p
1.23/350 = 0.0598 m
Physics 120
2013 March 15
— Midterm Exam 2 Solutions —
Formula Sheet
~·B
~ = A x Bx + Ay By + Az Bz = AB cos θ
A
~
~ = (Ay Bz − Az By )ı̂ + (Az Bx − A x Bz )̂ + (A x By − Ay Bx )k̂
A×B
A
~×B
~ = AB sin θ
√
If ax2 + bx + c = 0 then x = −b ±
~v = d~x
dt
For constant a:
For constant a:
Newton’s 2nd:
Friction:
K = 1 mv2
Z2
t2
J~ =
F~ dt
t1
∆U = −
Z
xf
F~ · d~x
xi
b2 − 4ac
2a
A
2
~a = d~v = d ~x
∆v = aavg ∆t
dt
dt2
∆x = 1 [vi + vf ] ∆t = vi ∆t + 1 a ∆t2
2
2
v2f = v2i + 2a ∆x
X
d~p
~p = m~v
F~n = m~a =
dt
n
fk = µk FN
fs,max = µs FN
Ug = mgy
Ug = −G Mm
r
Z x
f
W=
F~ · d~x
P = dW
dt
xi
X
mn~rn
n
~rcm = X
mn
B
A
n
ω = d2~θ
~ = d~
α
dt
dt2
∆ω = αavg ∆t
For constant α:
X
I=
mn~rn2
∆θ = ωi ∆t + 1 α ∆t2
2
X
τn = Iα
n
B
ω = dθ
dt
~L = ~r × ~p
n
~
~τ = d L
L = mvt r
L = Iω
K = 1 Iω2
2
dt
(In general θ, ω, α, τ are vectors, but most physics 120 problems only need their magnitudes.)
2
Uniform Circular Motion:
Fc = mv
r
q
SHM: Fs = −k ∆s
ω= k
Us = 1 k ∆s2
m
2
ω = 2π f
f = 1/T
r
g
Pendulum, small amplitudes: ω =
l
Universal Gravitation:
F12 = − Gm1 m2
G = 6.67 × 10−11 Nm2 /kg2
2
r12
Mearth = 5.09 × 1024 kg
rearth = 6.37 × 106 m (mean) gearth = 9.81 N/kg (mean)
22
Mmoon = 7.34 × 10 kg
rmoon = 1.74 × 106 m (mean) c = 3.0 × 108 m/s
Moments of Inertia (Rotational Inertias) about the centre of mass.
1
Uniform disk: I = 21 MR2 , Hoop: I = MR2 , Rod: 12
M`2 , Solid Sphere: I = 25 MR2 , Hollow Sphere: I = 32 MR2
6