Physics 122 Exam #2
Spring 2015
You are subject to the Bryn Mawr College honor code while taking this
exam. This is a closed book exam. You may use a simple calculator and a
pencil during this exam, as well as both sides of a single page of notes. Do
all questions on the exam. Do all work directly on the examination, in the
space provided. If you run out of room, you may use the back of the page.
Be sure to answer all parts of each question. Some problems will take longer
than others, but all will be weighted equally when graded. When providing
a sketch, please be as neat, accurate, and quantitative as possible. A sloppy
sketch indicates misunderstanding and will receive a reduced grade. The
final pages of this exam provide several potentially useful equations. You
may remove these page for easy reference.
This exam will be available between Friday, 3 April 2015 and Wednesday,
8 April 2015, at the reserve desk in the Collier Science Library during normal
library operating hours. I will collect all of the exams Thursday morning
when the library opens. Take the exam during a single 90 minute sitting
in the library.
Before you begin, write your name in the space below and at the
top of EVERY page of this exam, note the date in the space provided on
this page, and note the time. When you finish, note the time and return the
exam to the circulation desk. The difference between your start and finish
times should be no more that 90 minutes.
Name:
Date:
Start time:
Finish time:
m1
1. A block of mass m1 slides down the frictionless ramp shown below.
The block starts from rest at a height h above the bottom of the ramp.
At the bottom it collides elastically with a block of mass m2 = 2m1 .
Derive an expression for the speed of each block after the collision.
After the collision, m1 will travel back up the incline to a height h′ .
What is the ratio of the two heights, h′ /h?
h
h'
m2 = 2m1
The speed v1 of m1 at the bottom of the ramp, but before the collision can be found
using energy conservation,
m1 gh = m1 v12 /2
p
v1 = 2gh.
(2)
m1 v1 = m1 v1′ + m2 v2′ .
(3)
(1)
The momentum before and after the collision is conserved,
Also, for a one dimensional elastic collision the relative velocity before and after the
collision has the same magnitude, but opposite direction,
v1 = v2′ − v1′ .
(4)
To find v2′ I will solve (4) for v1′ and substitute into (3),
m1 v1 = m1 (v2′ − v1 ) + m2 v2′
2m1 v1 = (m1 + m2 )v2′
v2′ = 2m1 v1 /(m1 + m2 ).
(5)
(6)
(7)
Finally, I will substitute in v1 from (2) and m2 = 2m1 ,
v2′ = 2m1 v1 /3m1
2p
v2′ =
2gh.
3
Substituting (9) into (4) we get,
1p
2gh.
v1′ = −
3
(8)
(9)
(10)
Since m1 travels to the left after the collision, it will go some distance back up the
ramp. To find out how high I again use energy conservation,
1
m1 (v1′ )2
2
2
1
1p
gh′ =
2gh
−
2
3
1
gh′ = gh
9
h′
1
=
h
9
m1 gh′ =
(11)
(12)
(13)
(14)
2. A pendulum of length ℓ is held with its string horizontal, and then
released. The string runs into a peg a distance d below the pivot, as
shown below. What is the smallest value of d for which the string
remains taut at all times?
d
The radius of the circle after hitting the peg is r = ℓ−d. The height change from where
it is released to the top of the motion after hitting the peg is ℓ−2r = ℓ−2(ℓ−d) = 2d−ℓ.
Now we can use conservation of energy to find the velocity of the mass at the top of
the motion after it hits the peg
mg(2d − ℓ) = mv 2 /2.
In general, both gravity and tension pull down on the object at this point and keep it
moving in a circle. The minimum speed is when the tension at the top of the motion
is zero. In this case, just gravity provides the centripetal acceleration needed to keep
the object moving in a circle.
mg = mv 2 /r.
Putting in v and r from earlier we have
mg = m[2g(2d − ℓ)]/(ℓ − d).
Solving for d we find
d = 3ℓ/5.
3. Suppose a light weight golf ball is hurled at a heavy bowling ball, which
is initially at rest. Assume that the collision is head-on and elastic.
After the collision, which ball has the greater momentum? Which has
the greater kinetic energy? Why?
Before
golf
After
golf
bowling
After the collision the momentum of the bowling ball is greater.
Since the golf ball has a much smaller mass than the bowling ball, it will
reverse direction after the collision. To conserve momentum the bowling
ball must have as much momentum as the golf ball originally had plus
enough extra to counter the new momentum of the golf ball that is now
in the opposite direction.
After the collision the kinetic energy of the golf ball is bigger.
Since the golf ball is much less massive, its velocity will reverse and be
nearly equal in magnitude to the original velocity. So, the kinetic energy of
the golf ball will not change by much. That means that not much energy
will be transferred to the bowling ball.
Alternatively, we can consider the kinetic energy of the two balls after the
collision using K = p2 /2m. While the momentum (which appears in the
numerator) of the bowling ball after the collision is larger than that of the
golf ball, this difference is at most a factor of two. However, the mass
(which appears in the denominator) of the bowling ball is much larger than
that of the golf ball. This large mass ultimately makes the kinetic energy
of the bowling ball smaller.
4. A massless spring with spring constant k is placed between a block
of mass m and a block of mass 3m. Initially the blocks are at rest
on a frictionless surface and they are held together so that the spring
between them is compressed by an amount D from its equilibrium
length. The blocks are then released and the spring pushes them off
in opposite directions. The spring is not attached to either block so it
falls to the ground after the blocks move apart. Find the final speeds
of the two blocks.
before
after
v1
m
3m
v3
m
Using conservation of momentum we can write,
pbef ore = paf ter
0 = 3mv3 − mv1
v1 = 3v3 .
Using conservation of energy,
Kbef ore + Ubef ore = Kaf ter + Uaf ter
0 + 12 kD 2 = 21 mv12 + 21 3mv32 + 0
kD 2
= v12 + 3v32 .
m
Substituting v1 from above and solving for v3 , we find,
kD 2
= (3v3 )2 + 3v32
m
kD 2
= 12v32
m
r
kD 2
v3 =
.
12m
So,
v1 = 3
r
kD 2
.
12m
3m
Potentially useful equations
vx = v0x + ax t
x = x0 + v0x t + 21 ax t2
2
vx2 = v0x
+ 2ax (x − x0 )
r = rr̂
v = ṙr̂ + r θ̇θ̂
a = (r̈ − r θ̇2 )r̂ + (r θ̈ + 2ṙθ̇)θ̂
v2
= ω2r
r
v = rω
2π
ω = 2πf =
T
ac =
F = ma
Ff riction = µN
W = mg
Fs = −kx
GMa Mb
Fgrav =
r2
for q̈ + ω 2 q = 0,
q(t) = A sin(ωt + φ)
A · B = |A||B| cos θ
A · B = Ax Bx + Ay By + Az Bz
|A × B| = |A||B| sin θ
A × B = ı̂(Ay Bz − Az By ) − ̂(Ax Bz − Az Bx ) + k̂(Ax By − Ay Bx )
g = 10 m/s2
G = 6.673 × 10−11 m3 kg−1 s−2
2
for ax + bx + c = 0,
x=
−b ±
√
b2 − 4ac
2a
More potentially useful equations
p = mv
dp
F=
dtZ
∆p =
R=
F dt
1 X
mi ri
M i
Wba =
Z
a
b
F · dr
K = 21 mv 2
Wba = Kb − Ka
Ub − Ua = −
Z
b
a
F · dr
Uspring = 12 kx2
Ugrav = mgh
GMm
UG = −
r
p1 + p2 = p′1 + p′2
K1 + K2 = K1′ + K2′
v1 − v2 = v2′ − v1′
K + U = K′ + U ′
Wnc = ∆K + ∆U