Mar06 LeClair
PH 105-3
Exam 2
Solutions
Instructions: There are 8 problems below. Choose any 6 problems to solve. All questions have
equal point values.
Problem 1:
Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is
attached to one of them, and the blocks are pushed together with the spring between them (see
below). A cord initially holding the blocks together is burned; after this, the block of mass 3M
moves to the right with a speed of 2.00 m/s.
• Find the original elastic potential energy in the spring if M = 0.250 kg.
What we want in the end is to relate the initial energy (figure part a) stored in the spring to the
final energy which is purely kinetic. Thus at some point conservation of total energy is needed.
What we first have to realize is that once the cord has been burned, the spring pushes away both
masses, and the masses are moving apart in the final state. This means that each block has kinetic
energy after the cord is cut, not just the one with mass 3M. Therefore, we need to find the velocity
of the block of mass M before we proceed.
Finding the velocities after the cord has been cut can be most easily done by conservation of
momentum:
1
*
−
−
p before = *
p after
*
−
p before = 0
*
−
p after = 3Mv3M + MvM
∴ vM = −3v3M
= −6m/s
After we have both velocities, we can use conservation of energy:
Ki + U i = K f + U f
0 + Uspring = Kf + 0
1
1
2
2
+ (3M)v3M
Kf = MvM
2
2
Kf = (0.5)(0.25)(−6)2 J + (0.5)(3)(0.25)(2)2 J
Kf = 4.5J + 1.5J
= 6J
Uspring = 6J
Problem 2:
A bead slides without friction around a loop-the-loop (below). The bead is released from a height
h = 2.95R.
• What is its speed at point A? Answer in terms of R and g, the acceleration of gravity.
Since there is no friction present, we are only dealing with conservative forces, i.e., forces independent of the path taken. Basically, this means that we can use conservation of energy and make this
a very simple calculation.
Since the object starts out at rest at a height h off of the ground, its initial energy is purely potential:
Ei = Ki + Ui = mgh
When it reaches the top of the loop-the-loop, it has some velocity v and a height off the floor of
2R. The total energy is now:
2
1
mv 2 + mg(2R)
2
The most common mistake here was to forget that at point A on the loop, the height off of the
ground is not zero and miss the 2R. Equating E i and Ef ,
Ef = K f + U f =
1
mgh = mv 2 + 2mgR
2
1
mg(h − 2R) = mv 2
2
1
g(h − 2R) = g(2.95R − 2R) = 0.95gR = v 2
2
v 2 = 2(0.95)gR
p
v = 1.9gR
Note that if we had measured potential energy relative to the top of the loop instead of the floor,
we could just write down Ei = mg(h − 2R) and Ef = 21 mv 2 and the result is the same. It has to
be, gravitational potential energy only depends on height differences.
Problem 3:
Consider the “ballistic pendulum” shown below, an apparatus used to measure the speed of fast
moving projectiles. A bullet of mass m 1 = 7.0 × 10−3 kg (∼100 grains) is fired at the initially stationary block of wood (m2 = 2kg) hanging from light wires. The bullet, traveling with a velocity of
v1A , embeds itself in the block, and the entire block + bullet system swings up to some height h.
• If h = 0.5m, find v1A .
What we know in this problem are the masses and initial velocities of the bullet and block. What
we want to know is the final height the bullet + block swings up to. Moving the bullet + block to
a new height h implies we have gained a potential energy of U = (m 1 + m2 )gh, which has to be the
3
2.
same as the kinetic energy of the bullet + block system just after the collision K = 12 (m1 + m2 )vB
The velocity vB is the first thing we need to find then.
This is basically just a perfectly inelastic collision, i.e., a collision in which kinetic is not conserved.
We can’t just use conservation of energy between the initial state of the bullet and the final state
of the bullet + block, because the collision is inelastic – some mechanical energy is converted to
internal energy. We can use conservation of momentum to first relate v 1A and vB , however, and
then use conservation of energy after the collision.
Since the block + bullet represent a closed system with no external forces acting, momentum has
to be conserved :
*
−
−
p before = *
p after
*
−
p before = m1 v1A
*
−
p after = (m1 + m2 )vB
Equating the momentum before and after the collision, we can solve for v B :
vB =
m1
v1A
m1 + m 2
Note that this didn’t depend on what we called the collision, inelastic or otherwise: we just wrote
−
−
v i for every object in the system before and after, momentum is
down the momentum *
p i = mi *
always conserved.
Now that we have vB , we can use conservation of energy to find the maximum height the pendulum
swings to. The system is still isolated from external forces, so total energy (K + U ) is conserved
for the block + bullet system. The total energy just after collision (point B in the figure) is purely
kinetic, while the total energy at the maximum height (point C) is purely potential.
KB + U B = K C + U C
UB = K C = 0
so
K B = UC
The kinetic energy just after the collision can be found from v B above:
1
2
KB = mtot vB
2
1
2
= (m1 + m2 )vB
2
2
m1
1
v1A
= (m1 + m2 )
2
m1 + m 2
2
m21 v1A
=
2(m1 + m2 )
4
Finding UC is easier:
UC = mtot gh = (m1 + m2 )gh
Setting KB = UC and solving for v1A :
2
m21 v1A
= (m1 + m2 )gh
2(m1 + m2 )
2
m21 v1A
= 2(m1 + m2 )2 gh
m1 + m 2 2
2
gh
v1A
=2
m1
m1 + m 2 p
v1A =
2gh
m1
Using the numbers given, we find v1A ≈ 900m/s (∼3000 fps).
Problem 4:
In the “coefficient of restitution” lab, we dropped a golf ball from a height h and in one part measured the height after a certain number of bounces.
We presumed that we could model the collision between the ball and floor by saying that the veafter
, where ε
locity after rebound was related to the velocity before rebound by the formula ε = vvbefore
was a constant.
0
• What is the height of the golf ball after one bounce (h ) in terms of the starting height h and ε?
Neglecting air resistance, conservation of mechanical energy dictates that after being dropped from
a height h, the ball will reach the ground with a kinetic energy of:
1
mv 2
= mgh
2 before
The coefficient of restitution defined above gives us the velocity after the ball rebounds from the
ground once as vafter = εvbefore . The maximum height after rebound can then be calculated from
conservation of energy:
0
1
mv 2
2 after
1
2
= mε2 vbefore
2
1
2
= ε2 ( mvbefore
)
2
= ε2 (mgh)
mgh =
0
∴ h = ε2 h
or ε =
q
h1
h0
if you like. One can repeat the same analysis, and show that the height after the second
rebound is ε4 h, and in general hn = ε2n h.
5
Problem 5:
A block of mass 2.3 kg is attached to a horizontal spring that has a force constant of 3.0 × 10 3 N/m.
The spring is compressed 3.5 cm and is then released from rest.
• Calculate the speed of the block as it passes through the equilibrium position x = 0 if the
surface is frictionless.
Think about the initial and final states. Initially, the energy of the system is purely potential, stored
in the spring. In the final state, the energy is purely kinetic: at the equilibrium position, the spring
is neither compressed nor expanded (this defines equilibrium), and no energy is stored. Proceed
then with conservation of energy:
Ei = Ki + Ui,spring = Ui,spring
Ef = Kf + Uf,spring = Kf
∴ Ui,spring = Kf
Writing down the surviving energy terms (viz.,, K f and Ui,spring ):
1
Ui,spring = k(∆x)2
2
1
Kf = mv 2
2
Kf = Ui,spring
1
1
∴ mv 2 = k(∆x)2
2
2
r
k
∆x
v=
m
v = 1.26m/s
(You did convert the ∆x = 3.5cm into meters first, right? If not, you probably got 126 m/s ...)
Problem 6:
A 1000 kg car is traveling at 10 m/s and hits a 2000 kg SUV head-on. The SUV was at rest before
the collision, but left in neutral, such that the collision is elastic.
• What is the car’s final velocity (hint: magnitude and direction)?
As usual with a collision, conservation of momentum is the starting point. Let the car be “1” and
the SUV “2”. Then v1i = 10m/s, v2i = 0, m1 = 1000kg, and m2 = 2000kg. Conservation of
momentum before and after the collision gives:
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*
−
−
p before = *
p after
*
−
p before = m1 v1i + 0
*
−
p after = m1 v1f + m2 v2f
∴ m1 v1f + m2 v2f = m1 v1i
Since the collision is elastic, we know that kinetic energy is conserved as well:
KEi = KEf
1
1
1
2
2
2
m1 v1i
+ 0 = m1 v1f
+ m2 v2f
2
2
2
Now we have two equations with two unknowns (v 1f and v2f )
m1 v1f + m2 v2f = m1 v1i
1
1
1
2
2
2
m1 v1i
+ 0 = m1 v1f
+ m2 v2f
2
2
2
Solving for v1f and v2f ,
m1 − m 2
v1i
m1 + m 2
2m1
v1i
=
m1 + m 2
v1f =
v2f
This result has already been derived as equations 9.22 and 9.23 in the text. Plugging in the numbers
we have, the car’s final velocity is:
1000 − 2000
1
10
(10m/s) = −( )(10m/s) = − m/s
1000 + 2000
3
3
Where in this case the negative sign indicates that the final motion is the opposite direction of the
initial motion.
v1f =
Problem 7:
A block of mass m is released from the top of a frictionless incline plane of height h and angle θ,
which is sitting at the edge of a table of height H (see below). At the end of the incline, the mass
falls off of the table, and hits the floor at a distance R from the edge.
• What is the speed of the block just before it hits the floor in terms of the variables given
above?
Again, since there is no friction present, we are only dealing with conservative forces, i.e., forces
independent of the path taken. The essence of the problem is then that the initial potential energy
of the block at the top of the ramp is converted completely into kinetic energy just before it hits
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the floor.
Ki + U i = K f + U f
(1)
Ui = K f
1
= mv 2
2
(2)
(3)
The initial potential energy is just the total height off of the floor times the weight of the object
(mg) – the difference in height from initial to final states times mg, or mg(h + H). Note that R
and θ do not matter. So:
1
mv 2
2
v 2 = 2g(h + H)
p
v = 2g(h + H)
mg(h + H) =
Problem 8:
A pendulum consists of a mass m hanging from a cord of length L, shown below. Ignore the mass
of the cord. The pendulum is released from a position such that the cord makes an angle θ with
respect to the vertical.
0
• What will the speed of the mass be when it is at a new angle θ < θ?
Once again, we are starting from rest at a certain height, and we want to find the speed at some
lower height. This generally means conservation of energy ...
Only the gravitational force on the mass m is doing any work in this problem, so we need only
worry about converting gravitational potential energy to kinetic energy. By the way: this differs
8
0
from the practice exam, in that we are moving the pendulum from θ to θ , not from θ to 0.
We call the “equilibrium” position of the pendulum, when it is hanging perfectly vertical and at rest,
our reference position at which y = 0, and U = 0. If the pendulum is inclined at an angle θ, simple
geometry dictates that the mass m has moved upward by an amount h = L − L cos θ = L(1 − cos θ).
0
0
0
0
For an angle θ , then h = L − L cos θ = L(1 − cos θ )
Applying conservation of energy,
Ki + U i = K f + U f
0
1
0 + mgh = mv 2 + mgh
2
1
0
mv 2 = mg(h − h )
2
i
h
1 2
0
v = g (L − L cos θ) − L − L cos θ
2
h
i
0
v 2 = 2g L cos θ − L cos θ
q
v = 2gL(cos θ 0 − cos θ)
In the end, it doesn’t matter for this problem that the mass is part of a pendulum at all, except
that it constrains the path of the mass and therefore the change in height. The conservation of
energy equations are exactly the same as those for a falling mass. The trickier part of this problem
0
was getting the height difference correct from initial (θ) and final (θ ) states, or equivalently, noting
0
that the potential energy in the final (θ ) state is not zero.
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