Download Question Identical constant forces push two identical objects A and

MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
Fall Term 2007
Concept Questions Exam 2: Solutions
Question Identical constant forces push two identical objects A and B continuously from
a starting line to a finish line. If A is initially at rest and B is initially moving to
the right,
1.
2.
3.
4.
Object A has the larger change in momentum.
Object B has the larger change in momentum.
Both objects have the same change in momentum
Not enough information is given to decide.
Answer 1: Both objects have the same mass, are pushed the same distance, by the same
constant force, so they have the same acceleration. Since object B has an initial speed the
time interval needed to reach the finish is less than the corresponding time interval for
object A which started from rest. Therefore the change in velocity of object B is less than
the corresponding change in velocity for object A. Hence object A has a larger change in
momentum.
Concept Question: Ping-Pong Ball and Bowling Ball
Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same
momentum, and you exert the same force to stop each. How do the distances needed to
stop them compare?
1. It takes a shorter distance to stop the ping-pong ball.
2. Both take the same distance.
3. It takes a longer distance to stop the ping-pong ball.
Answer: 3. Both the initial momentum and the force acting on the two objects are equal.
Therefore the initial velocity and the acceleration of the ping-pong ball is greater than the
bowling ball by the ratio of the bowling ball mass to the ping-pong ball mass.
mping  pong v0, ping  pong  mbowling v0,bowling  px
mping  pong a ping  pong  mbowling abowling  Fx
Since both the force acting on each object and the change in momentum is the same, the
impulse acting on each ball is the same. Therefore, the time interval it takes to stop each
object is the same.
Since the displacement is equal to

a t 
x   vx,0  x  t
2 

The ratio
x ping  pong
xbowling

mbowling
mping  pong
hence the ping-pong has the greater displacement.
Question: Pushing Baseball Bat
The greatest acceleration of the center of mass will be produced by pushing with a force
F at
1.
2.
3.
4.
Position 1
Position 2
Position 3
All the same
Answer 4. The external force is equal to the total mass times the acceleration of the
center-of-mass. It doesn’t matter where the external force acts with regards to the centerof-mass acceleration.
Concept Question: Drop a stone from the top of a high cliff. Consider the earth and the
stone as a system. As the stone falls, the momentum of the system
1. increases in the downward direction.
2. decreases in the downward direction.
3. stays the same.
4. not enough information to decide.
Answer 3: The system is approximately isolated with no external forces acting on the
system so the momentum stays the same. (We are ignoring the effects of the sun
and moon), The forces between the earth and the stone are internal forces and hence
cancel in pairs.
Concept Question: Consider yourself and the Earth as one system. Now jump up. Does
the momentum of the system
1. Increase in the downward direction as you rise?
2. Increase in the downward direction as you fall?
3. Stay the same?
4. Dissipate because of friction?
Answer 3: (Same argument as in the previous question.) No external forces are acting on
the system so the momentum is unchanged.
Concept Question: Suppose you are on a cart, initially at rest on a track with very little
friction. You throw balls at a partition that is rigidly mounted on the cart. If the balls
bounce straight back as shown in the figure, is the cart put in motion?
1. Yes, it moves to the right.
2. Yes, it moves to the left.
3. No, it remains in place.
Answer: 2. Because all the balls bounce back to the right, then in order to conserve
momentum, the cart must move forward.
Concept Question: Compared to the amount of energy required to accelerate a car from
rest to 10 miles per hour, the amount of energy required to accelerate the same car from
10 mph to 20 mph is
1.
2.
3.
4.
the same
twice as much
three times as much
four times as much
Answer: 3. The energy increase in going from zero speed to speed v is (1 / 2)mv 2 . To go
from v to 2v is (1 / 2)m(2v)2  (1 / 2)mv 2  (3 / 2)mv 2 .
Concept Question: Consider two carts, of masses m and 2m, at rest on an air track. If
you push first one cart for 3 s and then the other for the same length of time, exerting
equal force on each, the kinetic energy of the light cart is
1. larger than
2. equal to
3. smaller than
the kinetic energy of the heavy car.
Answer: 1. The kinetic energy of an object can be written as
1 2 p2
mv 
2
2m
Because the impulse is the same for the two carts, the change in momentum is the same.
Both start from rest so they both have the same final momentum. Since the mass of the
lighter cart is smaller than the mass of the heavier cart, the kinetic energy of the light cart
is larger than the kinetic energy of the heavy cart.
Concept Question: A particle starts from rest at x  0 and moves to x  L under the
action of a variable force Fx (x) , shown in the figure. What is the particle's kinetic energy
at x  L / 2 and at x  L ?
1. Fmax L / 2 , and Fmax L
2. Fmax L / 4 , and 0
3. Fmax L , and 0
4. Fmax L / 4 , and Fmax L / 2
5. Fmax L / 2 , and Fmax L / 4
Answer 4: Since the particle starts from rest, the kinetic energy of the particle is equal tot
eh work done by the force. The work done by the force for a displacement x  x f  xi is
xf
equal to the integral W   Fx dx  K . The work is the area under the curve in the
xi
above diagram. So for the interval, x  0 to x  L / 2 , the area is
W  K  K(x  L / 2)  (1/ 2)Fmax (L / 2)  Fmax L / 4 .
For the interval form x  0 to x  L , the area is just twice the previous case, so
W  K  K(x  L)  Fmax L / 2 .
Concept Question: When a person walks, the force of friction between the floor and the
person's feet accelerates the person forward. The floor does
1. Positive work on the person.
2. Negative work on the person.
3. No work on the person.
Answer: 3. The friction is static and there is no displacement of the foot on the floor,
dr  0 , when the force is applied F . So the contribution to the work dW  F  dr  0 .
Keep in mind that a human being is not a rigid body. The correct energy transformation is
chemical energy is transformed into the motion of the muscles which is transformed into
kinetic energy of the center of mass.
Concept Question: A ball is given an initial horizontal velocity and allowed to fall under
the influence of gravity, as shown below.
The work done by the force of gravity on the ball is:
1. positive
2. zero
3. negative
Answer: 1. The force of gravity causes the ball to accelerate downward, so the
displacement has a component in the same direction as the force. Hence the work done is
positive. (The dot product of the ball's displacement and the downward force of gravity is
positive, dW  F  dr  0 .)
Concept Question: A comet is speeding along a hyperbolic orbit toward the sun.
While the comet is moving away from the sun, the work done by the sun on the comet is:
1. positive
2. zero
3. negative
Answer: 3. The displacement of the comet has a component in the opposite direction as
the force on the comet so the work done is negative. (The comet's acceleration is always
toward the sun; when the comet moves away from the sun, the work is negative,
dW  F  dr  0 ) .
Concept Question. Suppose you want to ride your mountain bike up a steep hill. Two
paths lead from the base to the top, one twice as long as the other. Compared to the
average force you would exert if you took the short path, the average force you exert
along the longer path is
1.
2.
3.
4.
5.
four times as small.
three times as small.
half as small.
the same.
undetermined-it depends on the time taken.
Answer: 3. The gravitational potential energy gained is the same in both cases and is
equal to the average force exerted times the distance traveled.
Concept Question. Consider the following sketch of potential energy for a particle as a
function of position. There are no dissipative forces or internal sources of energy.
If a particle travels through the entire region of space shown in the diagram, at which
point is the particle's velocity a maximum?
1.
2.
3.
4.
5.
a
b
c
d
e
Answer 1: The if the particle travels through the entire region then the energy must be at
least greater or equal to the largest potential energy which occurs at point d. The kinetic
energy is equal to the total energy minus the potential energy. So the smallest potential
energy has the largest kinetic energy hence when the particle is at the lowest minimum
point of the potential energy graph, the kinetic energy is maximum.
Concept Question: An object is dropped to the earth from a height of 10m. Which of the
following graphs of kinetic energy vs. time best represent the kinetic energy of the object
as it approaches the earth (neglect friction).
1.
2.
3.
4.
5.
a
b
c
d
e
Answer: 3. The velocity increases linearly with time (for constant acceleration from rest,
v y  gt ), and kinetic energy is proportional to v 2y so K is proportional to t 2 .
Concept Question: You lift a ball at constant velocity from a height hi to a greater
height h f . Considering the ball and the earth together as the system, which of the
following statements is true?
1.
2.
3.
4.
5.
6.
7.
The potential energy of the system increases.
The kinetic energy of the system decreases.
The earth does negative work on the system.
You do negative work on the system.
The source energy of the ball increases.
Two of the above.
None of the above.
Answer: 1. Here the positive work that you do produces an increase in the system’s
potential energy. Now the earth is in the system so it cannot do work on the system.
Concept Question: Consider the following sketch of potential energy for a particle as a
function of position. There are no dissipative forces or internal sources of energy.
What is the minimum total mechanical energy that the particle can have if you know that
it has traveled over the entire region of X shown?
1.
-8
2.
6
3.
10
4.
It depends on direction of travel
5.
Can’t say - Potential Energy uncertain by a constant
Answer 3: The mechanical energy is the sum of kinetic and potential energy. If the
mechanical energy were less that 10 J, then when the particle was at the position
corresponding to the point d on the figure, the PE is equal to 10 J so the kinetic energy
would be less than zero, but this is forbidden for a particle using just the principles of
classical mechanics. (In quantum mechanics, one would have a small but finite
probability of finding the particle in a classical forbidden region.)
Concept Question 2 The potential energy function U (x) for a particle with total
mechanical energy E is shown below.
The posiiton of the particle as a function of time is given by
x(t )  D cos(t )  D sin(t ) .
where D  0 . The particle first reaches the position 3 when
a)  t  0
b) t   / 4
c) t   / 2
d) t  3 / 4
e)  t  
f)  t  5 / 4
g)  t  3 / 2
h)  t  7  / 4
Answer d: The initial conditions associated with x(t )  D cos(t )  D sin(t ) are that
x0  D  0 . Taking a time derivative shows that vx (t )   D sin(t )   D cos(t ) , hence
vx ,0   D  0 . Thus the particle starts out with a positive x- coordinate and the initial x component of the velocity is positive therefore it is moving away from the origin. It
arrives at x(t )  D cos(t )  D sin(t )  0 when cos(t )   sin(t ) which first occurs at
cos(t )  3 / 4 .
Concept Question: In part (a) of the figure, an air track cart attached to a spring rests on
the track at the position xequilibrium and the spring is relaxed. In (b), the cart is pulled to the
position xstart and released. It then oscillates about xequilibrium. Which graph correctly
represents the potential energy of the spring as a function of the position of the cart?
Answer: 3. The cart starts at xstart with no kinetic energy, and so the spring's potential
energy is a maximum. Once released, the cart accelerates to the right and its kinetic
energy increases as the potential energy of the spring is converted into kinetic energy of
the cart. As the cart passes the equilibrium position, its kinetic energy is a maximum and
so the spring's potential energy is a minimum. Once to the right of xequilibrium, the cart
starts to compress the spring and it slows down as its kinetic energy is converted back to
potential energy of the recompressed spring. At the rightmost point it reaches, the cart
reverses its direction of travel. At that instant, it has no kinetic energy and the spring
again has maximum potential energy.
Concept Question: A block of inertia m is attached to a relaxed spring on an inclined
plane. The block is allowed to slide down the incline, and comes to rest. The coefficient
of kinetic friction of the block on the incline is µk. For which definition of the system is
the change in total energy (after the block is released) zero?
1.
2.
3.
4.
block
block + spring
block + spring + incline
block + spring + incline + Earth
Answer: 4 . Since there is dissipative forces acting between the block and the incline
plane some gravitational potential energy is dissipated into the block and incline plane
increasing the thermal energy of those objects.
Concept Question: Cart A is at rest. An identical cart B, moving to the right, collides
elastically with cart A. After the collision, which of the following is true?
1. Carts A and B are both at rest.
2. Cart B stops and cart A moves to the right with speed equal to the original speed
of cart B.
3. Cart A remains at rest and cart B bounces back with speed equal to its original
speed.
4. Cart A moves to the right with a speed slightly less than the original speed of cart
B and cart B moves to the right with a very small speed.
Answer: 2. Conservation of momentum implies
mvB,0  mvB, f  mv A, f ,
or
vB,0  vB, f  v A, f
(9.2.1)
.
The square of the initial speed is
v B,0 2  v B, f 2  v A, f 2  2v B, f v A, f
(9.2.2)
Note that when squaring the initial speed a cross term appears 2vB, f v A, f . The collision is
elastic so conservation of energy implies
1
1
1
mv B,0 2  mv B, f 2  mv A, f 2 ,
2
2
2
or
v B,0 2  v B, f 2  v A, f 2 .
(9.2.3)
Comparing Eq. (9.2.2) and Eq. (9.2.3) implies that the cross term in Eq. (9.2.2) must
vanish. Thus 2v B, f v A, f  0 . There are two possibilities: Cart B stops v B, f  0 and hence
by Eq. (9.2.1) cart A moves with the initial speed of cart B vB,0  v A, f . The other
possibility that v A, f  0 and hence vB,0  vB, f just reproduces the initial conditions.
Second argument: Center of mass reference frame
In the center of mass reference frame both carts approach each other with half the
original speed of Cart B in the lab frame. They bounce elastically with the same speeds.
So in the lab frame after the collision, Cart B is at rest and Cart A moves to the right with
speed equal to the original speed of cart B.
Concept Question: Cart A is at rest. An identical cart B, moving to the right, collides
elastically with cart A. They stick together. After the collision, which of the following is
true?
1.
2.
3.
4.
Carts A and B are both at rest.
Carts A and B moves to the right with speed greater than Cart B’s original speed.
Carts A and B move to the right with a speed less than cart B's original speed.
Cart B stops and cart A moves to the right with speed equal to the original speed
of cart B.
Answer 3: From conservation of momentum, mvB,0  2mv f . So v f  vB,0 / 2 . Thus they
move away with speed less one half the original speed of cart B.
Concept Question
Object A sits at the outer edge (rim) of a merry-go-round, and object B sits halfway
between the rim and the axis of rotation. The merry-go-round makes a complete
revolution once every thirty seconds. The magnitude of the angular velocity of Object B
is
1.
2.
3.
4.
half the angular speed of Object A .
the same as the angular speed of Object A .
twice the the angular speed of Object A .
impossible to determine
Answer: 2. All points in a rigid body rotate with the same angular velocity.
Concept Question: An object of mass m with velocity v moving in a straight line
collides head on elastically with an identical object. The second object is initially at rest
and is attached at one end to a string of length l and negligible mass. The other end of the
string is fixed (at the point S). After the collision the second object undergoes circular
motion with angular speed  . The kinetic energy of the second object after the collision
is
1. (1 / 4)ml 2 2
2. (1 / 2)ml 2 2
3. (3 / 4)ml 2 2
4. (1 / 4)ml 2
5. (1 / 2)ml 2
6. (1 / 4)ml
Answer: 2. The collision is elastic and the objects have equal masses so after the
collision the incoming object stops and the object on the string has velocity v . Since it is
then constrained to move in a circle, the radially inward tension force of the string
changes the direction of the velocity but not the magnitude. So the speed and the angular
1
velocity are related by v  l . The kinetic energy is then K  mv 2  (1 / 2)ml 2 2
2
Concept Question: A disk with mass M and radius R is spinning with angular speed 
about an axis that passes through the rim of the disk perpendicular to its plane. The
moment of inertia about cm is I cm  (1/ 2)mR2 . Its total kinetic energy is:
1. (1 / 4)mR 2 2
2. (1 / 2)mR 2 2
3. (3 / 4)mR 2 2
4. (1 / 4)mR 2
5. (1 / 2)mR 2
6. (1 / 4)mR
Answer: 3. The moment of inertia about the center of mass of a disk about an axis
perpendicular to its plane is given by I cm  (1/ 2)mR2 . The parallel axis theorem states
that the moment of inertia about an axis passing through a point on the rim perpendicular
to its plane is given
IP  mR 2  I cm  mR 2  (1 / 2)mR 2  (3 / 2)mR 2 .
The kinetic energy is given by
K
1
IP 2  (3 / 4)mR2 2 .
2
Concept Question: Consider two vectors rP,F  xˆi with x  0 and F  Fx ˆi  Fz kˆ with
Fx  0 and Fz  0 . The cross product rP,F  F points in the
1) + x-direction
2) -x-direction
3) +y-direction
4) -y-direction
5) +z-direction
6) -z-direction
7) None of the above directions
Answer: 4. We calculate the cross product noting that in a right handed choice of unit
ˆ   ˆj ,
vectors, ˆi ˆi  0 and ˆi  k

 
 

ˆ  xˆi  F ˆi  xˆi  F k
ˆ   xF ˆj
rP,F  F  xˆi  Fxˆi  Fz k
x
z
z
Since x  0 and Fz  0 , the direction of the cross product is in the  y -direction.
Concept Question
A fixed torque is applied to the shaft of the chrome inertial wheel. If the four weights on
the arms are slid out, the component of the angular acceleration along the shaft direction
will
1. increase.
2. decrease.
3. remain the same.
Answer: 2. torque about the central axis is proportional to the component of the angular
acceleration along that axis. The proportionality constant is the moment of inertia about
that axis. By pushing the weights out, the moment of inertia has increased. If the applied
torque is constant then the component of the angular acceleration must decrease.
Concept Question: In Experiment 5: Moment of Inertia, the figure below shows a plot of
angular velocity of the rotor as a function of time. The slope of the line B-C is equal to
1. the angular acceleration after the string has detached.
2. angular acceleration before the string has detached.
3. Neither of the above.
Answer: 1. The line between the points B and C on the above figure correspond to the
slowing down of the rotor after the string has detached. The negative angular acceleration
is due to the torsional friction in the rotor.
Concept Question
A figure skater stands on one spot on the ice (assumed frictionless) and spins around with
her arms extended. When she pulls in her arms, she reduces her rotational moment of
inertia and her angular speed increases. Assume that her angular momentum is constant.
Compared to her initial rotational kinetic energy, her rotational kinetic energy after she
has pulled in her arms must be
1. the same.
2. larger.
3. smaller.
Answer: 2. When she pulls her arms in there are no external torques so her angular
momentum is constant. The magnitude of her angular momentum about the rotation axis
passing through the center of the stool is L  I . Her kinetic energy is
K
1 2 1 L2 L2
I  I 2 
.
2
2 I
2I
Since L is constant and her moment of inertia decreases when she pulls her arms, her
kinetic energy must increase.
Concept Question: A tetherball of mass m is attached to a post of radius R by a string.
Initially it is a distance r0 from the center of the post and it is moving tangentially with a
speed v0 . The string passes through a hole in the center of the post at the top. The string
is gradually shortened by drawing it through the hole. Ignore gravity. Until the ball hits
the post,
1. The energy and angular momentum about the center of the post are constant.
2. The energy of the ball is constant but the angular momentum about the center of
the post changes.
3. Both the energy and the angular momentum about the center of the post, change.
4. The energy of the ball changes but the angular momentum about the center of the
post is constant.
Answer: 4. The tension force points radially in; so torque about central point is zero:
angular momentum about central point is constant.
 S  rS ,T  T  0
Lsystem
 Lsystem
S, f
S ,0
For all radial forces, (example gravitation), the angular momentum about the central
point is a constant of the motion.
A small displacement of the ball has a radially component inward so the work done by
tension is not zero.,
dW ext  T  dr  0 .
Hence the mechanical energy is not constant.
Concept Question: A tetherball of mass m is attached to a post of radius R by a string.
Initially it is a distance r0 from the center of the post and it is moving tangentially with a
speed v0 . The string wraps around the outside of the post. Ignore gravity. Until the ball
hits the post,
1. The energy and angular momentum about the center of the post are constant.
2. The energy of the ball is constant but the angular momentum about the center of
the post changes.
3. Both the energy of the ball and the angular momentum about the center of the
post, change.
4. The energy of the ball changes but the angular momentum about the center of the
post is constant.
Answer 2. The tension force points towards contact point; so torque about central point is
not zero: angular momentum about central point is not constant.
 S  rS ,T  T  0
Lsystem
 Lsystem
S, f
S ,0
Small displacement is always perpendicular to string since at each instant in time ball
undergoes instantaneous circular motion about string contact point with pole.
Therefore the tension force is perpendicular to the displacement
dW ext  T  dr  0 .
Hence mechanical energy is constant.
Concept Question: A streetcar is freely coasting (no friction) around a large circular
track. It is then switched to a small circular track.
When coasting on the smaller circle the streetcar's
1. mechanical energy is conserved and angular momentum about the center is
constant
2. mechanical energy is not conserved and angular momentum about the center is
constant
3. mechanical energy is not conserved and angular momentum about the center is
not constant
4. mechanical energy is conserved and angular momentum about the center is not
constant.
Answer: 4. Carefully draw a free body diagram for the streetcar while it is on the link of
track connecting the two circular tracks. In the vertical direction, the force of gravity at
all times balances the vertical component of the normal force of the track on the car, so
there is no acceleration in the vertical direction and we may ignore it for the rest of this
problem. Along the plane of motion, the wheels are guided by component of the normal
force in the plane. This force is always normal to the direction of motion, so no work is
done by this force. Therefore the streetcar's total energy does not change, and its speed
remains constant. Note that this force does exert a torque around the center of the circles
because it does not point radially inward, and therefore its angular momentum is not
constant!
Concept Question: Two disks are separated by a spindle of smaller diameter. A string is
wound around the spindle and pulled gently. Which positions of the string cause the
assembly to roll to the right?
1) Only A
2) Only B
3) Only C
4) A and B
5) B and C
6) A and B and C
7) None of the configurations shown
Answer: 1. When the string is pulled in direction A, the torque due to the pulling force is
out of the page and if there were no friction between the surfaces, the assembly would
spin counterclockwise and slide to the right. If there is some contact friction f s between
the surfaces, the torque due to the friction is into the page which will cause a clockwise
rotation. For small pulling forces, the frictional torque can exceed the pulling torque and
the assembly will roll to the right.
The force equation in the vertical direction yields F sin   N  mg  0 so the normal
force decreases as the angle  increases, according to N  mg  F sin  . Therefore the
kinetic fiction force decreases (since f K   N ) and can no longer supply enough torque
to keep the wheel spinning clockwise and rolling to the right. As the direction of the
pulling force F is shifted towards B, there will be a critical angle  c in which the
assembly will stop rotating, and the assembly will slide to the right with kinetic friction
opposing the motion. At any larger angle the assembly rotates counterclockwise and so
cannot roll to the right.
Concept Question: The angular momentum about the point O of a body of mass 
1.
is constant.
2.
changes throughout the motion because the speed changes.
3.
changes throughout the motion because the distance from O changes.
4.
changes throughout the motion because the angle  changes.
5.
Not enough information to decide.
6.
We had the angular momentum quiz last Friday so I don’t need to think about it
anymore.
Answer: 1. The torque about the point O due to the central force is zero,
0  r0  F1,2  0 ,
since the vector r0 and the vector F1,2 are anti-parallel. Therefore the change in angular
momentum is zero
0 
dL0
 0.
dt
For motion confined to the plane, the angular momentum must be constant.
Concept Question: The mechanical energy of a body of mass 
1.
2.
3.
4.
5.
is constant.
changes throughout the motion because the speed changes.
changes throughout the motion because the distance from O changes.
is not constant because the orbit is not zero hence the central force does work.
Not enough information to decide.
Answer: 1. There are no non-conservative forces so the non-conservative work is zero
hence mechanical energy is constant.