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Assignment 7 Solutions (PDF)

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Homework 7 Problems: Linear Momentum
1. An astronaut in her space suit has a total mass of 87.0 kg, including suit and oxygen tank.
Her tether line loses its attachment to her spacecraft while she’s on a spacewalk. Initially
at rest with respect to her spacecraft, she throws her 12.0-kg oxygen tank away from her
spacecraft with a speed of 8.00 m/s to propel herself back toward it.
a. [5 points] Determine the maximum distance she can be from the craft and still
return within 2.00 min.
b. [5 points] Explain in terms of Newton’s laws of motion why this strategy works.
Answer:
(a) If we specify the astronaut and the oxygen tank as the isolated system, then momentum is
conserved. Therefore, we can use momentum conservation to determine the velocity of
the astronaut.
(
)(
(
)
)
Therefore, after throwing her 12.0-kg oxygen tank away with a speed of 8.00 m/s, then
the astronaut moves at 1.28 m/s in the opposite direction. Therefore, the maximum
distance that the astronaut can travel in 2.00 min is
(
)(
)
(b) By Newton’s third law, when the astronaut exerts an action force on the oxygen tank, the
tank exerts a reaction force back towards the astronaut. By Newton’s 2nd law, this
reaction force on the astronaut accelerates her toward the spacecraft. Since there are no
other forces operating on the astronaut, by Newton’s 1st law, the astronaut will continue
to travel at constant velocity towards the spacecraft.
2. A pitcher throws a 0.14-kg baseball toward the batter so that it crosses home plate
horizontally and has a speed of 42 m/s just before it makes contact with the bat. The
batter then hits the ball straight back at the pitcher with a speed of 48 m/s. Assume the
ball travels along the same line leaving the bat as it followed before contacting the bat.
a. [5 points] What is the magnitude of the impulse delivered by the bat to the
baseball?
b. [5 points] If the ball is in contact with the bat for 0.0050 s, what is the magnitude
of the average force exerted by the bat on the ball?
Answer:
(a) To determine the magnitude of the impulse delivered by the bat to the baseball, we will
use the impulse-momentum theorem. Assuming that
and
,
the magnitude of the impulse is given as
(
) [(
)
(
)]
Therefore, the impulse delivered by the bat to the baseball as 12.6 kg m/s.
(b) The average force is computed through the impulse
3. [15 points] Two shuffleboard disks of equal mass, one orange and the other green, are
involved in a perfectly elastic glancing collision. The green disk is initially at rest and is
struck by the orange disk moving initially to the right at 5.00 m/s. After the collision, the
orange disk moves in a direction that makes an angle of 37.0вЃ° with the horizontal axis
while the green disk makes an angle of 53.0вЃ° with this axis. Determine the speed of each
disk after the collision.
Answer: Since this is a glancing collision, then momentum is conserved in the system.
Conservation of momentum in the x- and y-directions, respectively, yields
(
)
(
(
)
)
(
)
Since the masses of both shuffleboard disks are the same, we can reduce the above equations to
(
(
Solving (2) for
)
(
)
)
(
( )
)
( )
gives
(
)
(
( )
)
Substituting (3) into (1) gives
[
(
(
)
(
Solving for
gives
)
(
]
)
(
)
)
(
(
)
)
(
(
)
(
)
(
)
(
)
)
(
)
(
)
We use (3) to solve for
(
(
)
)
(
)
(
(
)
)
4. A block with mass m1 = 0.500 kg is released from rest on a frictionless track at a distance
h1 = 2.50 m above the top of a table. It then collides elastically with an object having
mass m2 = 1.00 kg that is initially at rest on the table.
a. [3 points] Describe the motion of the blocks from the point at which
is
released until both blocks land on the ground. Describe the energy and
momentum transformations that occur throughout the motion of both blocks.
b. [3 points] Determine the velocities of the two objects just after the collision.
c. [3 points] How high up the track does the 0.500 kg object travel back after the
collision?
d. [3 points] How far away from the bottom of the table does the 1.00-kg object
land, given that the height of the table is h2 = 2.00 m?
e. [3 points] How far away from the bottom of the table does the 0.500-kg object
eventually land?
Answer:
(a) When block
is released, the gravitational force performs work on the block, which
converts the initial gravitational potential energy of the block to kinetic energy.
Therefore, just before the collision, block
has kinetic energy and momentum. Since
the collision is an elastic collision, kinetic energy and momentum are conserved for the
two blocks. This implies that there will be a transfer of momentum and kinetic energy to
block
, which causes it to fall off the table. As block
falls off the table, the
gravitational force continues to do work on the block, leading to an increase of kinetic
energy. Since the change in momentum of block
is equal and opposite to the change
in momentum of block
, this implies that block
will move to the left immediately
after the collision with the kinetic energy that is left over from the collision. Since the
track is frictionless, block
moves up the incline, conserving mechanical energy.
Therefore, it will reach a maximum height up the track and then fall down, exiting the
track. Since mechanical energy is conserved during this part of the motion, the kinetic
energy after the collision will be the same kinetic energy in which block
leaves the
track. After leaving the track, block
will follow the same motion as block
.
(b) This problem can be split into 3 stages: (1) the motion of m1 down the frictionless track
just before it contacts m2; (2) the elastic collision between m1 and m2; and (3) the motion
of m1 and m2 off of the table.
Because the track is frictionless, then the motion of block m1 in stage (1) conserves
energy. The velocity near the bottom of the frictionless track can be determined by the
conservation of energy
в€љ (
в€љ
)(
)
In stage 2, block m1 (which has an initial velocity of 7 m/s) undergoes an elastic collision
with block m2, in which kinetic energy and momentum is conserved. Using momentum
and kinetic energy conservation gives (and assuming that
) gives
( )
(
Solving (2) for
)
( )
gives
( )
Substituting (3) into (1) and solving for
gives
(
(
)
(
)
)
(
)
(
)(
)
To determine the velocity on block m2, we solve (3) for
(
)
Therefore, after the collision, block m2 has a speed of 4.66 m/s and block m1 has a speed
of 2.33 m/s.
(c) In stage 3, block m2 falls off the table with an initial speed of 4.66 m/s and block m1 goes
up the incline with an initial speed of 2.33 m/s. To determine the maximum height that
m1 reaches, we use mechanical energy conservation
(
)
(
)
(d) Because mechanical energy is conserved for block m1, then the speed at which it goes up
the incline will be the speed that it leaves the table. Therefore, block m1 falls off the table
with an initial speed of 2.33 m/s. To determine the horizontal range of m1, we use the
kinematics equation (2) in the vertical and horizontal direction (note that
)
в€љ
(
в€љ
в€љ
)(
)
(
)
(e) Since mechanical energy is conserved, the speed of block m1 as it leaves the table will be
the same as the speed of block m1 after the collision. To determine the horizontal range of
m2, we use the kinematics equation (2) in the horizontal direction
(
)(
)
Bonus: [6 points] For an elastic collision between two objects, we can write the momentum and
kinetic energy conservation equations as
( )
( )
(
Show mathematically that these conditions imply
steps for full credit.
). Show all of your
Answer: Rewriting (2) gives
(
)
(
)
( )
Rewriting (1) gives
(
)
(
)
( )
Dividing (3) by (4) gives
(
)
(
(
(
)
)(
(
(
)
)
(
)
(
)
)(
(
)
)
)
(
)
(
)
This expression states that the relative velocity of the two objects before the collision equals the
negative of the relative velocity of the two objects after the collision. An example of this is seen
in Problem 4. In Problem 4, the relative velocity of the objects before the collision is 7 m/s,
whereas after the collision, the relative velocity of the objects is -7 m/s.
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