Download Problem 7.54 A Ball Hits a Wall Elastically

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Due: 11:59pm on Wednesday, October 14, 2015
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Problem 7.54
A 64.0­kg skier starts from rest at the top of a ski slope of height 61.0 m .
Part A
If frictional forces do −1.03×104 J of work on her as she descends, how fast is she going at the bottom of the slope?
Take free fall acceleration to be g = 9.80 m/s2 .
ANSWER:
v
= 29.6 m/s Correct
Part B
Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.24. If the patch is
of width 60.0 m and the average force of air resistance on the skier is 180 N , how fast is she going after crossing the
patch?
ANSWER:
v
= 15.9 m/s Correct
Part C
After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.0 m into it before coming
to a stop. What is the average force exerted on her by the snowdrift as it stops her?
ANSWER:
F
= 4060 N Correct
A Ball Hits a Wall Elastically
A ball of mass m moving with velocity v i⃗ strikes a vertical wall. The angle between the ball's initial velocity vector and the
wall is θi as shown on the diagram, which depicts the situation as seen from above. The duration of the collision between
the ball and the wall is Δt, and this collision is completely elastic. Friction is negligible, so the ball does not start spinning.
In this idealized collision, the force exerted on the ball by the wall is parallel to the x axis.
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Part A
What is the final angle θf that the ball's velocity vector makes with the negative y axis?
Express your answer in terms of quantities given in the problem introduction.
Hint 1. How to approach the problem
Relate the vector components of the ball's initial and final velocities. This will allow you to determine θf in terms
of θi .
Hint 2. Find the y component of the ball's final velocity
What is v f y , the y component of the final velocity of the ball?
Express your answer in terms of quantities given in the problem introduction and/or v ix and v iy , the x
and y components of the ball's initial velocity.
Hint 1. How to approach this part
There is no force on the ball in the y direction. From the impulse­momentum theorem, this means that
the change in the y component of the ball's momentum must be zero.
ANSWER:
vf y
= −v i cos(θi )
Hint 3. Find the x component of the ball's final velocity
What is v f x , the x component of the ball' final velocity?
Express your answer in terms of quantities given in the problem introduction and/or v ix and v iy , the x
and y components of the ball's initial velocity.
Hint 1. How to approach this problem
Since energy is conserved in this collision, the final speed of the ball must be equal to its initial speed.
ANSWER:
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vf x
= −v i sin(θi )
Hint 4. Putting it together
Once you find the vector components of the final velocity in terms of the initial velocity, use the geometry of
similar triangles to determine θf in terms of θi .
ANSWER:
θf
= θi
Correct
Part B
What is the magnitude F of the average force exerted on the ball by the wall?
Express your answer in terms of variables given in the problem introduction and/or v ix .
Hint 1. What physical principle to use
Use the impulse­momentum theorem, J
case, only one force is acting, so ∣∣J ∣∣⃗ ⃗ = p f⃗ − p i⃗ , along with the definition of impulse, J
. Putting everything together, F
= F Δt
=
p f⃗ −p i⃗ Δt
⃗ ⃗ = ∑ F Δt
. In this
.
Hint 2. Change in momentum of the ball
The fact that θf = θi implies that the y component of the ball's momentum does not change during the
collision. What is Δpx , the magnitude of the change in the ball's x momentum?
Express your answer in terms of quantities given in the problem introduction and/or v ix .
ANSWER:
Δp
x
= 2mv i sin(θi )
ANSWER:
F
= 2mvix
Δt
Correct
Pucks on Ice
Two hockey players, Aaron and Brunnhilde, are pushing two pucks on a frictionless ice rink. The pucks are initially at rest
on the starting line. Brunnhilde is pushing puck B, which has a mass three times as great as that of puck A, which Aaron is
pushing. The players exert equal constant forces of magnitude F on their pucks, directed horizontally, towards the finish
line. They start pushing at the same time, and each player pushes his or her puck until it crosses the finish line, a distance d away.
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Part A
Which puck reaches the finish line first?
Hint 1. Compute the relative acceleration of the pucks
If aA and aB are the magnitudes of the accelerations of pucks A and B, respectively, what is the value of the
ratio aA /aB ?
ANSWER:
aA /aB
= 3
ANSWER:
Both pucks reach the finish line at the same time.
Puck A reaches the finish line first.
Puck B reaches the finish line first.
More information is needed to answer this question.
Correct
Part B
Let KA be the magnitude of the kinetic energy of puck A at the instant it reaches the finish line. Similarly, KB is the
magnitude of the kinetic energy of puck B at the (possibly different) instant it reaches the finish line. Which of the
following statements is true?
Hint 1. Determine the simplest way to answer this question
There are several possible approaches to this problem. Which is the simplest?
Choose the best option.
ANSWER:
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Use F
= ma
Use d =
1
2
(force equals mass times acceleration) to find the acceleration of each puck.
at
2
(relating distance traveled to acceleration and time) to find the time to the finish line.
Use the work­energy theorem.
Apply conservation of momentum and energy.
Hint 2. Work done on puck A
Find WA , the work done on puck A over the distance d.
ANSWER:
WA
= Fd
Hint 3. Work done on puck B
Find WB , the work done on puck B over the distance d.
ANSWER:
WB
= Fd
ANSWER:
KA
= KB
KA
< KB
KA
> KB
You need more information to decide.
Correct
Part C
Let pA be the magnitude of the momentum of puck A at the instant it reaches the finish line. Similarly, pB is the
magnitude of the momentum of puck B at the (possibly different) instant it reaches the finish line. Which of the
following statements is true?
Choose the best option.
Hint 1. Method 1: Compute the ratio of the pucks' velocities
The momentum of an object is the product of its mass and velocity. From the problem introduction, you know
m
that m
A
B
=
1
3
. Find v A /v B , the ratio of the velocity of puck A at the instant it reaches the finish line to the
velocity of puck B at the (possibly different) instant it reaches the finish line.
Hint 1. How to find the final velocities
You can easily compute the ratio using the (already determined) fact that the final kinetic energy of both
pucks is the same. Write the kinetic energy of each puck in terms of its velocity (for example, 1
KA =
mA vA
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2
2
). Set these expressions equal, and use the known ratio of the masses.
ANSWER:
v A /v B
= 1.73
Hint 2. Method 2: Use the impulse­momentum theorem
The impulse­momentum theorem states that
.
Δp = F Δt
You are given that both forces are the same, and you have compared the times in an earlier part.
ANSWER:
pA
= p
pA
< p
pA
> p
B
B
B
You need more information to decide.
Correct
± The Impulse­Momentum Theorem
Learning Goal:
To learn about the impulse­momentum theorem and its applications in some common cases.
Using the concept of momentum, Newton's second law can be rewritten as
⃗ ΣF =
⃗ ⃗ where ΣF is the net force F net acting on the object, and dp ⃗ dt
, (1)
dp ⃗ dt
is the rate at which the object's momentum is changing.
If the object is observed during an interval of time between times t1 and t2 , then integration of both sides of equation (1)
gives
∫
t2
t1
⃗ ΣF dt = ∫
t2
t1
dp ⃗ dt
dt. (2)
→
The right side of equation (2) is simply the change in the object's momentum p2
→
− p1
. The left side is called the impulse of
⃗ the net force and is denoted by J . Then equation (2) can be rewritten as
→
→
⃗ J = p2 − p1
.
This equation is known as the impulse­momentum theorem. It states that the change in an object's momentum is equal to
⃗ the impulse of the net force acting on the object. In the case of a constant net force F net
acting along the direction of
motion, the impulse­momentum theorem can be written as
F (t 2 − t 1 ) = mv 2 − mv 1
. (3)
Here F , v 1 , and v 2 are the components of the corresponding vector quantities along the chosen coordinate axis. If the
motion in question is two­dimensional, it is often useful to apply equation (3) to the x and y components of motion
separately.
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The following questions will help you learn to apply the impulse­momentum theorem to the cases of constant and varying
force acting along the direction of motion. First, let us consider a particle of mass m moving along the x axis. The net force
F is acting on the particle along the x axis. F is a constant force.
Part A
The particle starts from rest at t
t > 0.
= 0
. What is the magnitude p of the momentum of the particle at time t? Assume that
Express your answer in terms of any or all of m, F , and t.
ANSWER:
p
= F (t)
Correct
Part B
The particle starts from rest at t
t > 0.
= 0
. What is the magnitude v of the velocity of the particle at time t? Assume that Express your answer in terms of any or all of m, F , and t.
ANSWER:
v
= F (t)
m
Correct
Part C
The particle has momentum of magnitude p1 at a certain instant. What is p2 , the magnitude of its momentum
\texttip{\Delta t}{Deltat} seconds later?
Express your answer in terms of any or all of \texttip{p_{\rm 1}}{p_1}, \texttip{m}{m}, \texttip{F}{F}, and
\texttip{\Delta t}{Deltat}.
ANSWER:
\texttip{p_{\rm 2}}{p_2} = F\left({\Delta}t\right)+p_{1}
Correct
Part D
The particle has momentum of magnitude \texttip{p_{\rm 1}}{p_1} at a certain instant. What is \texttip{v_{\rm 2}}{v_2},
the magnitude of its velocity \texttip{\Delta t}{Deltat} seconds later?
Express your answer in terms of any or all of \texttip{p_{\rm 1}}{p_1}, \texttip{m}{m}, \texttip{F}{F}, and
\texttip{\Delta t}{Deltat}.
ANSWER:
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\texttip{v_{\rm 2}}{v_2} = \large{\frac{F\left({\Delta}t\right)+p_{1}}{m}}
Correct
Let us now consider several two­dimensional situations.
A particle of mass \texttip{m}{m} is moving in the positive x direction at speed \texttip{v}{v}. After a certain constant force is
applied to the particle, it moves in the positive y direction at speed 2v.
Part E
Find the magnitude of the impulse \texttip{J}{J} delivered to the particle.
Express your answer in terms of \texttip{m}{m} and \texttip{v}{v}. Use three significant figures in the numerical
coefficient.
Hint 1. How to approach the problem
This is a two­dimensional situation. It is helpful to find the components \texttip{J_{\mit x}}{J_x} and
\texttip{J_{\mit y}}{J_y} separately and then use the Pythagorean theorem to find \texttip{J}{J}.
Hint 2. Find the change in momentum
Find \Delta p_x, the magnitude of the change in the x component of the momentum of the particle.
Express your answer in terms of \texttip{m}{m} and \texttip{v}{v}.
ANSWER:
\Delta p_x = m v
ANSWER:
\texttip{J}{J} = \sqrt{5}mv
Correct
Part F
Which of the vectors below best represents the direction of the impulse vector \vec{J}?
ANSWER:
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1
2
3
4
5
6
7
8
Correct
Part G
What is the angle \texttip{\theta }{theta} between the positive y axis and the vector \vec{J} as shown in the figure?
ANSWER:
26.6 degrees
30 degrees
60 degrees
63.4 degrees
Correct
Part H
If the magnitude of the net force acting on the particle is \texttip{F}{F}, how long does it take the particle to acquire its
final velocity, 2v in the positive y direction?
Express your answer in terms of \texttip{m}{m}, \texttip{F}{F}, and \texttip{v}{v}. If you use a numerical
coefficient, use three significant figures.
ANSWER:
\texttip{t}{t} = \large{{\frac{mv}{F}}\sqrt{5}}
Correct
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So far, we have considered only the situation in which the magnitude of the net force acting on the particle was either
irrelevant to the solution or was considered constant. Let us now consider an example of a varying force acting on a
particle.
Part I
A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. A varying force F(t)=6.00t^2­4.00t+3.00 is acting on
the particle between t=0.00 seconds and t=5.00 seconds. Find the speed \texttip{v}{v} of the particle at t=5.00
seconds.
Express your answer in meters per second to three significant figures.
Hint 1. Use the impulse­momentum theorem
In this case, v_1=0 and v_2=v. Therefore,
\large{\int_{0.00}^{5.00}F\,dt=\Delta mv}.
Hint 2. What is the correct antiderivative?
Which of the following is an antiderivative \large{\int(6.00t^2­4.00t+3.00)dt}?
ANSWER:
6.00t^3­4.00t^2+3.00t 6.00t­4.00 2.00t^3­2.00t^2+3.00t 12.00t­4.00
ANSWER:
\texttip{v}{v} = 43.0 m/s Correct
Ballistic Pendulum
In a ballistic pendulum an object of mass \texttip{m}{m} is fired with an initial speed \texttip{v_{\rm 0}}{v_0} at a pendulum
bob. The bob has a mass \texttip{M}{M}, which is suspended by a rod of length \texttip{L}{L} and negligible mass. After the
collision, the pendulum and object stick together and swing to a maximum angular displacement \texttip{\theta }{theta} as
shown .
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Part A
Find an expression for \texttip{v_{\rm 0}}{v_0}, the initial speed of the fired object.
Express your answer in terms of some or all of the variables \texttip{m}{m}, \texttip{M}{M}, \texttip{L}{L}, and
\texttip{\theta }{theta} and the acceleration due to gravity, \texttip{g}{g}.
Hint 1. How to approach the problem
There are two distinct physical processes at work in the ballistic pendulum. You must treat the collision and the
following swing as two separate events. Identify which physical law or principle applies to each event, write an
expression to describe the collision, write an expression to describe the swing, and then relate the two
expressions to find \texttip{v_{\rm 0}}{v_0}.
Hint 2. Determine which physical laws and principles apply
Which of the following physical laws or principles can best be used to analyze the collision between the object
and the pendulum bob? Which can best be used to analyze the resulting swing?
A. Newton's first law
B. Newton's second law
C. Newton's third law
D. Conservation of mechanical energy
E. Conservation of momentum
Enter the letters corresponding to the correct answer, with a letter first for the collision and then a
second letter for the swing separated by a comma.
ANSWER:
E,D
Hint 3. Describe the collision
Compose an expression that describes the collision between the object and the pendulum bob. Put this
expression in the form v_0 = \cdots.
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Express your answer in terms of some or all of the variables \texttip{m}{m}, \texttip{M}{M}, \texttip{v_{\rm
0}}{v_0}, \texttip{v}{v}, \texttip{L}{L}, and \texttip{\theta }{theta} and the acceleration due to gravity,
\texttip{g}{g}.
Hint 1. Identify the type of collision
Is the collision between the object and the pendulum bob an elastic or inelastic collision?
ANSWER:
elastic
inelastic
Hint 2. Find the momentum before the collision
Compose an expression for \texttip{p_{\rm before}}{p_before}, the momentum of the object and pendulum
bob before the collision when the object moves with speed \texttip{v_{\rm 0}}{v_0}.
Express your answer in terms of some or all of the variables \texttip{m}{m}, \texttip{M}{M},
\texttip{v_{\rm 0}}{v_0}, \texttip{v}{v}, \texttip{L}{L}, and \texttip{\theta }{theta} and the acceleration
due to gravity, \texttip{g}{g}.
Hint 1. Momentum
The momentum of an object of mass \texttip{m}{m} moving with speed \texttip{v}{v} is given by
mv.
ANSWER:
\texttip{p_{\rm before}}{p_before} = m v_{0}
Hint 3. Find the momentum after the collision
Compose an expression for \texttip{p_{\rm after}}{p_after}, the momentum of the object and pendulum
bob after the collision when they move with speed \texttip{v}{v}.
Express your answer in terms of some or all of the variables \texttip{m}{m}, \texttip{M}{M},
\texttip{v_{\rm 0}}{v_0}, \texttip{v}{v}, \texttip{L}{L}, and \texttip{\theta }{theta} and the acceleration
due to gravity, \texttip{g}{g}.
ANSWER:
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\texttip{p_{\rm after}}{p_after} = \left(m+M\right) v
ANSWER:
\texttip{v_{\rm 0}}{v_0} = \large{\frac{m+M}{m} v}
Hint 4. Describe the swing
Compose an expression that describes the motion of the object and the pendulum bob after the collision. Put
this expression in the form v = \cdots.
Express your answer in terms of some or all of the variables \texttip{m}{m}, \texttip{M}{M}, \texttip{v_{\rm
0}}{v_0}, \texttip{v}{v}, \texttip{L}{L}, and \texttip{\theta }{theta} and the acceleration due to gravity
\texttip{g}{g}.
Hint 1. Identify the energy at the bottom of the swing
What is the mechanical energy \texttip{E_{\rm bottom}}{E_bottom} of the object and pendulum bob just
after the collision but while they are still located at the bottom of the swing? Assume that the height of
the pendulum bob and object is zero at this location.
Express your answer in terms of some or all of the variables \texttip{m}{m}, \texttip{M}{M},
\texttip{v_{\rm 0}}{v_0}, \texttip{v}{v}, \texttip{L}{L}, and \texttip{\theta }{theta} and the acceleration
due to gravity, \texttip{g}{g}.
Hint 1. Mechanical energy
The mechanical energy of a system is the total kinetic and gravitational potential energy of the
system. The kinetic energy \texttip{K}{K} of an object of mass \texttip{m}{m} moving with speed
\texttip{v}{v} is
\large{K = \frac{1}{2}mv^2}.
The gravitational potential energy \texttip{U}{U} of an object of mass \texttip{m}{m} a height
\texttip{y}{y} above some reference point is
U = mgy.
Hint 2. Find the gravitational potential energy at the bottom of the swing
What is the gravitational potential energy \texttip{U_{\rm bottom}}{U_bottom} of the object and
pendulum bob at the bottom of the swing? Keep in mind that the height of the pendulum bob and
object is zero at this location.
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ANSWER:
U_{\rm bottom}= mgL
U_{\rm bottom}= 0
U_{\rm bottom}= ­mgL
Hint 3. Find the kinetic energy at the bottom of the swing
What is the kinetic energy \texttip{K_{\rm bottom}}{K_bottom} of the object and pendulum bob at
the bottom of the swing?
Express your answer in terms of some or all of the variables \texttip{m}{m}, \texttip{M}{M},
\texttip{v_{\rm 0}}{v_0}, \texttip{v}{v}, \texttip{L}{L}, and \texttip{\theta }{theta} and the
acceleration due to gravity, \texttip{g}{g}.
ANSWER:
\texttip{K_{\rm bottom}}{K_bottom} = \large{\frac{1}{2} \left(m+M\right) v^{2}}
ANSWER:
\texttip{E_{\rm bottom}}{E_bottom} = \large{\frac{1}{2} \left(m+M\right) v^{2}}
Hint 2. Identify the energy at the top of the swing
What is the mechanical energy \texttip{E_{\rm top}}{E_top} of the object and pendulum bob at the top of
the swing when it has reached its maximum angular displacement?
Express your answer in terms of some or all of the variables \texttip{m}{m}, \texttip{M}{M},
\texttip{v_{\rm 0}}{v_0}, \texttip{v}{v}, \texttip{L}{L}, and \texttip{\theta }{theta} and the acceleration
due to gravity, \texttip{g}{g}.
Hint 1. Mechanical energy
The mechanical energy of a system is the total kinetic and gravitational potential energy of the
system. The kinetic energy \texttip{K}{K} of an object of mass \texttip{m}{m} moving with speed
\texttip{v}{v} is
\large{K = \frac{1}{2}mv^2}.
The gravitational potential energy \texttip{U}{U} of an object of mass \texttip{m}{m} a height
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\texttip{y}{y} above some reference point is
U = mgy.
Hint 2. Find the height at the top of the swing
What is the height \texttip{h}{h} of the object and pendulum bob at the top of the swing, when they
have reached their maximum displacement? Keep in mind that the pendulum has a length
\texttip{L}{L} and swings through an angle \texttip{\theta }{theta}.
Express your answer in terms of some or all of the variables \texttip{L}{L} and \texttip{\theta
}{theta}.
ANSWER:
\texttip{h}{h} = L \left(1 ­ {\cos}\left({\theta}\right)\right)
Hint 3. Find the gravitational potential energy at the top of the swing
What is the gravitational potential energy \texttip{U_{\rm top}}{U_top} of the object and pendulum
bob at the top of the swing?
Express your answer in terms of some or all of the variables \texttip{m}{m}, \texttip{M}{M},
\texttip{v_{\rm 0}}{v_0}, \texttip{v}{v}, \texttip{L}{L}, and \texttip{\theta }{theta} and the
acceleration due to gravity, \texttip{g}{g}.
ANSWER:
\texttip{U_{\rm top}}{U_top} = \left(m+M\right) g L \left(1 ­ {\cos}\left({\theta}\right)\right)
Hint 4. Find the kinetic energy at the top of the swing
What is the kinetic energy \texttip{K_{\rm top}}{K_top} of the object and pendulum bob at the top
of the swing?
ANSWER:
\large{K_{\rm top} = \frac {1}{2} m v_0^2}
\large{K_{\rm top} = \frac {1}{2} m v^2}
\large{K_{\rm top} = \frac {1}{2} (m + M) v^2}
K_{\rm top} = 0
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ANSWER:
\texttip{E_{\rm top}}{E_top} = \left(m+M\right) g L \left(1 ­ {\cos}\left({\theta}\right)\right)
ANSWER:
\texttip{v}{v} = \sqrt{2 g L \left(1 ­ {\cos}\left({\theta}\right)\right)}
Hint 5. Relating the two physical processes
By applying conservation of momentum to the collision, you found an expression for \texttip{v_{\rm 0}}{v_0}:
v_0 = \left ( {m+M}\over {m} \right ) v.
By applying conservation of mechanical energy to the swing, you found an expression for \texttip{v}{v}:
v= \sqrt{2gL[1 ­ \cos(\theta)]}.
Combine these two expression into just one expression for \texttip{v_{\rm 0}}{v_0}.
ANSWER:
\texttip{v_{\rm 0}}{v_0} = \large{\frac{\left(m+M\right)\sqrt{2gL\left(1­{\cos}\left({\theta}\right)\right)}}{m}}
Correct
The ballistic pendulum was invented during the Napoleonic Wars to aide the British Navy in making better
cannons. It has since been used by ballisticians to measure the velocity of a bullet as it leaves the barrel of a gun.
In Part B you will use your expression for \texttip{v_{\rm 0}}{v_0} to compare the initial speeds of bullets fired from
9.0­{\rm mm} and .44­caliber handguns.
Part B
An experiment is done to compare the initial speed of bullets fired from different handguns: a 9.0 {\rm mm} and a .44
caliber. The guns are fired into a 10­{\rm kg} pendulum bob of length \texttip{L}{L}. Assume that the 9.0­{\rm mm} bullet
has a mass of 6.0 {\rm g} and the .44­caliber bullet has a mass of 12 {\rm g} . If the 9.0­{\rm mm} bullet causes the
pendulum to swing to a maximum angular displacement of 4.3^\circ and the .44­caliber bullet causes a displacement of
10.1^\circ , find the ratio of the initial speed of the 9.0­{\rm mm} bullet to the speed of the .44­caliber bullet,
(v_0)_{9.0}/(v_0)_{44}.
Express your answer numerically.
Hint 1. How to approach the problem
Use your expression from Part A to set up the ratio (v_0)_{9.0}/(v_0)_{44}. Try to cancel as many terms as
possible before plugging in your numbers to solve for a numeric answer.
ANSWER:
(v_0)_{9.0}/(v_0)_{44} = 0.85
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Correct
Police officers in the United States commonly carry 9.0­{\rm mm} handguns because they are easier to handle,
having a shorter barrel than typical .44­caliber guns. Not only does the .44­caliber bullet have more mass than the
9.0­{\rm mm} one, its passage through a longer gun barrel means that it also moves faster as it leaves the barrel,
which makes the .44­caliber Magnum a particularly powerful handgun. A .44­caliber bullet can travel at speeds
over 1000 {\rm {miles \;per\; hour}} (1600 {\rm kilometers\;per\;hour}).
Exercise 8.46
A 0.148 {\rm kg} glider is moving to the right on a frictionless, horizontal air track with a speed of 0.870 {\rm m/s} . It has a
head­on collision with a 0.290 {\rm kg} glider that is moving to the left with a speed of 2.11 {\rm m/s} . Suppose the collision
is elastic.
Part A
Find the magnitude of the final velocity of the 0.148 {\rm kg} glider.
ANSWER:
3.08 m/s Correct
Part B
Find the direction of the final velocity of the 0.148 {\rm kg} glider.
ANSWER:
to the right
to the left
Correct
Part C
Find the magnitude of the final velocity of the 0.290 {\rm kg} glider.
ANSWER:
9.61×10−2 m/s Correct
Part D
Find the direction of the final velocity of the 0.290 {\rm kg} glider.
ANSWER:
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to the right
to the left
Correct
Score Summary:
Your score on this assignment is 102%.
You received 6.13 out of a possible total of 6 points.
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