Download A Ball Hits a Wall Elastically

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 θ i 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.
Part A
What is the final angle θ f 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 θ f
in terms of θ i .
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(θ i )
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?
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Chapter 8_assignment 7
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:
vf x
= −v i sin(θ i )
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 θ f in terms of θ i .
ANSWER:
θf
= θi
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
⃗
= p f⃗ − p i⃗
In this case, only one force is acting, so ∣∣J ∣∣⃗
, along with the definition of impulse, J
= F Δt
. Putting everything together, F
=
⃗
⃗
= ∑ F Δt
p f⃗ −p i⃗
Δt
.
.
Hint 2. Change in momentum of the ball
The fact that θ f = θ i 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(θ i )
ANSWER:
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Chapter 8_assignment 7
F
= sin(θ f )
2mv i
Δt
A Relation Between Momentum and Kinetic Energy
Part A
A cardinal (Richmondena cardinalis) of mass 4.00×10−2 kg and a baseball of mass 0.140 kg have the same
kinetic energy. What is the ratio of the cardinal's magnitude pc of momentum to the magnitude pb of the
baseball's momentum?
Hint 1. How to approach the problem
Recall that the kinetic energy of an object (of mass m and speed v ) is given by K =
1
2
mv
2
, and the
magnitude of the momentum by p = mv . Combining these equations into a single expression can then
be used to eliminate v , giving an expression of the kinetic energy in terms of the momentum instead of
the velocity. We can then use this relation, along with the assumptions, to find the ratio of the momenta pc /pb in terms of the masses.
Hint 2. Find a relationship between kinetic energy and momentum
Select the general expression for the kinetic energy K of an object with mass m and momentum p.
ANSWER:
1
2
pm 2
1
p
2
m
pm
p
2
m
1
2
(
p
m
)
2
ANSWER:
pc
pb
= 0.535
Part B
A man weighing 730 N and a woman weighing 440 N have the same momentum. What is the ratio of the man's
K
K
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Chapter 8_assignment 7
kinetic energy Km to that of the woman Kw ?
Hint 1. How to approach the problem
As in the previous part, an expression for the momentum must be found in terms of the kinetic energy.
Then the ratio of the kinetic energies Km /Kw must be found in terms of the weights, instead of the
masses.
Hint 2. Find a relationship between momentum and kinetic energy
Select the general expression for the momentum p of an object with mass m and kinetic energy K .
ANSWER:
K
m
Km
2Km
−−
√
K
m
−
−
−
−
√Km
−
−
−
−
−
√2Km
ANSWER:
Km
Kw
= 0.603
Exercise 8.8
Force of a Baseball Swing. A baseball has mass 0.148 kg .
Part A
If the velocity of a pitched ball has a magnitude of 42.0 m/s and the batted ball's velocity is 51.0 m/s in the
opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by
the bat.
Express your answer to three significant figures and include the appropriate units.
ANSWER:
P
= 13.8 kg⋅m
s
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Chapter 8_assignment 7
Part B
If the ball remains in contact with the bat for 2.5 ms , find the magnitude of the average force applied by the
bat.
Express your answer to two significant figures and include the appropriate units.
ANSWER:
F
= 5.5 kN
Exercise 8.11
At time t = 0 , a 2750 kg rocket in outer space fires an engine that exerts an increasing force on it in the +x ­
2
direction. This force obeys the equation F x = At , where 1.25 s is time, and has a magnitude of 781.25 N when t
= 1.25 s .
Part A
Find the SI value of the constant A.
ANSWER:
A
= 500 N/s 2 Part B
What impulse does the engine exert on the rocket during the 1.50 s interval starting 2.00 s after the engine is
fired?
ANSWER:
p
= 5810 kg ⋅ m/s Part C
By how much does the rocket's velocity change during this interval?
ANSWER:
Δv
= 2.11 m/s 5/17
Chapter 8_assignment 7
Exercise 8.15
To warm up for a match, a tennis player hits the 58.0 g ball vertically with her racket.
Part A
If the ball is stationary just before it is hit and goes 5.00 m high, what impulse did she impart to it?
ANSWER:
p
= 0.574 kg ⋅ m/s Exercise 8.18
A 66.5 kg astronaut is doing a repair in space on the orbiting space station. She throws a 2.30 kg tool away from
her at 3.60 m/s relative to the space station.
Part A
With what speed will she begin to move?
ANSWER:
v
= 0.125 m/s Part B
In what direction will she begin to move?
ANSWER:
opposite to the direction in which she throws the tool
to the direction in which she throws the tool
Exercise 8.23
kg
N/cm
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Chapter 8_assignment 7
Two identical 1.70 kg masses are pressed against opposite ends of a spring of force constant 1.85 N/cm ,
compressing the spring by 15.0 cm from its normal length.
Part A
Find the maximum speed of each mass when it has moved free of the spring on a smooth, horizontal lab table.
ANSWER:
v max
= 1.11 m/s Exercise 8.25
A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20 g bullets at 965 m/s . The mass of
the hunter (including his gun) is 71.5 kg , and the hunter holds tight to the gun after firing it.
Part A
Find the recoil speed of the hunter if he fires the rifle horizontally.
ANSWER:
v
= 5.67×10−2 m/s Part B
Find the recoil speed of the hunter if he fires the rifle at 58.0 ∘ above the horizontal.
ANSWER:
v
= 3.00×10−2 m/s Exercise 8.32
Two skaters collide and grab on to each other on frictionless ice. One of them, of mass 70.0 kg , is moving to the
right at 2.00 m/s , while the other, of mass 65.0 kg , is moving to the left at 2.80 m/s .
Part A
What are the magnitude of the velocity of these skaters just after they collide?
ANSWER:
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Chapter 8_assignment 7
v
= 0.311 m/s Part B
What are the direction of the velocity of these skaters just after they collide?
ANSWER:
to the left
to the right
Exercise 8.38
Two cars collide at an intersection. Car A, with a mass of 1900 kg , is going from west to east, while car B , of
mass 1500 kg , is going from north to south at 17 m/s . As a result of this collision, the two cars become
enmeshed and move as one afterwards. In your role as an expert witness, you inspect the scene and determine
that, after the collision, the enmeshed cars moved at an angle of 60 ∘ south of east from the point of impact.
Part A
How fast were the enmeshed cars moving just after the collision?
Express your answer using two significant figures.
ANSWER:
v
= 8.7 m/s Part B
How fast was car A going just before the collision?
Express your answer using two significant figures.
ANSWER:
v
A
= 7.7 m/s 8/17
Chapter 8_assignment 7
Exercise 8.40
To protect their young in the nest, peregrine falcons will fly into birds of prey (such as ravens) at high speed. In one
such episode, a 570 g falcon flying at 20.0 m/s hit a 1.60 kg raven flying at 8.0 m/s . The falcon hit the raven at
right angles to its original path and bounced back at 5.0 m/s . (These figures were estimated by the author as he
watched this attack occur in northern New Mexico.)
Part A
By what angle did the falcon change the raven's direction of motion?
Express your answer using two significant figures.
ANSWER:
θ
= 48 ∘ Part B
What was the raven's speed right after the collision?
Express your answer using two significant figures.
ANSWER:
v
= 12 m/s Exercise 8.43
A 11.0 g rifle bullet is fired with a speed of 370 m/s into a ballistic pendulum with mass 7.00 kg , suspended from
a cord 70.0 cm long.
Part A
Compute the initial kinetic energy of the bullet;
ANSWER:
K
= 753 J Part B
Compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the
pendulum.
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Chapter 8_assignment 7
ANSWER:
K
= 1.18 J Part C
Compute the vertical height through which the pendulum rises.
ANSWER:
h
= 1.72 cm ± Traffic Accident Analysis
Consider the following two­car accident: Two cars of equal mass m collide at an intersection. Driver E was traveling
eastward, and driver N, northward. After the collision, the two cars remain joined together and slide, with locked
wheels, before coming to rest. Police on the scene measure the length d of the skid marks to be 9 meters. The
coefficient of friction μ between the locked wheels and the road is equal to 0.9.
Each driver claims that his speed was less than 14
meters per second (about 31 mph). A third driver, who
was traveling closely behind driver E prior to the collision,
supports driver E's claim by asserting that driver E's
speed could not have been greater than 12 meters per
second. Take the following steps to decide whether driver
N's statement is consistent with the third driver's
contention.
Part A
Let the speeds of drivers E and N prior to the collision be denoted by v e and v n , respectively. Find v 2 , the
square of the speed of the two­car system the instant after the collision.
Express your answer terms of v e and v n .
Hint 1. How to approach the problem
In general, you can find v 2 either by using conservation of energy or by finding the individual
⃗
components of the velocity v using conservation of momentum, depending on which quantity is
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Chapter 8_assignment 7
conserved. What is conserved in this collision?
ANSWER:
energy only
momentum only
both energy and momentum
Hint 2. Find the x and y components of velocity
The velocity of the two­car system immediately after the collision may be written as
^
^
v ⃗ = vx i + vy j
,
where the x and y directions are the eastward and northward directions, respectively. Find v x and v y .
Express the two components, separated by a comma, in terms of v e and v n .
Hint 1. Conservation of momentum in two dimensions
Recall that conservation of linear momentum may be expressed as a vector equation,
⃗
p initial
= p f⃗ inal .
This means that each vector component of linear momentum is conserved separately. Also recall
that
⃗
p initial
= ∑ m i v i⃗ ,
i
where mi is the mass of the ith component with initial velocity v i⃗ , and similarly, p f⃗ inal = m f inal v f⃗ inal .
Hint 2. Find the initial momentum
Find the components pe and pn of the initial momentum of the two­car system.
Express your answers, separated by a comma, in terms of m, v n , and v e .
ANSWER:
p
e
, pn (initial) = ,
mv e mv n
Hint 3. Find the final momentum
Write the components of the final momentum of the two­car system.
Express your answers, separated by a comma, in terms of m, v x , and v y , the components
of the final velocity of the two linked cars.
ANSWER:
pe
, pn (final) = ,
2mv x 2mv y
ANSWER:
vx
, v y = ,
0.5v e 0.5v n
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Chapter 8_assignment 7
Hint 3. The magnitude of the velocity vector
Recall that the square of the magnitude of a vector is given by the Pythagorean formula:
2
2
2
v = vx + vy .
ANSWER:
v
2
= ve
2
+v n
2
4
Part B
What is the kinetic energy K of the two­car system immediately after the collision?
Express your answer in terms of v e , v n , and m.
Hint 1. Definition of kinetic energy
The kinetic energy K of an object of mass m and speed v is given by the formula K =
1
2
mv
2
.
Hint 2. The mass of the system
After the collision, the mass of the two­car system is 2m .
ANSWER:
K
= (v e
2
+v n
2
)m
4
Part C
Write an expression for the work W f ric done on the cars by friction.
Express your answer symbolically in terms of the mass m of a single car, the magnitude of the
acceleration due to gravity g , the coefficient of sliding friction μ , and the distance d through which the
two­car system slides before coming to rest.
Hint 1. Definition of work
For a constant applied linear force F ⃗ , the work W required to move an object through a straight­line
⃗
⃗
⃗
W = F ⋅ d = F d cos θ , where θ is the angle between the direction of the
displacement d is given by
force and the direction of the displacement.
Hint 2. Magnitude of the frictional force
The magnitude of the frictional force Ff ric is given by Ff ric = μn , where n is the magnitude of the
normal force exerted on the cars by the road. In this problem, n is simply equal to the combined weight
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Chapter 8_assignment 7
of the two cars. (Why? Because aside from gravity and the normal force, there are no other forces acting
in a direction perpendicular to the road surface, and the cars are not accelerating in this direction.)
Hint 3. Direction of the frictional force
The sliding frictional force is always directed opposite to the direction of motion. Quantitatively, the
angle θ between the frictional force and the straight­line displacement of the two cars is π radians.
ANSWER:
W f ric
= −2mgμd
Part D
Using the information given in the problem introduction and assuming that the third driver is telling the truth,
determine whether driver N has reported his speed correctly. Specifically, if driver E had been traveling with a
speed of exactly 12 meters per second before the collision, what must driver N's speed have been before the
collision?
Express your answer numerically, in meters per second, to the nearest integer. Take g , the magnitude of
the acceleration due to gravity, to be 9.81 meters per second per second.
Hint 1. How to approach the problem
Use the work­energy theorem, Kf inal = Kinitial + W nc , to relate the kinetic energy of the two­car
system immediately after the collision (now Kinitial for this part of the motion), to the nonconservative
work done by friction in bringing the two cars finally to rest.
Hint 2. The final kinetic energy
Note that Kf inal
= 0
, since the cars are finally at rest.
ANSWER:
vn
= 22 m/s If you believe the report by the third driver that the speed of driver E's car was less than or equal to 12
meters per second, then driver N's speed just obtained is the minimum speed that driver N could have had
before the collision. So, even if you do not know that driver E's car was traveling at exactly 12 meters per
second before the collision, it is still evident that the driver of car N was not reporting his speed accurately.
Also, we have assumed that neither driver brakes before or during the collision. Including this factor makes
the analysis somewhat more involved in real situations.
Exercise 8.47
Blocks A (mass 2.00 kg ) and B (mass 14.00 kg ) move on a frictionless, horizontal surface. Initially, block B is
at rest and block A is moving toward it at 5.00 m/s . The blocks are equipped with ideal spring bumpers. The
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Chapter 8_assignment 7
collision is head­on, so all motion before and after the collision is along a straight line. Let +x be the direction of
the initial motion of A.
Part A
Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. Find the maximum energy.
ANSWER:
Ub
= 21.9 J Part B
Find the velocity of A.
ANSWER:
vA
= 0.625 m/s Part C
Find the velocity of B.
ANSWER:
vB
= 0.625 m/s Part D
Find the velocity of each block after they have moved apart.
Find the velocity of A.
ANSWER:
vA
= ­3.75 m/s Part E
Find the velocity of B.
14/17
Chapter 8_assignment 7
ANSWER:
vB
= 1.25 m/s Problem 8.04
Part A
A stationary 1.67­kg object is struck by a stick. The object experiences a horizontal force given by F = at ­ bt 2,
where t is the time in milliseconds from the instant the stick first contacts the object. If a = 1500 N/(ms) and b
= 20 N/(ms)2, what is the speed of the object just after it comes away from the stick at t = 2.74 ms?
ANSWER:
22 m/s
3.7 m/s
3.3 m/s
25 m/s
Problem 8.23
Part A
A 900­kg car traveling east at 15.0 m/s collides with a 750­kg car traveling north at 20.0 m/s. The cars stick
together. Assume that any other unbalanced forces are negligible.
(a) What is the speed of the wreckage just after the collision?
ANSWER:
12.2 m/s Part B
(b) In what direction does the wreckage move just after the collision?
ANSWER:
15/17
Chapter 8_assignment 7
48.0 ° N of E Problem 8.26
Part A
On a frictionless horizontal table, two blocks (A of mass 2.00 kg and B of mass 3.00 kg) are pressed together
against an ideal massless spring that stores 75.0 J of elastic potential energy. The blocks are not attached to
the spring and are free to move free of it once they are released from rest. The maximum speed achieved by
each block is closest to:
ANSWER:
6.71 m/s (A), 4.47 m/s (B)
5.00 m/s (A), 6.12 m/s (B)
6.12 m/s (A), 5.00 m/s (B)
4.47 m/s (A), 6.71 m/s (B)
5.48 m/s for both
Problem 8.80
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg , was stopped
at a red light when it was hit from behind by car A, of mass 1500 kg . The cars locked bumpers during the collision
and slid to a stop. Measurements of the skid marks left by the tires showed them to be 7.25 m long, and inspection
of the tire tread revealed that the coefficient of kinetic friction between the tires and the road was 0.650.
Part A
What was the speed of car A just before the collision?
ANSWER:
v
= 51.6 mph Part B
By how many mph was car A exceeding the speed limit 35.0 mph ?
ANSWER
Δv
= 16.6 mph 16/17