Download Homework 10 - Physics | Oregon State University

Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004
This print-out should have 26 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before making your selection. The due time is
Central time.
Chapter 9 problems.
001 (part 1 of 1) 0 points
An object of mass m is moving with speed
v0 to the right on a horizontal frictionless
surface, as shown, when it explodes into two
2
pieces. Subsequently, one piece of mass m
5
v0
to the left.
moves with a speed v2/5 =
2
v0
m
The speed k~v3/5 k of of the other piece the
object is
v0
.
2
v0
2. k~v3/5 k = .
3
7 v0
.
3. k~v3/5 k =
5
3 v0
4. k~v3/5 k =
.
2
1. k~v3/5 k =
5. k~v3/5 k = 2 v0 . correct
Explanation:
The horizontal component of the momentum is conserved, so
2
3
m v2/5 + m v3/5
5
5
2 ³ v0 ´ 3
0 + m v0 = m −
+ m v3/5
5
2
5
1
3
mv0 = − m v0 + m v3/5
5
5
3
6
v = v0
5 3/5 5
k~v3/5 k = 2 v0 .
1
engine pulling the car. It begins to rain. The
rain falls straight down and begins to fill the
train car.
The speed of the car
1. stays the same.
2. decreases. correct
3. increases.
Explanation:
Using Newton’s second law, we have
dP
= 0,
dt
since no external forces act on the train in the
horizontal direction. With no rain, the train
will move at a constant velocity; however,
when it starts to rain, and the rain starts to
fill the car, the mass of the train changes.
Thus,
dm
dv
m
= −v
.
dt
dt
dm
is positive; i.e., the mass of the
Since
dt
train is increasing with accumulating rain,
dv
should be negative; i.e., the speed of the
dt
train should decrease.
003 (part 1 of 2) 5 points
Consider the set up of a ballistic pendulum where m1 = 11.4 g, m2 = 3.97 kg,
h = 0.243 m.
The acceleration of gravity is 9.8 m/s2 .
0 + m v0 =
002 (part 1 of 1) 0 points
An open train car moves with speed 19 m/s
on a flat frictionless railroad track, with no
m1+ m 2
v1
m1
m2
vf
h
Find the final velocity of the system (m1 +
m2 ) immediately after the collision and before
the pendulum starts to swing upwards.
Correct answer: 2.18238 m/s.
Explanation:
From the setup, the final velocity of the collision is the initial velocity of the subsequent
Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004
2
swing. During the swinging process the total
energy is conserved
after he catches the ball back. That is, when
the man throws the ball away,
Ei = E f
M v1 = m v
1
(m1 + m2 ) vf2 = (m1 + m2 ) g h .
2
Therefore
p
2gh
q
= 2 (9.8 m/s2 ) (0.243 m)
vf =
and when the man catches the ball bounced
back from the wall,
(M + m) V1 = M v1 + m v = 2 m v .
Solving for V1 , we get,
= 2.18238 m/s .
V1 =
004 (part 2 of 2) 5 points
Find v1 , the initial speed of m1 .
Correct answer: 762.188 m/s.
Explanation:
The linear momentum is conserved in a
collision
pi = p f
m1 v1 = (m1 + m2 ) vf
Therefore
m1 + m 2
vf
m1
(0.0114 kg) + (3.97 kg)
(2.18238 m/s)
=
(0.0114 kg)
= 762.188 m/s .
v1 =
005 (part 1 of 2) 0 points
A 83 kg man holding a 0.336 kg ball stands
on a frozen pond next to a wall. He throws
the ball at the wall with a speed of 15.7 m/s
(relative to the ground) and then catches the
ball after it rebounds from the wall.
Hint: Ignore the projectile motion of the
ball, and assume that it loses no energy in its
collision with the wall.
How fast is he moving after he catches the
ball?
Correct answer: 0.126601 m/s.
Explanation:
Let M and m be the mass of the man and
the ball respectively. Let v be the speed
of the ball. Using momentum conservation
twice, we can get the speed the man moves
2mv
M +m
006 (part 2 of 2) 0 points
Note: Remember that the ball is thrown with
a speed of 15.7 m/s relative to the ground
each time it is thrown.
How many times does the man have to go
through this process before his speed reaches
at least 2.7 m/s relative to the ground?
Correct answer: 21 .
Explanation:
If he repeats this n times, he’ll move at a
speed
2mnv
Vn =
.
M +m
If he is to reach V , he should repeat this
µ
¶
M +m
n=V
2mv
times.
007 (part 1 of 4) 3 points
A massless spring with force constant
k =441 N/m is fastened at its left end to
a vertical wall, as shown below.
The acceleration of gravity is 9.8 m/s2 .
k
MC MD
C D
Initially, block C (mass mC =3.57 kg) and
block D (mass mD =2.86 kg) rest on a horizontal surface with block C in contact with
the spring (but not compressing it) and with
block D in contact with block C. Block C is
then moved to the left, compressing the spring
Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004
a distance of 0.4 m, and held in place while
block D remains at rest as shown in below.
d
C
µ
D
Determine the elastic energy U stored in
the compressed spring.
Correct answer: 35.28 J.
Explanation:
Should use the expression for the energy
stored in a spring:
1
U = k x2
2
1
= (441 N/m) (0.4 m)2
2
= 35.28 J .
008 (part 2 of 4) 3 points
Block C is then released and accelerates to the
right, toward block D. The surface is rough
and the coefficient of friction between each
block and the surface is µ =0.421. The two
blocks collide instantaneously, stick together,
and move to the right. Remember that the
spring is not attached to block C.
Find the speed vC of block C just before it
collides with block D.
Correct answer: 4.05759 m/s.
Explanation:
Apply conservation of energy or workenergy theorem to this problem. Kinetic en1
ergy of block C is K = mC vC2 ; work done (or
2
energy dissipated by) friction is Wf = µ Fn d.
Therefore we have
1
mC vC2 = U − µ mC g d .
2
Solving for vC
s
2
(U − µ mC g d)
vC =
mC
= 4.05759 m/s .
Correct alternate
Z solution
Computing
Fnet dx, where Fnet = k x −
µ Fn , to find the kinetic energy, and then
computing the speed.
3
009 (part 3 of 4) 2 points
Find the speed vf of blocks C and D just after
they collide.
Correct answer: 2.25282 m/s.
Explanation:
In this configuration, we have the conservation of momentum:
mC vC = (mC + mD ) vf .
Solving for vf
mC vC
mC + m D
= 2.25282 m/s .
vf =
010 (part 4 of 4) 2 points
Find the horizontal distance the blocks move
before coming to rest.
Correct answer: 0.615054 m.
Explanation:
The blocks come to rest when all their
kinetic energy has been dissipated; i.e.,
∆KE =Work done by frictional force
1
(mC + mD ) vf2 = µ (mC + mD ) g `
2
Solving for `
`=
vf2
2µg
(2.25282 m/s)2
=
2 (0.421) (9.8 m/s2 )
= 0.615054 m .
Alternate Solution:
X
F = ma
P
F
a=
m
µ (mC + mD ) g
=
(mC + mD )
= µg
= 4.1258 m/s2
v 2 = v02 + 2 a `
v 2 − v02
`=
2a
= 0.615054 m .
Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004
m v 2 k x2 m vi2
+
−
2
2
2
(0.0055 kg) (177.182 m/s)2
=
2
(1400 N/m) (0.012 m)2
+
2
(0.0055 kg) (315 m/s)2
−
2
= (86.3323 J) + (0.1008 J)
− (272.869 J)
= −186.436 J
|∆E| = 186.436 J .
=
011 (part 1 of 2) 0 points
A bullet of mass 5.5 g moving with an initial
speed 315 m/s is fired into and passes through
a block of mass 2.85 kg, as shown in the figure.
The block, initially at rest on a frictionless,
horizontal surface, is connected to a spring of
force constant 1400 N/m.
v
If the block moves a distance 1.2 cm to the
right after the bullet passed through it, find
the speed v at which the bullet emerges from
the block.
Correct answer: 177.182 m/s.
Explanation:
From conservation of energy
1
1
M V02 = k x2 .
2
2
4
013 (part 1 of 2) 0 points
Consider the general case of collisions of two
masses m1 = m with m2 = 2m along a frictionless horizontal surface. Denote the initial
and the final center of mass momenta to be
picm = p1 + p2 ,
and
pfcm = p01 + p02 .
And the initial and final kinetic energies to be
Hence,
V0 =
r
k x2
= 0.265964 m/s .
M
Then from momentum conservation
m v0 = M V0 + m v ,
or
M V0
v = v0 −
m
(2.85 kg) (0.265964 m/s)
= (315 m/s) −
(5.5 g) (0.001 kg/g)
= 177.182 m/s .
Ki = K 1 + K 2 ,
and
Kf = K10 + K20 .
v1
m1
v2
m2
For an elastic collision which pair of statements is correct?
1. picm < pfcm , Ki < Kf
2. picm > pfcm , Ki = Kf
012 (part 2 of 2) 0 points
Find the magnitude of the energy lost in the
collision.
Correct answer: 186.436 J.
Explanation:
The energy lost is
∆E = Kf + Uf − Ki
3. picm > pfcm , Ki > Kf
4. picm = pfcm , Ki > Kf
5. picm > pfcm , Ki < Kf
6. picm = pfcm , Ki = Kf correct
Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004
7.
picm
<
pfcm ,
with the magnitude of the momentum conserved.
Note: In the CM frame the total momentum is still zero if each velocity changes sign.
Hence, the final velocity of m2 in the CM
frame is
Ki = Kf
8. picm < pfcm , Ki > Kf
9. picm = pfcm , Ki < Kf
Explanation:
Momentum and energy are always conserved in elastic collision.
014 (part 2 of 2) 0 points
Now consider an elastic head-on collision
again with initial velocity of m1 to be v1 and
of m2 to be v2 = 0.
Find the final speed v2f of m2 .
1. v2f =
1
v1
3
(CM )
v2,f
(CM )
lab
= v2,f + VCM = 2 VCM
v2,f
2m1 v1
2
=
= v1 .
m1 + m 2
3
v2,f = 2vcm − v2,i .
3
v1
4
2
4. v2f = v1
7
3
5. v2f = v1
5
1
6. v2f = v1
2
v1
For the present case,v2,i = 0, vcm =
, so
3
2
v2,f = v1
3
7. v2f = 2 v1
1
v1
5
2
9. v2f = v1 correct
3
1
10. v2f = v1
4
Explanation:
The CM velocity is given by
P
mi vi
m1 v1
=
VCM = Pi
m1 + m 2
i mi
8. v2f =
as v2 = 0. Thus, in the CM frame, m2 has a
velocity
= v2 − VCM = −VCM =
m1 v1
m1 + m 2
To sum up, we have the following general
expression,
3. v2f =
(CM )
= VCM =
lab
The velocity v2,f
(v2f ) in the lab frame is
2. v2f = v1
v2
5
−m1 v1
.
m1 + m 2
After a one dimensional elastic collision, each
body changes the direction of its momentum
015 (part 1 of 3) 0 points
Two balls with masses m1 and m2 are on
the x-axis. Ball m1 has an initial velocity
v1 > 0 along the positive x-axis and ball m2
is initially at rest. The balls collide elastically
and remain on the x-axis after the collision.
If m1 = m2 , what is the final velocity v10 of
the ball m1 ?
1. v10 = +v1
2. +2 v1 < v10 < ∞
3. v10 = −v1
4. v10 = 0 correct
5. v10 = +2 v1
Explanation:
Basic Concepts: Conservation of Energy,
where v2 = 0:
1
1
1
2
2
m1 v1 2 = m1 v10 + m2 v20
2
2
2
Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004
Conservation of Momentum, where v2 = 0:
m1 v1 =
Solving for
v10
m1 v10
+ m2 v20
3. +2 v1 < v10 < ∞
4. v10 = −v1 correct
v20
and yields:
µ
¶
m1 − m 2
0
v1 =
v1
m1 + m 2
¶
µ
2 m1
0
v1
v2 =
m1 + m 2
5. v10 = +2 v1
(1)
(2)
All solutions can be determined using the
above equations (1) and (2). Think of playing pool. You hit the cue ball with velocity v1
dead center on the 8 ball v2 = 0. The cue ball
stops and the 8 ball proceeds with the same
velocity as the cue ball originally had. Part 1:
Using eqn (1), where m2 = m1 = m,
¶
µ
m−m
0
v1 = 0
v1 =
m+m
Part 2: Using eqn (2), where m2 = m1 = m,
µ
¶
2m
0
v2 =
v1 = +v1
m+m
016 (part 2 of 3) 0 points
0
If m1 = m2 , what is the final velocity v2 of
the ball m2 ?
1. v20 = −v1
2. +2 v1 < v20 < ∞
3. v20 = +2 v1
4. v20 = 0
5. v20 = +v1 correct
Explanation:
Explanation:
Think of throwing a golf ball with velocity v1 against a concrete wall v2 = 0. The
golf ball hits the wall, bounces back (reverses
direction) and has the same speed as it originally had (elastic collision). Part 3: Using
m1
→ 0,
eqn (1), where 1 À
m2
!
Ãm
1
−
1
2
v1 = −v1
v10 = m
m1
m2 + 1
Part 4: Using eqn (2), where 1 À
v20
=
2.
v10
= +v1
Ã
m1
2m
2
m1
m2
+1
!
m1
→ 0,
m2
v1 = 0
018 (part 1 of 1) 0 points
A 22.5 g bullet is fired horizontally into a 1 kg
wooden block resting on a horizontal surface
( µ = 0.257 ). The bullet goes through the
block and comes out with a speed of 186 m/s.
The acceleration of gravity is 9.8 m/s2 .
If the block travels 5.64 m before coming to
rest, what was the initial speed of the bullet?
Correct answer: 422.893 m/s.
Explanation:
Let m be the mass of the bullet and M the
mass of the block. From momentum conservation, we have,
mv1 = mv + M v2
From energy conservation, we have,
017 (part 3 of 3) 0 points
In the limit, when m1 ¿ m2 , what is the final
velocity v10 of the ball m1 ?
1. v10 = 0
6
Thus we have,
1
M v22 = µM gd
2
p
v2 = 2µgd
v1 = v +
Mp
M
v2 = v +
2µgd
m
m
Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004
11 m/s
60
◦
87 kg
524 kg
9.8 m/s2
What is the speed of the two blocks immediately after the collision?
Correct answer: 9.43372 m/s.
Explanation:
9.8 m/s2
611 kg
h
019 (part 1 of 2) 5 points
A(n) 524 kg mass is sliding on a horizontal
frictionless surface with a speed of 11 m/s
when it collides with a 87 kg mass initially at
rest, as shown in the figure. The masses stick
together and slide up a frictionless track at
60◦ from horizontal.
The acceleration of gravity is 9.8 m/s2 .
To what maximum height h above the horizontal surface will the masses slide?
Correct answer: 4.54056 m.
Explanation:
The kinetic energy K of the combined system is
K=
m21
1
2
(m1 + m2 ) v12
=
v2
2
2 (m1 + m2 ) 1
At the maximum height, all the kinetic energy
is converted to potential energy, and so
(m1 + m2 ) g h = K =
Let : v12
m1
m2
v1
g
= speed of blocks after collision
= 524 kg ,
= 87 kg ,
= 11 m/s , and
= 9.8 m/s2 .
Therefore, the speed of the combined masses
is
m1
v1
m1 + m 2
(524 kg)
=
(11 m/s)
(524 kg) + (87 kg)
= 9.43372 m/s .
v12 =
m21
v2
2 (m1 + m2 ) 1
Solving for h, we get
·
¸2
v12
m1
h=
2 g m1 + m 2
(11 m/s)2
=
2 (9.8 m/s2 )
·
¸2
(524 kg)
×
(524 kg) + (87 kg)
= 4.54056 m .
The initial momentum of block 1 is m1 v1 .
When m1 sticks with m2 , the resulting momentum p12 is equal to that initial momentum; thus,
p12 ≡ (m1 + m2 ) v12 = m1 v1 .
021 (part 1 of 2) 0 points
A 30-06 caliber hunting rifle fires a bullet of
mass 0.0158 kg with a velocity of 284 m/s to
the right. The rifle has a mass of 6.24 kg.
What is the recoil speed of the rifle as the
bullet leaves the rifle?
Correct answer: 0.719103 m/s.
Explanation:
From conservation of momentum ∆p = 0
0 = ∆pb + ∆pr
or
020 (part 2 of 2) 5 points
7
0 = m b vb + m r vr .
Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004
Hence
mb vb
mr
(0.0158 kg) (284 m/s)
=−
(6.24 kg)
= 0.719103 m/s .
V =−
022 (part 2 of 2) 0 points
If the rifle is stopped by the hunter’s shoulder
in a distance of 3.88 cm, what is the magnitude of the average force exerted on the
shoulder by the rifle?
Correct answer: 41.5819 N.
Explanation:
The average force is
Fav =
∆pr
.
∆t
The time of contact is
∆t =
2∆s
= 0.107912 s ,
vr
since the average velocity from the guns initial
velocity (v = vr ) to when it stops (v = 0) is
vr
v =
assuming constant de-acceleration.
2
Then the average force is
Fav
mr vr
=
∆t
(6.24 kg) (0.719103 m/s)
=
(0.107912 s)
= 41.5819 N .
023 (part 1 of 2) 5 points
Given: G = 6.67259 × 10−11 N m2 /kg2
Two hypothetical planets of masses 3.8 ×
23
10 kg and 6.5 × 1023 kg and radii 3.6 × 106 m
and 5.9 × 106 m, respectively, are at rest when
they are an infinite distance apart. Because
of their gravitational attraction, they head
toward each other on a collision course.
When their center-to-center separation is
6.5 × 108 m, find their relative velocity.
Correct answer: 459.459 m/s.
Explanation:
8
At infinite separation the potential energy
U is zero, and at rest the kinetic energy K is
zero. Since energy is conserved we have
0=
1
1
G m 1 m2
m1 v12 + m2 v22 −
.
2
2
d
The initial momentum is zero and momentum
is conserved, so
0 = m 1 v1 − m 2 v2 .
Combine these two equations to find
s
2G
v1 = m 2
d (m1 + m2 )
= 6.5 × 1023 kg
s
2 (6.67259 × 10−11 N m2 /kg2 )
×
(6.5 × 108 m) (1.03 × 1024 kg)
= 289.95 m/s
s
2G
v2 = m 1
d (m1 + m2 )
= 3.8 × 1023 kg
s
2 (6.67259 × 10−11 N m2 /kg2 )
×
(6.5 × 108 m) (1.03 × 1024 kg)
= 169.509 m/s .
where m1 + m2 = (3.8 × 1023
(6.5 × 1023 kg) = 1.03 × 1024 kg.
The relative velocity is
kg) +
vr = v1 − (−v2 )
= v1 + v2
= 289.95 m/s + 169.509 m/s
= 459.459 m/s .
024 (part 2 of 2) 5 points
Find the total kinetic energy of the planets
just before they collide.
Hint: Both energy and momentum are conserved.
Correct answer: 1.73487 × 1030 J.
Explanation:
From conservation of energy the total kinetic energy of the system just before the
Answer, Key – Homework 10 – David McIntyre – 45123 – May 10, 2004
planets collide is the opposite of the gravitational potential energy calculated for a distance d = r1 + r2 :
G m 1 m2
r1 + r 2
= (6.67259 × 10−11 N m2 /kg2 )
(3.8 × 1023 kg) (6.5 × 1023 kg)
×
(3.6 × 106 m) + (5.9 × 106 m)
= 1.73487 × 1030 J .
Ktot =
025 (part 1 of 2) 0 points
For this problem, which consists of 2 parts,
we assume that we are on Planet-I. The
radius of this planet is R =4290 km, the
gravitational acceleration at the surface is
gI =2.48 m/s2 , and the gravitational constant
G = 6.67259 × 10−11 N m2 /kg2 . The mass of
Planet-I is not given. Not all the quantities
given here will be used.
Suppose a rocket of mass m =5900 kg is
projected vertically upward from the surface of this planet. It stops at a point
h =20935.2 km from the surface of the planet.
Caution: Here the gravitational acceleration decreases as the rocket travels away from
Planet-I.
Determine the kinetic energy (in Joules) of
the rocket immediately after it is fired off.
Correct answer: 5.20959 × 1010 J.
Explanation:
For r > R, the potential energy of the
GM m
GM
rocket is U = −
. Since gI =
,
r
R2
m gI R 2
U = −
. Apply conservation of enr
ergy: E = Ki + Ui = Kf + Uf
⇒ Ki = Uf − Ui = −m gI R2 /r −
(−m gI R) = mgI R(1 − 1/n).
026 (part 2 of 2) 0 points
Find the minimum initial velocity of the
rocket such that it will escape the gravitational field of Planet-I.
Note: You can work on this part even if
you did not get Part 1.
Correct answer: 4612.85 m/s.
Explanation:
9
For the minimum velocity case, the final
kinetic energy of the rocket is zero. Since
at the end the rocket will be infinitely far
away from Planet-I, the final potential energy
of the rocket is also zero. Conservation of
energy implies Ki + Ui = Kf + Uf = 0. Thus
Mm
Ki = 0 − U i = G
= m gI R
R
p
⇒ vi = 2 g R