# PHYSICS 111 HOMEWORK SOLUTION #8 March 24, 2013

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```PHYSICS 111 HOMEWORK
SOLUTION #8
March 24, 2013
0.1
A particle of mass m moves with momentum of magnitude p.
2
p
• a) Show that the kinetic energy of the particle is: K = 2m
(Do this on paper. Your instructor may ask you to turn in this
work.)
• b) Express the magnitude of the particle’s momentum in terms
of its kinetic energy and mass.
a)
By definition, the momentum of a partcile moving with velocity ~v is : p~ = m~v .
Its magnitude is p = mv.
Kinetic energy is :
K
=
=
=
mv 2
2
p 2
m( m
)
2
p2
2m
b)
K=
mv 2
2 ,
therefore v =
q
2K
m
p
= mv
r
2K
m
2K
=
m2
m
√
=
2Km
= m
r
2
0.2.
0.2
An object has a kinetic energy of 239 J and a momentum of magnitude
27.3 kgm/s. Find the speed and the mass of the object.
• Let’s use the expressions of problem 1
K
p
v
mv 2
2mv
v
=
2
2K
=
p
2 × 239
=
27.3
= 17.51m/s
=
• The mass cam be then obtained from momentum as :
m
p
v
27.3
=
17.51
= 1.56kg
=
0.3
At one instant, a 17.0-kg sled is moving over a horizontal surface of
snow at 3.10 m/s. After 7.00 s has elapsed, the sled stops. Use a
momentum approach to find the magnitude of the average friction
force acting on the sled while it was moving.
• The sled started with a speed of 3.10 m/s and will come to stop after
7.00s. This change in speed or momentum is due to friction.
3
We can rewrite Newton’s second law by taking momentum change into
consideration as follows:
X
F~i
= m~a
d~v
dt
d(m~v )
dt
d~
p
dt
= m
=
=
+y
~
N
moving right
f~
+x
m~g
If we project this on the horizontal surface , the only force that remains
is friction f. In average:
−f
f
4
pf − pi
∆t
vf − vi
= m
∆t
0 − 3.10
= 17 ×
7
= 7.53N
=
0.4.
0.4
A 45.2-kg girl is standing on a 159-kg plank. Both originally at rest
on a frozen lake that constitutes a frictionless, flat surface. The girl
begins to walk along the plank at a constant velocity of 1.48 ~i m/s
relative to the plank.
• a) What is the velocity of the plank relative to the ice surface?
• b) What is the girl’s velocity relative to the ice surface?
a)
The system {girl+plank} is in a frictionless environment for which momentum
should be conserved during motion.
We adopt the following notations :
• the girl has mass m and velocity ~v relative to surface
~ relative to surface
• the plank has mass M and velocity V
• the girl’s velocity relative to the plank is ~vgrl/plk = 1.48~i
At rest vector momentum is ~0, during the motion this momentum is
~ . We should have : m~v + M~v = ~0 or consequently m~v = −M V
~.
m~v + M V
This already indicates that the girl and the plank are moving in opposite directions. On the other hand, the girl’s velocity relative to surface is an addition
of her velocity relative to the plank and the velocity of the plank relative to
the surface :
~
~v = ~vgrl/plk + V
~
m~v = −M V
~ ) = −M V
~
m(~vgrl/plk + V
~ = m~vgrl/plk
−(m + M )V
~ = − m ~vgrl/plk
V
m+M
45.2
= −
× 1.48~i
159 + 45.2
= −0.33~i
The plank is moving with speed 0.33 m/s in the opposite direction.
5
b)
The girl’s velocity relative to the ice surface is:
~v
v
6
~
= ~vgrl/plk + V
= 1.48~i − 0.33~i
=
1.15~i
=
1.15m/s
0.5.
0.5
Two blocks of masses m and 3m are placed on a frictionless, horizontal
surface. A light spring is attached to the more massive block, and the
blocks are pushed together with the spring between them as shown
in the figure below. A cord initially holding the blocks together is
burned; after that happens, the block of mass 3m moves to the right
~3m = 2.60~i m/s
with a speed of V
• a)What is the velocity of the block of mass m? (Assume right
is positive and left is negative.)
• b) Find the system’s original elastic potential energy, taking m
= 0.460 kg.
• c) Is the original energy in the spring or in the cord?
• e) Is the momentum of the system conserved in the burstingapart process?
• f) Explain how that is possible considering there are large forces
acting.
• g)Explain how that is possible considering there is no motion
beforehand and plenty of motion afterward?
7
a)
Momentum is conserved and we have :
~3m
m~vm + 3mV
~vm
= ~0
=
=
3m ~
V3m
m
~3m
−3V
−
= −3 × 2.60~i
= −7.80~i
|~vm | =
7.80m/s
b)
Total energy is conserved, gravitation potential energy doesnt change but elastic potential energy changes after the spring is released.
Uelf − Ueli
0 − Uelf
Ueli
−(Kf − Ki )
1
1
2
2
+ 3mV3m
)
= −( mvm
2
2
m 2
2
=
(v + 3V3m
)
2 m
0.460
=
(7.802 + 3 × 2.602 )
2
= 18.7J
=
c)
the original energy is in the spring.
d)
A force had to be exerted over a distance to compress the spring, transferring
energy into it by work. The cord exerts force, but over no distance.
e)
the momentum of the system is conserved in the bursting-apart process and
that’s what we used in the first question.
f)
The forces on the two blocks are internal forces, which cannot change the
momentum of the system the system is isolated.
8
0.6.
e)
Even though there is motion afterward, the final momenta are of equal magnitude in opposite directions so the final momentum of the system is still zero.
0.6
After a 0.390-kg rubber ball is dropped from a height of 1.70 m, it
bounces off a concrete floor and rebounds to a height of 1.55 m.
• a) Determine the magnitude and direction of the impulse delivered to the ball by the floor.
• b) Estimate the time the ball is in contact with the floor to be
0.07 seconds. Calculate the average force the floor exerts on the
ball.
a)
Hitting the floor then bouncing up will cause a momentum change , the impulse
delivered to the ball by the floor is just this momentum change.
Impulse
=
∆M omentum
= m~vaf ter − m~vbef ore
Falling from a 1.70 m will give the speed at the moment the ball hits the floor:
2
vbef
ore − 0
=
2a∆H
2
vbef
ore
=
vbef ore
=
=
2g∆H
p
2g∆H
√
2 × 9.81 × 1.70
=
=
5.77m/s
−5.77~j
~vbef ore
Similarly, Bouncing up to 1.55m will give the speed at the moment the ball
bounces up:
√
vaf ter =
2 × 9.81 × 1.55
=
~vaf ter
=
5.51m/s
5.51~j
9
Finally,
Impulse
=
∆M omentum
=
=
m~vaf ter − m~vbef ore
0.390(5.51~j + 5.77~j)
=
4.40~j
The impulse delivered amounts to 4.40 kg.m/s with direction up.
b)
P~
In problem 2 we rewrote Newton’s 2nd Law as:
Fi =
The average force the floor exerts on the floor is then
F~
~
dP
dt
.
∆P
∆t
4.40~
=
j
0.07
= 62.9~j
=
with maginutude 62.9 N and direction up.
0.7
A tennis player receives a shot with the ball (0.060 0 kg) traveling
horizontally at 59.5 m/s and returns the shot with the ball traveling
horizontally at 37.5 m/s in the opposite direction. (Assume the initial
direction of the ball is in the x direction.)
• a) What is the impulse delivered to the ball by the tennis racquet?
• b)What work does the racquet do on the ball?
a)
We will use the same procedure as in problem 6.
The impulse delivered to the ball is :
10
0.8.
Impulse
=
∆M omentum
= m~vaf ter − m~vbef ore
= 0.060(37.5~i − (−59.5)~i)
=
5.82~i kg.m/s or N.s
b)
The work done by the raquet on the ball can be calculated from the kinetic
enrgy change as:
W
=
∆K
1
=
m(vf2 − vi2 )
2
1
=
× 0.06(37.52 − 59.52 )
2
= −64J
0.8
A 1 240.0 kg car traveling initially with a speed of 25.000 m/s
in an easterly direction crashes into the back of a 8 100.0 kg
truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.
• a)What is the velocity of the truck right after the collision?
• b) What is the change in mechanical energy of the cartruck
system in the collision?
• c) Account for this change in mechanical energy.
11
a)
Conservation of momentum of the {car+truck} system can be expressed as:
~ci + M V
~ti
mV
~tf
V
~cf + M V
~tf
mV
m ~
~cf ) + V
~ti
(Vci − V
=
M
1240
=
(28 − 18)~i + 20~i
8100
= 21.0716~i
=
The truck will keep moving east with speed of 21.0716 m/s
b)
The change in mechanical energy is computed through the kinetic energy
change as no potential energy change takes place:
∆E
=
=
=
1
1
2
m(Vcf
− Vci2 ) + M (Vtf2 − Vti2 )
2
2
1
1
2
× 1240 × (18 − 252 ) + × 8100(21.0722 − 202 )
2
2
−8301J
c)
Most of the energy was transformed to internal energy with some being carried
away by sound.
0.9
A 9.6-g bullet is fired into a stationary block of wood having mass m
= 4.90 kg. The bullet imbeds into the block. The speed of the bulletplus-wood combination immediately after the collision is 0.603 m/s.
What was the original speed of the bullet? (Express your answer with
four significant figures.)
We use momentum conservation and we take into account that the bullet and
the block have the same speed after collision
~i
m~vi + M V
12
~f
= m~vf + M V
0.10.
m~vi + 0
vi
~f
(m + M )V
m+M
=
Vf
m
0.0096 + 4.90
0.603
=
0.0096
= 308.4m/s
=
0.10
A neutron in a nuclear reactor makes an elastic, head-on collision
with the nucleus of a carbon atom initially at rest.
• a) What fraction of the neutron’s kinetic energy is transferred to
the carbon nucleus? (The mass of the carbon nucleus is about
12.0 times the mass of the neutron.)
• b) The initial kinetic energy of the neutron is 1.1010−13 J. Find
its final kinetic energy and the kinetic energy of the carbon
nucleus after the collision.
a
Let’s adopt the following notations :
• for the neutron, mass m, vi and vf are the initial and final velocity
respectively.
• for the atom , mass M , Vi and Vf are the initial and final velocity
respectively.
The fraction of the neutron’s kinetic energy that’s transferred to the carbon
nucleus is just :
1
2
2 M VF
1
2
2 mvi
Conservation of momentum on the head-on collision gives:
mvi
vi − v f
= mvf + M Vf
M
=
Vf
m
= 12Vf
13
Conservation of kinetic energy gives :
1
mv 2
2 i
vi2 − vf2
and
vi2 − vf2
= vi + v f
vi − vf
1
1
mv 2 + M Vf2
2 f 2
M 2
=
V
m f
= 12Vf2
=
12Vf2
12Vf
= Vf
=
We get the set of equations :
vi − vf = 12Vf
and
vi + vf = Vf
to finally obtain
vi =
13
Vf
2
and the fraction transfered to the atom:
1
2
2 M VF
1
2
2 mvi
=
12 × (
=
0.284
2 2
)
13
b)
The kinetic energy of the carbon nucleus is:
1
M VF2
2
=
0.284Kiinitial
=
0.284 × 1.10 × 10−13
=
3.12 × 10−14 J
The final kinetic energy of the neutron is :
Kif inal
14
=
1.10 × 10−13 − 3.12 × 10−14
=
7.88 × 10−14 J
0.11.
0.11
Two automobiles of equal mass approach an intersection. One vehicle
is traveling with velocity 13.0 m/s toward the east and the other is
traveling north with speed v2 . Neither driver sees the other. The
vehicles collide in the intersection and stick together, leaving parallel
skid marks at an angle of 61.5◦ north of east. Determine the initial
speed v2i of the northward-moving vehicle.
The speed limit for both roads is 35 mi/h and the driver of the
northward-moving vehicle claims he was within the speed limit when
the collision occurred. Is he telling the truth?
15
~vf
~v1i
~v2i
Momentum conservation :
m~v1i + m~v2i
=
2m~vf
~v1i + ~v2i
=
2~vf
We can project the last equation on the West-East Direction to get:
v1i = 2vf cos 61.5
and on the South-North direction to get :
v2i = 2vf sin 61.5
v2i
v1i
16
=
tan 61.55
0.12.
v2i
=
v1i tan 61.55
=
13 tan 61.5
=
23.9m/s
=
53.5mile/hour
Second driver is definitely lying about his claim.
0.12
An object of mass 2.99 kg, moving with an initial velocity of 5.01~i,
collides with and sticks to an object of mass 2.31 kg with an initial
velocity of -2.96~j m/s. Find the final velocity of the composite object.
Momentum conservation of the two-object system will give us the final velocity
after collision:
~f
(m1 + m2 )V
~f
V
=
m1~v1i + m2~v2i
1
=
(m1~v1i + m2~v2i )
m1 + m2
1
=
(2.99 × 5.01~i − 2.31 × 2.96~j)
2.99 + 2.31
= 2.83~i − 1.29~j m/s
0.13
A billiard ball moving at 5.20 m/s strikes a stationary ball of the
same mass. After the collision, the first ball moves at 4.67 m/s at an
angle of 26.0◦ with respect to the original line of motion. Assuming
an elastic collision (and ignoring friction and rotational motion), find
the struck ball’s velocity after the collision.
17
~v1i
at rest
~v1f
α
β
~v2f
Momentum is again conserved :
m~v1i
=
~v1i
m~v1f + m~v2f
= ~v1f + ~v2f
Projections on the x and y-axis:
18
v1i
=
v1f cos α + v2f x
v2f x
=
v1i − v1f cos α
=
5.20 − 4.67 cos 26
=
1.00263
0.13.
and
0
=
v1f sin α + v2f y
v2f y
=
−v1f sin α
=
−4.67 sin 26
=
−2.04719
Speed of the struck ball is:
v2f
=
p
1.0032 + 200472
=
2.97m/s
and making an angle β of:
β
=
arctan
−2.047
1.003
= −63.9◦
N.B: α + |β| = 90◦ . This can be derived easily by looking at conservation
of monentum and kinetic energy:
~v1i
= ~v1f + ~v2f
2
v1i
=
2
2
v1f
+ v2f
+ 2~v1f · ~v2f
and
1
mv 2
2 1i
2
v1i
1
1
mv 2 + mv 2
2 1f 2 2f
2
2
= v1f
+ v2f
=
This implies ~v1f · ~v2f = 0 or α + |β| = 90◦ .
19
0.14
An unstable atomic nucleus of mass 1.80 × 10−26 kg initially at rest
disintegrates into three particles. one of the particles of mass 5.16 ×
10−27 kg, moves in the y direction with a speed of 6.00 × 106 m/s.
Another particle of mass 8.46 × 10−27 kg, moves in the x direction
with a speed of 4.00 × 106 m/s.
• Find the velocity of the third particle
• Find the total kinetic energy increase in the process.
20
0.14.
~v1
~v2
at rest
?
~v3
Momentum is conserved :
~0
~v3
= m1~v1 + m2~v2 + m3~v3
1
= −
(m1~v1 + m2~v2 )
m3
1
= −
(0.516 × 6 × 106~j + 0.846 × 4 × 106~i) × 10−26
0.438 × 10−26
= −7.73 × 106~i − 7.07 × 106~j
its maginitude is v3 =
√
7.732 + 7.072 = 10.47 × 106 m/s
b)
The kinetic energy increase of the system is then computed as:
∆Ki =
1
1
1
m1 v12 + m2 v22 + m3 v32
2
2
2
21
=
=
22
0.516 × 62 + 0.846 × 42 + 0.438 × 10.472
× 101 2 × 10−26
2
4.01 × 10−13 J
```