# PHY302K Sai HW6 solutions ```Oslund (hno56) – Homework 6 – sai – (58015)
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001 (part 1 of 2) 10.0 points
Two blocks, A and B, are connected by a
compressed spring. The mass of block A is
twice that of block B. Immediately after
release, compare the net force on block A
to the net force on block B. The table is
frictionless and the mass of the spring is zero.
1. FA,net and FA,net are both directed toward
the left with FA,net larger than FA,net , but not
twice as large. Block A is pulling block B.
1
Note: This is correct only when the mass
of the spring can be neglected as mentioned
in the problem.
002 (part 2 of 2) 10.0 points
Which of the following best describes the situation after the blocks are released?
1. The total momentum of the system of
blocks is conserved because there is no net
external force. correct
2. The total momentum of the system of
blocks is not conserved because there is an
external velocity acting on the system.
2. FA,net is equal in magnitude and opposite
in direction to FB,net , and not zero. correct
3. The total momentum of the system is not
conserved because there are external forces;
e.g., gravity.
3. FA,net will be in the opposite direction
and half as large as FB,net .
4. Momentum for any body is always conserved.
4. There is no net force on block A or block
B.
5. The momentum of block A is conserved
and the momentum of block B is conserved.
However, the total momentum of the system
of blocks is not conserved.
5. FA,net is equal in magnitude and direction
to FB,net , and not zero.
6. FB,net will be in the opposite direction
and half as large as FA,net .
Explanation:
FA,net is the force on block A from the
spring, it’s reaction force is the force acting
on the spring by block A, which we denote
as FS,A . They have the same magnitude but
they are in opposite directions. Similarly,
FB,net has a reaction force FS,B . The total
force on the spring
! S,net = F
! S,A + F
! S,B
F
! A,net − F
! B,net
= −F
&quot;
!
! A,net + F
! B,net
=− F
= mspring &middot; a
=0 &middot; a
= 0.
!
! B,net must have the
So that FA,net and F
same magnitude but in opposite directions.
6. The momentum of block A is conserved
and the momentum of block B is conserved.
It follows that the total momentum of the
system of blocks is also conserved.
Explanation:
Neither block A nor B conserves its momentum since there is a force on each of them.
However, once we consider the two blocks and
the spring as a whole system, its momentum
is conserved since there is no external force.
The contribution of momentum from the
spring is zero since its mass is zero.
003 10.0 points
A 62.8 kg astronaut is on a space walk when
the tether line to the shuttle breaks. The
astronaut is able to throw a 11.0 kg oxygen
tank in a direction away from the shuttle with
a speed of 13.7 m/s, propelling the astronaut
back to the shuttle.
Oslund (hno56) – Homework 6 – sai – (58015)
Assuming that the astronaut starts from
rest, find the final speed of the astronaut after
throwing the tank.
Explanation:
Let toward the spacecraft be positive.
Let : m1 = 62.8 kg ,
m2 = 11.0 kg , and
v2,f = −13.7 m/s .
The wrench has the same initial speed as the
astronaut, and vi = 0 m/s, so
(m1 + m2 ) !vi = m1 !vf,1 + m2 !vf,2
0 = m1 !vf,1 + m2 !vf,2
m2 vf,2
m1
(11 kg) (−13.7 m/s)
=−
62.8 kg
vf,1 = −
= 2.39968 m/s .
004 10.0 points
A spring is compressed between two cars on a
frictionless airtrack. Car A has four times the
mass of car B, MA = 4 MB , while the spring’s
mass is negligible. Both cars are initially at
rest. When the spring is released, it pushes
them away from each other.
Which of the following statements correctly
describes the velocities, the momenta, and the
kinetic energies of the two cars after the spring
is released? Note: Velocities and momenta are
given below as vectors.
1
1. !vA = − !vB
3
2
!pA = − !pB
3
4
KA = KB
3
2. !vA = −4 !vB
!pA = −16!pB
KA = 64 KB
1
3. !vA = − !vB
4
2
!pA = −!pB
1
KA = KB correct
4
4. !vA = −!vB
!pA = −!pB
KA = KB
1
5. !vA = + !vB
5
4
!pA = + !pB
5
4
KB
KA =
25
1
6. !vA = + !vB
4
!pA = +!pB
KA = 4 KB
1
7. !vA = − !vB
4
!pA = −!pB
KA = 4 KB
8. !vA = −!vB
!pA = −4!pB
KA = 16 KB
9. !vA = −2 !vB
!pA = −8!pB
KA = 16 KB
1
10. !vA = − !vB
2
!pA = −2!pB
KA = KB
Explanation:
Let : MA = 4 MB .
There are no external forces acting on the
cars, so their net momentum !pA + !pB is conserved. The initial net momentum is obviously zero, hence after the spring is released,
!pA + !pB = 0; i.e., !pA = −!pB .
The velocities and the kinetic energies follow from the momenta:
!v =
!p
.
M
Oslund (hno56) – Homework 6 – sai – (58015)
So given MA = 4 MB , it follows that
1
!vA = − !vB .
4
Likewise,
K=
!p2
1
M !v 2 =
.
2
2M
Since !pA = −!pB ,
KA =
3
Explanation:
Solution: Denote the smaller mass initially moving to the right as m and the larger
mass initially moving to the left as 4m. By
assumption of the problem, before the collision
!vm = +v ı̂
and !v4m = −v ı̂
where ı̂ is the unit vector in the positive x direction. Hence, the particle’s initial momenta
are respectively
1
KB .
4
!pm = +m v ı̂
005 (part 1 of 2) 10.0 points
Two particles of masses m and 4 m are
moving toward each other along the x-axis
with the same speed v. They undergo a headon elastic collision and rebound along the xaxis.
v
v
and !p4m = −4 m v ı̂,
and the total momentum of the two particles
is
!ptotal = !pm + !p4m = −3 m v ı̂.
In magnitude,
&quot;!ptotal &quot; = 3 m v .
m
4m
Determine the magnitude of the total momentum of the system &quot;!ptotal &quot; at the instant
when the two particles are touching each
other; i.e., at the moment of collision.
006 (part 2 of 2) 10.0 points
Determine the final speed v4m of the heavier
object.
1. v4m =
2
v
3
1. &quot;!ptotal &quot; = 4 m v
2. v4m = 2 v
2. &quot;!ptotal &quot; = 5 m v
3. v4m = v
3. &quot;!ptotal &quot; = ∞
4. v4m =
4. &quot;!ptotal &quot; = 0
5. v4m = 4 v
5. &quot;!ptotal &quot; = 2 m v
6. v4m =
6. &quot;!ptotal &quot; = 3 m v correct
7. v4m = 3 v
7. &quot;!ptotal &quot; = 8 m v
8. v4m = ∞
8. &quot;!ptotal &quot; = m v
9. &quot;!ptotal &quot; = 6 m v
10. &quot;!ptotal &quot; = 7 m v
9. v4m =
1
v
3
3
v
2
1
v correct
5
10. v4m = 0
Explanation:
Oslund (hno56) – Homework 6 – sai – (58015)
From the relative velocities formula,
!
!
vm − v4m = v4m
− vm
.
(1)
From conservation of momentum,
!
!
m vm + 4 m v4m = m vm
+ 4 m v4m
.
(2)
Combining equations (1) and (2) gives
m vm + 4 m v4m
!
!
= m (v4m
+ v4m − vm ) + 4 m v4m
, or
m vm + 4 m v4m − m (v4m − vm )
!
!
= m v4m
+ 4 m v4m
, so
!
v4m
m vm + 4 m v4m − m (v4m − vm )
=
m +4m
m v + 4 m (−v) − m (−v − v))
=
m+ 4m
1
= − v .
5
Alternative Solution: The compact form
of the head-on elastic collision formula, can
be derived from coordinate transformations
between the lab frame and the center-of-mass
frame, since
What is the kinetic energy of the second
ball after the collision?
Explanation:
Basic Concepts
For an elastic, head-on collision, both the
linear momentum and the kinetic energy are
conserved.
The balls have the same mass m and v2 = 0,
so by conservation of momentum,
m1 v1 + m2 v2 = m1 v1! + m2 v2!
m v1 = m v1! + m v2!
v1 = v1! + v2!
v1! = v1 − v2! .
By conservation of energy,
1
1
1
1
m1 v12 + m2 v22 = m1 v1!2 + m2 v2!2
2
2
2
2
1
1
1
m v12 = m v1!2 + m v2!2
2
2
2
v12 = v1!2 + v2!2
v12 = (v1 − v2! )2 + v2!2
!
v4m
= 2 vcm − v4m
v12 = v12 − 2 v1 v2! + v2!2 + v2!2
0 = −2v1 v !2 + 2 v2!2
Part one implies that
vcm
ptotal
=
m+ 4m
3mv
=−
5m
3
= − v.
5
This gives
4
0 = −v1 + v2!
v2! = v1
and
v1! = v1 − v2! = 0 .
Thus
1
m v2!2
2
1
= m v12
2 \$
#
1
=
(0.842 kg) (0.684 m/s)2
2
= 0.196967 J .
K2! =
!
v4m
= 2 vcm − v4m
\$
#
3
= 2 − v − (−v)
5
1
= − v .
5
007 (part 1 of 2) 10.0 points
A billiard ball of mass 0.842 kg and speed
0.684 m/s makes a head-on elastic collision
with an identical billiard ball at rest.
008 (part 2 of 2) 10.0 points
What is the magnitude of its linear momentum?
Oslund (hno56) – Homework 6 – sai – (58015)
Explanation:
Since
5
Solution:
p!2 = m v2! = m v1
= (0.842 kg) (0.684 m/s)
= 0.575928 kg m/s .
009 (part 1 of 2) 10.0 points
A 42.1 kg student runs down the sidewalk and
jumps with a horizontal speed of 4.03 m/s
onto a stationary skateboard. The student
and skateboard move down the sidewalk with
a speed of 3.81 m/s.
a) Find the mass of the skateboard.
Explanation:
Basic Concept:
mb!vi,b + ms!vi,s = (mb + ms )!vf
mb!vi,b = (mb + ms )!vf
since vi,s = 0 m/s.
Given: Let to the right be positive:
(mb + ms )vf
mb
(42.1 kg + 2.43097 kg)(6.52 m/s)
=
42.1 kg
= 6.89648 m/s
vi,b =
011 10.0 points
Use 5.98 &times; 1024 kg as the mass of the Earth.
A 1080 kg meteorite has a speed of 147 m/s
just before colliding head-on with the Earth.
Determine the recoil speed of the Earth.
Correct answer: 2.65485 &times; 10−20 m/s.
Explanation:
From conservation of momentum ∆p = 0
m v + (m + Me ) V = 0 .
Therefore
mv
[M + m]
(1080 kg) (147 m/s)
=
[(5.98 &times; 1024 kg) + (1080 kg)]
= 2.65485 &times; 10−20 m/s .
!&quot;=
&quot;V
mb = 42.1 kg
vi,b = 4.03 m/s
vf = 3.81 m/s
Solution:
mb vi,b = mb vf + ms vf
mb (vi,b − vf )
ms =
vf
(42.1 kg)(4.03 m/s − 3.81 m/s)
=
3.81 m/s
= 2.43097 kg
012 10.0 points
A(n) 5 g bullet is fired into a 255 g block that
is initially at rest at the edge of a frictionless
table of height 1.4 m. The bullet remains in
the block, and after impact the block lands
2.5 m from the bottom of the table.
The acceleration of gravity is 9.8 m/s2 .
1.4 m
010 (part 2 of 2) 10.0 points
b) How fast would the student have to jump
to have a final speed of 6.52 m/s?
2.5 m
Find the initial speed of the bullet.
Explanation:
Given:
vf = 6.52 m/s
Explanation:
Oslund (hno56) – Homework 6 – sai – (58015)
Given : m1 = 5 g ,
m2 = 255 g ,
∆y = −1.4 m ,
∆x = 2.5 m .
Explanation:
Given : m1
v1
m2
v2
m3
v3
and
The bullet-block combination is a projectile,
with the vertical motion defining the time:
1
∆y = − g (∆t)2
%2
−2 ∆y
∆t =
g
m1 v1 + m2 v2 + m3 v3 = (m1 + m2 + m3 ) v !
Since
m! = m1 + m2 + m3
= 6 kg + 12 kg + 7 kg
= 25 kg ,
!
= ∆x
g
−2 ∆y
Applying conservation of momentum to the
collision,
m1 v1 = (m1 + m2 ) v !
(m1 + m2 ) v !
v1 =
m1
&amp;
(m1 + m2 ) ∆x
g
=
m1
−2∆y
%
9.8 m/s2
(5 g + 255 g) (2.5 m)
=
5g
−2 (−1.4 m)
= 243.208 m/s .
013 10.0 points
Three carts of masses 6 kg, 12 kg, and 7 kg
move on a frictionless horizontal track with
speeds of 8 m/s, 2 m/s, and 2 m/s, as shown in
figure. The carts stick together after colliding.
8 m/s
6 kg
= 6 kg ,
= 8 m/s ,
= 12 kg ,
= 2 m/s ,
= 7 kg , and
= −2 m/s .
From conservation of momentum
Horizontally, the constant velocity is v ! from
just after the collision, so
∆x = v ∆t
∆x
v! =
∆t &amp;
6
2 m/s
2 m/s
12 kg
7 kg
we find that
m1 v1 + m2 v2 + m3 v3
v! =
m1 + m2 + m3
(6 kg) (8 m/s) (12 kg) (2 m/s)
=
+
25 kg
25 kg
(7 kg) (−2 m/s)
+
25 kg
= 2.32 m/s .
toward the right.
014 (part 1 of 3) 10.0 points
Two small spheres of mass 417 g and 601 g are
suspended from the ceiling at the same point
by massless strings of equal length 10.7 m .
The lighter sphere is pulled aside through an
angle of 48◦ from the vertical and let go.
The acceleration of gravity is 9.8 m/s2 .
417 g
.
10
7
m
48◦
Find the final velocity of the three carts.
9.8 m/s2
Before
601 g
Oslund (hno56) – Homework 6 – sai – (58015)
7
After lighter sphere is let go and collides with
the heavier sphere at the bottom of its swing,
the two spheres immediately bind together.
What is the speed of the combined system
just after the collision?
θf
After
9.8 m/s2
At what speed will the lighter mass m1 hit
the heavier mass m2 ?
Explanation:
Let :
m1 = 417 g ,
m2 = 601 g ,
θi = 48◦ ,
# = 10.7 m ,
vi = before collision ,
Vf = after collision , and
g = 9.8 m/s2 .
The velocity just before the collision vi can
be determined by energy conservation. When
particle 1 is at its initial condition, it is at rest
and displaced by an angle θi from the vertical.
The total energy is all potential and is given
by
Ui = m1 g # (1 − cos θi )
where # (1 − cos θi ) is the distance above the
lowest point. Just before the collision, the en1
2
ergy of sphere 1 is all kinetic energy, m1 v1i
.
2
Equating the two energies gives
1
2
m1 v1i
= m1 g # (1 − cos θi ) .
2
Solving for V1i gives
v1i = {2 g # [1 − cos θi ]}1/2
'
= 2 (9.8 m/s2 ) (10.7 m)
&times; [1 − cos(48◦ )]}1/2
= 8.33006 m/s .
015 (part 2 of 3) 10.0 points
Explanation:
This is a perfectly inelastic collision. The
speed of the two spheres after collision is determined by momentum conservation
m1 v1i = (m1 + m2 ) Vf
(1)
where m1 is the mass and v1i is the initial
velocity of sphere 1 just before the collision,
m2 is the mass of sphere 2, and Vf is the
velocity of the combined spheres just after
the collision. Note: v2i = 0.
Now, calculating Vf of the composite system from Eq. (1),
m1
Vf =
v1i
(2)
m1 + m2
(417 g)
(8.33006 m/s)
=
(417 g) + (601 g)
= 3.41222 m/s .
016 (part 3 of 3) 10.0 points
What is the maximum angle of deflection of
the two bound objects?
Explanation:
The composite system has an initial energy
1
(m1 + m2 ) Vf2
2
(
)2
m1
1
2
v1i
= (m1 + m2 )
2
m1 + m2
m21
[2 g # (1 − cos θi )] .
=
2 (m1 + m2 )
The composite system comes to rest at an
angular displacement θf , where the potential
energy equals the initial kinetic energy of the
composite system,
(m1 +m2 ) g # (1 − cos θf )
1
= (m1 + m2 ) Vf2
2
m21
[2 g # (1 − cos θi )] .
=
2(m1 + m2 )
Oslund (hno56) – Homework 6 – sai – (58015)
Therefore
cos θf = 1 −
#
m1
m1 + m2
\$2
Solving for θf we have
*
#
θf = cos−1 1 −
m1
m1 + m2
[1 − cos θi ]
\$2
[(− cos θi )
+
(
1 − (0.409627)2
)
◦
&times; (1 − cos 48 )
= cos
−1
= 19.1816◦ .
Alternative Solution: From conservation of energy after the collision, we have
8
1
1
3. v1! = v1 , v2! = v1
2
2
1
3
4. v1! = − v1 , v2! = v1
2
2
1
1
5. v1! = − v1 , v2! = v1
2
2
6. v1! = 0, v2! = v1 correct
7. v1! = v1 , v2! = −v1
8. 0 &lt; v2! &lt; v1! , v1! &lt; 2v2!
9. v1! = −2v1 , v2! = 0
10. v1! = 2 v1 , v2! = −v1
1
(m1 + m2 ) g h = (m1 + m2 ) Vf2 ,
2
Explanation:
Basic Concepts: Conservation of Energy,
where v2 = 0:
where Vf has been determined in Eq. (2).
Therefore
1
1
1
2
2
m1 v1 2 = m1 v1! + m2 v2!
2
2
2
2
1 Vf
h=
2 g
1 (3.41222 m/s)2
=
2 (9.8 m/s2 )
= 0.594042 m .
#−h
Since cos θ =
, we have
#
\$
#
−1 # − h
θ = cos
#
#
\$
−1 (10.7 m) − (0.594042 m)
= cos
(10.7 m)
◦
= 19.1816 .
017 10.0 points
You are playing pool. You hit the cue ball
with velocity v1 dead center on another ball.
All balls including the cue ball have the same
mass.
What is the final velocity of the cue ball v1!
and the final velocity of the other ball v2! ?
1. v1! = v1 , v2! = 0
2. v1! = 0, v2! = 0
Conservation of Momentum, where v2 = 0:
m1 v1 = m1 v1! + m2 v2!
Solving for v1! and v2! yields:
#
\$
m1 − m2
!
v1 =
v1
m1 + m2
#
\$
2 m1
!
v2 =
v1
m1 + m2
(1)
(2)
All solutions can be determined using the
above equations (1) and (2). Part 1: Using eqn (1), where m2 = m1 = m,
#
\$
m−m
!
v1 =
v1 = 0
m+m
Part 2: Using eqn (2), where m2 = m1 = m,
\$
#
2m
!
v1 = +v1
v2 =
m+m
018 (part 1 of 2) 10.0 points
A 1790 kg car skidding due north on a level
frictionless icy road at 212.52 km/h collides
with a 3168.3 kg car skidding due east at
Oslund (hno56) – Homework 6 – sai – (58015)
# 3 \$#
\$
138 km/h in such a way that the two cars
10 m
1h
&times;
stick together.
km
3600 s
= 1.21451 &times; 105 kg m/s , and
θ
pf y = m1 v1
vf
138 km/h
= (1790 kg) (212.52 km/h)
# 3 \$#
\$
N
10 m
1h
3168.3 kg
&times;
km
3600 s
5
212.52 km/h 1790 kg
= 1.0567 &times; 10 kg m/s .
At what angle (−180◦ ≤ θ ≤ +180◦ ) East
of North do the two coupled cars skid off at?
Explanation:
Let : m1
v1
m2
v2
= 1790 kg ,
= 212.52 km/h ,
= 3168.3 kg , and
= 138 km/h .
θ
pf = mf vf
p2
N
m2 v2
p1
m1 v1
Basic Concepts: Momentum Conserva1
tion,
K = mv 2 ,
!p = m!v .
2
During the collision, the total momentum
of the two car system will be conserved
!pf = !p1 + !p2
= m1 !v1 + m2 !v2
pf x ı̂ + pf y ̂ = m1 v1 ̂ + m2 v2 ı̂ .
Looking at the x and y components of momentum,
pf x = m2 v2
= (3168.3 kg) (138 km/h)
9
Since we are asked to find the angle from the
\$
pf x
tan θ =
pf y
\$
#
1.21451 &times; 105 kg m/s
=
1.0567 &times; 105 kg m/s
= 1.14935#
\$
1.21451 &times; 105 kg m/s
θ = arctan
1.0567 &times; 105 kg m/s
= 48.9749◦ .
#
019 (part 2 of 2) 10.0 points
How much kinetic energy is lost in the collision?
Correct answer: 2.83339 &times; 106 J.
Explanation:
To find the energy lost in the collision, we
need to know the total kinetic energy of the
system before and after the collision. (There
is no change in potential energy since the cars
remain on level ground.) The initial and final
energies are given by
Ei = Ki =
1
1
m1 v12 + m2 v22
2
2
1
(1790 kg) (59.0333 m/s)2
2
1
+ (3168.3 kg) (38.3333 m/s)2
2
= 5.44684 &times; 106 J , and
Ef = Kf
1
= (m1 + m2 ) vf2 .
2
=
Oslund (hno56) – Homework 6 – sai – (58015)
To find vf , we need to go back to the momentum equation;
!pf
,
m1 + m2
or in component form,
pf x
vf x =
,
m1 + m2
pf y
vf y =
.
m1 + m2
Explanation:
Let : M = 7 kg ,
v = 9.63 m/s ,
θ = 33◦ .
!vf =
and
and
The y component of the momentum is unchanged. The x component of the momentum
is changed by
Now that we know the components of !vf , we
can find its magnitude
∆Px = −2 M v cos θ .
Therefore, using impulse formula,
vf2 = vf2x + vf2y .
We can substitute this into the equation for
the final kinetic energy to get
2
10
2
1 pf x + pf y
2 m1 + m2
= 2.61345 &times; 106 J .
Ef =
∆P
∆t
−2 M v cos θ
=
∆t
−2 (7 kg) (9.63 m/s) cos 33◦
=
0.318 s
! &quot; = 355.565 N .
&quot;F
F =
The energy lost is then
Note: The direction of the force is in negative
x direction, as indicated by the minus sign.
Elost = −∆E
= Ei − Ef
= (5.44684 &times; 106 J) − (2.61345 &times; 106 J)
= 2.83339 &times; 106 J .
020 10.0 points
A 7 kg steel ball strikes a wall with a speed
of 9.63 m/s at an angle of 33◦ with the normal
to the wall. It bounces off with the same
speed and angle, as shown in the figure.
9. 6
3m
y
/s
◦
7 kg
33
x
33◦
7 kg
9. 6
/s
3m
If the ball is in contact with the wall for
0.318 s, what is the magnitude of the average
force exerted on the ball by the wall?
021 10.0 points
If you throw a raw egg against a wall, you’ll
break it, but if you throw it with the same
speed into a sagging sheet it won’t break.
Why?
1. The sheet is much slicker than the wall.
2. The breaking egg causes a larger impact
time, decreasing the force.
3. The impact time when the egg strikes a
sagging sheet is long, so the impact force is
small. correct
4. The velocity of the egg decreases faster in
the sheet than on the wall.
Explanation:
Although the impulses may be the same for
the two cases, the times of impact are not.
When the egg strikes the wall, impact time
is short and impact force is correspondingly
large. But when the egg strikes the sagging
Oslund (hno56) – Homework 6 – sai – (58015)
sheet, impact time is long and the force correspondingly small.
022 10.0 points
A(n) 4.3 lb hammer head, traveling at 5 ft/s
strikes a nail and is brought to a stop in
0.00079 s.
The acceleration of gravity is 32 ft/s2 .
What force did the nail receive?
Applying impulse,
! ∆t = m!vf − m!vi = m!vf
F
since vi = 0 m/s.
m vf
∆t
(0.33 kg) (12 m/s)
=
0.21 s
= 18.8571 N .
F =
Explanation:
Let : W = 4.3 lb ,
vo = 5 ft/s , and
t = 0.00079 s .
Impulse is the change in momentum:
F ∆t = m ∆v = m (vf − vo ) .
11
024 (part 1 of 3) 10.0 points
The force shown in the force-time diagram
acts on a 622 g object.
Fx (N)
The hammer’s final velocity is 0, so
Ft=
W
(0 − vo )
g
Thus the force the hammer felt was
−W vo
F =
gt
(it had to be a negative force to stop it) and
by Newton’s Third Law of Motion, the force
F =
W vo
(4.3 lb) (5 ft/s)
=
gt
(32 ft/s2 ) (0.00079 s)
= 850.475 lb .
023 10.0 points
A football punter accelerates a 0.33 kg football from rest to a speed of 12 m/s in 0.21 s.
What constant force does the punter exert
on the ball?
Explanation:
Given : m = 0.33 kg ,
vf = 12 m/s, and
∆t = 0.21 s .
2
1
t (s)
0
0
1
2
3
4
5
Find the impulse of the force.
Explanation:
The impulse is the area under the F versus t
graph; i.e, the sum of the area of the rectangle
plus the triangle, so
Impulse = (1 N) (1 s) +
1
(1 N) (2 s)
2
= 2 Ns.
025 (part 2 of 3) 10.0 points
Find the final velocity of the object if it is
initially at rest.
Explanation:
Oslund (hno56) – Homework 6 – sai – (58015)
Given : m = 622 g = 0.622 kg ,
vi = 0 m/s .
and
Impulse = ∆p
F ∆t = m (vf − vi )
m vf = F ∆t + m vi
F ∆t
vf =
+ vi
m
(2 N s)
+ (0 m/s)
=
(622 g)
= 3.21543 m/s .
026 (part 3 of 3) 10.0 points
Find the final velocity of the object if it is
initially moving along the x axis with average
velocity of −15 m/s.