07prob_11

07prob_11 - Assigment#7 Page 1 of 4 UC Davis Physics 9A-C...

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Unformatted text preview: Assigment #7, 5/13/11 Page 1 of 4 UC Davis Physics 9A-C Assignment #7 Cole [8.7]


 Δ p Δ t = 0.045 kg ( ) 25 m ( ) 2 × 10 − 3 s 
=

563
N.
The
weight
of
the
ball
is
less
than
half
a
Newton,
so
the
 weight
is
not
significant
while
the
ball
and
club
are
in
contact. [8.20]

a)
This
process
happens
in
the
center
of
mass
frame
where

Í p
=
 0,
both
before
and
after
 the
collision.

 m A ′ v A + m B ′ v B = .

õ

 
 ′ v A = − m B ′ v B m A = − 3 
kg 1 
kg 1.2 
m/s ( ) 
=

-3.6
m/s 
b)

This
is
an
inelastic
"explosion",
but
it
uses
potential
energy
stored
in
the
spring
so
we
can
 conserve
total
energy,
 Eø
=
E f .

The
potential
energy
of
the
spring
appears
as
kinetic
energy
of
 the
masses:

 U
=
K A 

+
K B 

=

 (1/2)(1.00
kg)(3.60
m/s) 2 
+
(1/2)(3.00
kg)(1.200
m/s) 2 
=
8.64
J . [8.27]

As
the
rain
hits
the
car,
the
water
experiences
an
impulse
in
the
direction
of
travel
of
the
 car
which
speeds
the
water
up
so
it
travels
horizontally
with
the
car,
and
the
car
experiences
and
 equal
but
opposite
impulse
which
slows
the
car
down
.

Conserving
momentum Mvø

=

(m
+
M)
v

õ v = 2.4 × 10 4 kg 4 m/s ( ) 2.7 × 10 4 kg 
=
 3.56
m/s. 
 [8.35]

a)

The
small
car
places
a
force
 F 
on
the
large
car,
and
by
Newton's
3rd
law,
the
large
car
 must
place
an
equal
but
opposite
force
on
the
small
car.

 F = Δ p t 
õ

They
experience
equal
but
 opposite
momentum
changes.

õ

 m s Δ v s = m L Δ v L 

õ
 Δ v s = m L m s Δ v L = 3000 1200 Δ v L = 2.5 Δ v L b)

The
acceleration
is
proportional
to
the
velocity
change;
hence,
the
small
car
has
2.5
times
the
 acceleration
of
the
large
car.

The
forces
experienced
by
the
small
car
passengers
are
2.5
times
 that
of
the
large
car.

This
difference
can
be
mitigated
by
building
crumple
zones
into
cars
so
the
 any
passengers
wearing
seat
belts
stop
more
slowly.

However,
note
that
massive
cars
have
a
big
 safety
advantage
just
by
virtue
of
their
large
mass. [8.37]

The
initial
momentum
of
the
car
must
be
the
 x-component
of
the
final
momentum
 as
the
truck
had
no
initial 
x-component
of
momentum,
so
 
Í p x 

=
m car v car 

=

(m car 
+
m truck )v 
sin
24° v car = m car + m truck ( ) m car v sin24 ° = 2850 950 16 m/s ( ) sin24 ° 
=
19.5
m/s.
 For
the
 y- direction:

Í p y...
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This note was uploaded on 09/28/2011 for the course PHY 9A taught by Professor Svoboda during the Spring '08 term at UC Davis.

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07prob_11 - Assigment#7 Page 1 of 4 UC Davis Physics 9A-C...

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