07quiz_11

07quiz_11 - Physics 9A-C 5/13/2011 Quiz week #7 Name: Last...

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Unformatted text preview: Physics 9A-C 5/13/2011 Quiz week #7 Name: Last 4 ID: A
rifle
bullet
of
mass
5.0
g
strikes
and
embeds
itself
in
a
block
of
mass
0.995
kg
that
rests
on
a
 frictionless
horizontal
surface
and
is
attached
to
an
ideal
massless
spring.

The
impact
 compresses
the
spring
10.0
cm.

Calibration
of
the
spring
shows
that
a
force
of
0.80
N
 compresses
the
spring
0.20
cm.

a)
(2
pts)

What
is
the
spring
force
constant
k? Solution:

 vø k

is
found
from
the
calibration
information:

 F = kx ′ 

õ
 
k
=
0.8
N/0.002
m
=
400
N/m 

 b)
(5
pts)
What
is
the
initial
speed
of
the
bullet? There
are
two
time
scales
here.

Because
of
its
extreme
speed,
the
bullet
slams
into
the
block
and
 is
stopped

very
quickly.


But,

because
of
its
large
inertia
(initially
it
is
at
rest),
the
block
 responds
by
compressing
the
spring
on
a
much
longer
time
scale. 
The
bullet
embeds
itself
in
the
block
which
is
a
completely
inelastic
collision.

Though
energy
is
 not
conserved,
momentum
is: mvø

=
(m+M)v'

õ

 v = m+M v' m The
spring
then
compresses,
and
because
the
spring
exerts
a
conservative
force
on
the
block,
we
 can
conserve
energy. 1 2 ( m + M ) v′ 2 = 1 kx 2 

õ

 v′ = 2 õ

 v = 
 k 400 
N/m x 
=
 ( 0.1
m ) 
=
2
m/s m+M 1
kg 
 m+M 1
kg v' = ( 2 
m/s ) =

400
m/s m 0.005
kg c)
(3
pts)
If
the
bullet
takes
2.0
ms
to
embed
itself
in
the
block,
what
average
force
does
the
 bullet
exert
on
the
block? Fblock = 
 impulse − Δpbullet − m ( v′ − v ) − ( 0.005 
kg ) ( 2 
m/s-400
m/s ) = = = 
=
+
995
N t t t 0.002
s Total
points
=
10
pts. ...
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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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