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1/13/20
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AP Physics 1: Homework
Unit 4A.2: Impulse - Safety & Bouncing
Reasoning Tasks
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Physics
AP
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40mg Ys
5mg (-3%-54)=-40 Mgmrs
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of
each
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ended
Area of first
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Area of
25
that
Sene rio
since
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:
35
"
given
was
I
following
defined
be
each
getting -40mg Vs
up
as
mass
time
A- MN,
and
These
-
calculations
graph
Time
.
the
al
momentum
velocities
.
3 seconds
time
we
that
was
asked
are
have been
the
incorrect
.
In
order
calculate
must
have
S g.
are
only
things
Should
,
-
-
to
seemed
.
-
.
-
-
6N
calculated
final limit
the
4/1/41
:
A- Neocons
two
I
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)=2④
first
can
momentum
multiplied
very
2 N 's
student
the
the
to
find
area
force
by
impulse
the
under
the
for and did not find area
different Impulse during
.
at
even
the
impulse
got
the
.
that
The
vs
.
student
exact
though these
first 2. seconds
3
during the fong
second one right actually
white
Force
a
graph
the
time
on
seconds
,
is
No hemi
Unit
4A 2
:
.
Impulse safety & Bouncing
Concept Questions
Impulse
g)
This
not
is
the
change
larger force
that of
is
-
the
smaller
F
to
x
St
for
time
change
in
could
cause
the
Impulse
was
smaller
gained
force An
and
.
to
this
of change
boulder
the
.
Mass and
proportional
Problem
to
one
Meath od
under the curve
b) Impulse
Fet
-
-
or
F
83.37g
=
is
inversely
of
finding
.
a
Nmg=
62)
5N 's
about
box
momentum
lighter
-
=
mmtygmsa
)=@6hgFor6N5
-
8-6=-2
will push
it
graph
o.o 's
o.SN
before
momentum
s
.
.
=
or
f-800mg )(8
=
.
I Impulse
ChangeinM0mentUm:-2lhgmT
8.16ms
1 Newton
is
to the
-
Kang) ( 2.0 's) lounges
-
-
10=230
added
to
caused
a
force
while
left
.
in
momentum
the force
c) since
the
left
much
would
than
the
be
neg q magnitude
.
of
is
direction
Mgk
isonomy
direction
F
magnitude
Fet
the momentum ,
Eos
=
-
greater impulse
66÷;F
IN
-
change
AV
60%)
p
240
after
(5800mg) ( 8.60%1=1611549)
s
5 Nees
c)
b)
-
GON)l2s)=240
-
5mg
momentum
Impulse
Eggman N
after
5800+315=6 "
you sxo.SN
,
momentum
)
b) @ omg (4.0ms) -8mg 't
In
smaller
means
why
3D a)
6115mg
a) (2.0mg ) (3. OF
also
.
.
go min
area
.
on
gained 315mg
are
-
one
greater inertia
mass
=
or
32) Impulse
cars
the
because of its
cars
more
,
due to
car
being
up
which is
velocity
in
the
heavier
the
has
momentum
in
.
end
3,59%9 g3ol.im?n
using
approx
50N x
momentum
MeV
5--0.060
0.060
-
change
back
31,32
,
is
-
of
which
it
,
.
:
impulse
greater
its
are
terms
mass
that
massive
more
a
other
force
the
another
Solving 19,62
49) a) total
graphical
velocity
larger
that
than
less
pebble because of
of the
mass
be
will
velocity for
in
ride
assume
can
means
change
both
that it will
push back the
to
rate
to
This
applied
is
we
which
.
pebble 's
the lighter pebble
causes
for
safer
is
is
than
speed
out
fact
the
boulder
the
less
equal
the same
also
is
for
2
momentum
the
force
same
time
in
overall that
15) It
b) The
change
though
even
the
objects and
these
applied
boulder's
the
of change of
rate
because
of
both
that
larger force
the
be equal to the
to
than
of change of
rate
momentum
smaller
the much
overall
an
smaller
Since
.
the smaller
impulse than
example of
,
force
the
with
occurs
be much
could
will
because
case
that
time
the
equal
) The
Il
always
in
P :S
Physics
AP
9,11 , 12,15
:
Fet
-
Vela
1/12/20
-
be
larger
.
would
: