Momentum 03/10/23
inertia
generally described as object's resistance to motion
-
momentum -
tendency of object...
5 momentum
:
rector units: kg.m/s (no name)
m mass
=
I
*
=
velocity vector mr
=
-- and Ihave same direction
-force is required to change a
body's momentum
-something big/slow can have the same momentum as sat small/fast
(equivalentmomental -> P mt m = =
=
Change in Momentum
1P 1 (m.v) =
1P 1m.
change:
=
m
I 14 Ir
change:
=
m.
both 1P 1m. Ar
change:
=
Impolse allows us to consider the time over which a force is exerted on an object
the force implied, the force
longer a is more
5 Est = 5:FnetAt
Fret At 1E
or
=
units of impulse momentum!! - -
a
larger force acts for a short period of time
a smaller force acts for a
longer period of time
Newton's
original and law =
, Example 5.8kg mass is accelerated from vestto 3.5m/s in 2.55. Find fetfor mass
FAt 14 mAv
=
=
m(vf vi)
=
-
f=
w-vi) (35ms
-om/s)
ON
-
=
Example 1.3
kg ball
coming straight to 75kg player at13m/s [right] who kicks itin
direction 22 ms [left]
opposite at
m/argf ofROON. How long are his
foot - ball in contact?
e↑ itetts
At 0.03795
-
=
Eisle to avoid the
Collisions 03/28/23
it
A b.w
=
= f.St 5 SP
=
=
areas
t
ball force
Example a
2.0kg atrestis acted upon by a thatvaries
my time as shown.
II
FFttE=
find
ma
~0.5.0, 2.0
t
Egg
Area 5 15
- ↑
=
2.5ms
=
Vi =
e
inertia
generally described as object's resistance to motion
-
momentum -
tendency of object...
5 momentum
:
rector units: kg.m/s (no name)
m mass
=
I
*
=
velocity vector mr
=
-- and Ihave same direction
-force is required to change a
body's momentum
-something big/slow can have the same momentum as sat small/fast
(equivalentmomental -> P mt m = =
=
Change in Momentum
1P 1 (m.v) =
1P 1m.
change:
=
m
I 14 Ir
change:
=
m.
both 1P 1m. Ar
change:
=
Impolse allows us to consider the time over which a force is exerted on an object
the force implied, the force
longer a is more
5 Est = 5:FnetAt
Fret At 1E
or
=
units of impulse momentum!! - -
a
larger force acts for a short period of time
a smaller force acts for a
longer period of time
Newton's
original and law =
, Example 5.8kg mass is accelerated from vestto 3.5m/s in 2.55. Find fetfor mass
FAt 14 mAv
=
=
m(vf vi)
=
-
f=
w-vi) (35ms
-om/s)
ON
-
=
Example 1.3
kg ball
coming straight to 75kg player at13m/s [right] who kicks itin
direction 22 ms [left]
opposite at
m/argf ofROON. How long are his
foot - ball in contact?
e↑ itetts
At 0.03795
-
=
Eisle to avoid the
Collisions 03/28/23
it
A b.w
=
= f.St 5 SP
=
=
areas
t
ball force
Example a
2.0kg atrestis acted upon by a thatvaries
my time as shown.
II
FFttE=
find
ma
~0.5.0, 2.0
t
Egg
Area 5 15
- ↑
=
2.5ms
=
Vi =
e
