Download Purdue Physics - Purdue University

PHYSICS 149: Lecture 18
• Chapter 7: Linear Momentum
– 7.2 Momentum
– 7.3 The Impulse-Momentum Theorem
– 7.4 Conservation of Momentum
Lecture 18
Purdue University, Physics 149
1
ILQ 1
Two objects are known to have the same
kinetic energy. Do these two objects
necessarily have the same momentum?
A) Yes
B) No
Lecture 18
The example of the ice skaters
starting from rest and pushing off
of each other...the total momentum
is zero because they travel in
opposite directions but the kinetic
energy is not zero (they start with
none and gain a bunch)
Purdue University, Physics 149
2
ILQ 2
An astronaut is taking a space walk when his
tether breaks.
breaks He has a big drill
drill. How can he
get back to the spaceship?
A) Face away from the ship and
throw the drill forward.
B) Face the ship and throw
the drill forward.
C) Throw the drill towards
the tether.
Lecture 18
Purdue University, Physics 149
3
Momentum
• Momentum is a vector quantity.
quantity
• Units: kg⋅m/s, N⋅s
– Unit conversion: 1 kg⋅m/s = 1 N⋅s
• Momentum is denoted by p.
p
• Momentum is sometimes referred to as “linear
momentum”
t ” to
t distinguish
di ti
i h it ffrom angular
l momentum.
t
Lecture 18
Purdue University, Physics 149
4
Momentum is Conserved
• Momentum is “Conserved” meaning it can not be
created nor destroyed
– Can be transferred
• Total Momentum does not change with time
• Momentum is a VECTOR
3 Conservation Laws in one!
Lecture 18
Purdue University, Physics 149
5
Momentum is Conserved
• Define Momentum p = m v
–
–
–
–
m1 Δv1 = -m2 Δv2
Δp1 = -Δp2
Δp1 + Δp2 = 0
Σpinitial = Σpfinal
Momentum is conserved
p1f-p1i+p2f-p2i=0
p1f+p2f=p1i+p2i
• Example: Jane pushes Fred so he is going 2.0 m/s.
If Fred is twice as heavy as Jane,
Jane how fast does
Jane end up moving?
– Σpinitial = Σpfinal
– 0 = mFred vFred + mJane vJane
– vJane = -mFred vFred / mJane = -4 m/s
Lecture 18
Purdue University, Physics 149
6
Momentum Transfer
• During an interaction, momentum is transferred from one
bodyy to another,, but the total momentum of the two is
unchanged.
G
G
G
G
G
G
Δp2 = −Δp1 or
p1i + p2i = p1 f + p2 f
• Consider a collision between two spaceships.
G
v1i
G
v1 f
G
v2 i
G
G G
G
G G
F
Δv1 = a1Δt = 12 Δt ⇒ Δp1 = m1Δv1 = F12 Δt
m1
G
G
F21 = − F12
Lecture 18
G
v2 f
G
G
G G
G
G
F21
Δv2 = a2 Δt =
Δt ⇒ Δp2 = m2 Δv2 = F21Δt
m2
G
G
G
G
G
G
⇒ Δp2 = m2 Δv2 = F21Δt = (− F12 )Δt = −m1Δv1 = −Δp1
G
G
Thus, Δp2 = −Δp1
Purdue University, Physics 149
7
Impulse
• Impulse is the product of (average) force and time interval
during which the force acts.
G G
I = FΔt
• The change
g in momentum of an object
j
is equal
q
to the
impulse (recall a collision between two spaceships).
G G G
Δp = I (= FΔt )
• Impulse is a vector quantity.
quantity
• Units: N⋅s, kg⋅m/s
– Unit conversion: 1 N⋅s = 1 kg⋅m/s
• Impulse is often denoted by II.
Lecture 18
Purdue University, Physics 149
8
Impulse Changes Momentum
• Define Impulse I = FΔt
– Change in momentum requires Force acting
over a time or Impulse
– Recall:
mΔv = FΔt
– Or
Δp = FΔt
Δp = I
Impulse
I = FΔt
Momentum p = mv
Lecture 18
Purdue University, Physics 149
9
ILQ
By what factor does an object’s kinetic energy
change if its speed is doubled? By what factor
does the momentum change?
A)) 4,, 2
B) 2, 2
C) 0.5, 4
G
G
p = mv
Lecture 18
1 2
K = mv
2
Purdue University, Physics 149
10
Impulse and Momentum
I ≡ FaveΔt = pf - pi = Δp
• For single object …
– F = 0 ⇒ momentum conserved (Δp = 0)
• For collection of objects …
– ptotal = Σp
– Internal forces: forces between objects in system
– External forces: all other forces
– ΣFext = 0 ⇒ total momentum conserved ((Δp
ptot = 0))
– Fext = mtotal a
Lecture 18
Purdue University, Physics 149
11
Conservation of Momentum
•
Momentum is a vector quantity, so both the magnitude and the
direction of the total momentum at the beginning and end of the
interaction must be the same.
•
The total momentum of a system
y
is the vector sum of the
momenta of each object in the system.
Internal interactions do not change the total momentum of a
system.
External interactions can change the total momentum of a system.
•
•
•
During a collision, the interaction often takes place during a
sufficiently
ffi i l short
h amount off time
i
that
h externall fforces may b
be
ignored; conservation of momentum may be used.
Lecture 18
Purdue University, Physics 149
12
ILQ
Two identical balls are dropped from the same height onto the floor. In
each case they have velocity v downward just before hitting the floor. In
case 1 the ball bounces back up, and in case 2 the ball sticks to the
fl
floor
without
ih
b
bouncing.
i
IIn which
hi h case iis the
h magnitude
i d off the
h iimpulse
l
given to the ball by the floor the biggest?
A) Case 1
Δ mΔ = maΔt
Δp=mΔv
m Δt = FΔt = I
B) Case 2
C) The same
Lecture 18
Bouncing Ball
Sticky Ball
|I| = |Δp|
|I| = |Δp|
= |mv
| final – m vinitial|
= |mv
| final – m vinitial|
= |m( vfinal - vinitial)|
= |m(0 - vinitial)|
=2mv
= mv
Purdue University, Physics 149
13
ILQ
Two identical balls are dropped from the same height
onto the floor. In case 1 the ball bounces back up, and in
case 2 the ball sticks to the floor without bouncing.
bouncing In
both cases of the above question, the direction of the
impulse given to the ball by the floor is the same. What
is this direction?
A) Upward
B) Downward
G
G
G
G
I = Δp = mv2 − mv1
time
Lecture 18
Purdue University, Physics 149
14
Momentum is a Vector
A car with mass 1200 kg is driving north at 40 m/s, and turns east driving
30 m/s. What is the magnitude of the car’s change in momentum?
A) 0
B) 12,000
C) 36,000
D) 48,000
E) 60,000
pinitial = m vinitial = (1200 kg) x 40 m/s = 48,000 kg m/s North
pfinal = m vfinal = (1200 kg) x 30 m/s = 36,000 kg m/s East
North-South:
Pfinal – Pinitial = (0 – 48000) = -48,000 kg m/s
East-West:
Pfinal – Pinitial = (36000 - 0) = +36,000 kg m/s
30 m/s
40 m/s
Magnitude:
sqrt(P2North + P2East ) = 60,000 kg m/s
Lecture 18
Purdue University, Physics 149
15
ILQ
You drop an egg onto A) the floor B) a thick piece of
foam rubber. In both cases, the egg does not bounce.
In which case is the impulse greater?
A) Floor
B) Foam
I = Δp
C) the same
Same change in momentum
In which case is the average force greater
A) Floor
B)) Foam
C) the same
Lecture 18
Δp = F Δt
F = Δp/Δt
smaller Δt = large F
Purdue University, Physics 149
16
Example: Time to Burn
For a safe re-entry into the Earth's atmosphere the
pilots of a space capsule must reduce their speed
from 2.6 × 104 m/s to 1.1 × 104 m/s. The rocket
engine
g
p
produces a backward force on the capsule
p
of 1.8 × 105 N. The mass of the capsule is 3800 kg.
For how long must they fire their engine?
[Hint: Ignore the change in mass of the capsule due
to the expulsion of exhaust gases.]
Lecture 18
Purdue University, Physics 149
17
Example: Time to Burn
vi=2.6×104 m/s
vf=1.1×104 m/s
F=1.8×105 N
M=3800
M
3800 Kg
Δt=?
FΔ t = m Δ v
m Δv m ( v f − v i )
Δt =
=
=
F
F
3800kg ( 2.6 − 1.1) × 104 m / s
1N
×
= 317 s
=
5
2
1.8 × 10 N
1 Kgm / s
Lecture 18
Purdue University, Physics 149
18
Example
• A fright train is being assembled in switching yard. Car 1
has a mass of m1=65 ×103 kg and moves at a velocity
v01=0.80
0 80 m/s.
/ Car
C 2h
has a mass off m2=92
92 ×10
103 kg
k and
d
moves at a velocity v02=1.3 m/s and couples to it.
Neglecting friction, find the common velocity vf of the cars
after
ft they
th become
b
coupled.
l d
m2
m2
Lecture 18
v02
m1
m1
v01
vf
Purdue University, Physics 149
initial
final
19
Example
• Note the velocity of car 1 increases while the velocity of car 2
decreases.
• The
Th acceleration
l ti and
dd
deceleration
l ti arise
i b
because th
the cars
exert internal forces on each other.
• Momentum conservation allows us to determine the change in
velocity without knowing what the internal forces are.
• Note that often momentum is conserved but the kinetic energy
is not! (Kinetic energy is conserved only in elastic collisions)
m2
v02
m2
Lecture 18
m1
m1
v01
vf
Purdue University, Physics 149
initial
f l
final
20
Example
• A fright train is being assembled in switching yard. Car 1
has a mass of m1=65 ×103 kg and moves at a velocity
v01=0.80
0 80 m/s.
/ Car
C 2h
has a mass off m2=92
92 ×10
103 kg
k and
d
moves at a velocity v02=1.3 m/s and couples to it.
Neglecting friction, find the common velocity vf of the cars
after
ft they
th become
b
coupled.
l d
Apply conservation of momentum:
Pi=m
m1vo1+m
m2v02
Pf=(m
(m1+m
m2)vf
(m1+m2)vf = m1v01+m2v02
m1v01 + m2 v02
vf =
=
m1 + m2
(65 ×103 kg )(0.8
)(0 8m / s ) + (92 ×103 kg )(1.3
)(1 3m / s )
= 1.1m / s
3
3
(65 ×10 kg + 92 ×10 kg )
Lecture 18
Purdue University, Physics 149
21
ILQ
At the instant a bullet is fired from a gun, the bullet and
the gun have equal and opposite momenta
momenta. Which object
has the greater kinetic energy? Or are they the same
A) The same
Apply conservation of
B) The gun
momentum:
C) The bullet
Pi=0
0
Pf=M
MgunVgun+m
mbulletvbullet
Vgun=-(mbullet/Mgun)vbullet
K gun
⎛ mbullet
1
1
2
= M gunVgun = M gun ⎜
⎜M
2
2
⎝ gun
Lecture 18
2
⎞ 2
mbullet
K bullet
⎟⎟ vbullet =
M gun
⎠
Purdue University, Physics 149
22
Impulse-Momentum Theorem
• The “total” impulse on an object is equal to the change in
the object’s momentum during the same time interval.
G
= Itotal
• Example:
– Today’s automobiles are designed with a number of safety
features, all designed to make the impact last longer (Δt) and so
applied force weaker (F). That way we can minimize injury upon
collision.
collision
pi = mvi, pf = mvf = 0 Æ |Δp| = |pf – pi| = |–mvi| = |F|Δt
The longer Δt,
Δt the weaker |F|
|F|.
Lecture 18
Purdue University, Physics 149
23
Newton’s Second Law
• The net force is the rate of change of momentum
momentum.
• This relation is valid even when mass is not constant (just
like a rocket engine).
g )
Lecture 18
Purdue University, Physics 149
24
Newton’s Second Law
G
G
∑ F = ma
G
G
I = ∑ F Δt = Δp
G
G
Δp
∑ F = Δt
Lecture 18
New
formulation of
N2L
Purdue University, Physics 149
25
ILQ
•
A 3.0 kg object is initially at rest. It then
receives an impulse of magnitude 15 N⋅s
N⋅s. After
the impulse, the object has
A)
B)
C)
D)
a momentum of magnitude 5.0 kg⋅m/s.
a speed of 45
45.0
0 m/s.
m/s
a momentum of magnitude 15.0 kg⋅m/s.
a speed of 7
7.5
5 m/s
m/s.
Lecture 18
Purdue University, Physics 149
26
ILQ
•
An 18-wheeler and a Volkswagen Beetle are
rolling along with the same momentum
momentum. If you
exert the same force with the brakes to stop
each one,, which takes a longer
g time to bring
g to
rest?
a)
b))
c)
d)
18-wheeler
Volkswagen
g beetle
same for both
impossible to say
Lecture 18
Purdue University, Physics 149
27
Example
• A pole-vaulter of mass 60.0 kg vaults to a height of 6.0 m
before dropping to thick padding placed below to cushion
her fall.
(a) Find the speed with which she lands.
– Gravity (conservative force) is the only force acting on her before
g, so we mayy use the conservation of mechanical energy
gy
landing,
(recall Chapter 6).
– 0 + mgh = ½ mv2 + 0
Æ v = sqrt(2gh) = sqrt(2 ⋅ 9.8m/s2 ⋅ 6.0m) = 10.8 m/s
(b) If the padding brings her to a stop in a time of 0.50 s,
what is the average force on her body due to the padding
d i that
during
h time
i
iinterval?
l?
– Use the impulse-momentum theorem
– Δp = pf – pi = mvf – mvi = 60.0 kg [0 – (–10.8 m/s)]
= 648 kg⋅m/s, upward
F = Δp/Δt = 648 kg⋅m/s / 0.50 s = 1,296 N, upward
Lecture 18
Purdue University, Physics 149
28