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Chapter 9
Linear Momentum and
Collisions
Momentum
Linear momentum of object: mass×velocity
p  mv
Magnitude: p=mv, Direction: same as velocity
Quantity of motion, ability to acting force
Newton’s second law of motion (original edition)
The changing rate of momentum of an object
is equal to the net force applied on it.
dv
dp d  mv 

m
 ma
F
dt
dt
dt
2
Impulse
Integral form of Newton’s second law
tf
dp
F
  p  p f  p i   Fdt ①
ti
dt
Change in momentum is the integral of the net
force over the time, the integral is called impulse
J 

tf
ti
Fdt
 J  Ft
if F is constant

Eq.① is called impulse-momentum theorem
momentum ~ impulse
energy ~ work
1) any force
3) J
2) vector
F?
3
Projectile problem
Example1: Object m is fired with v at angle 45°to
the horizontal, no air friction. Determine:
y
a) Initial momentum
b) Final momentum
v
c) Impulse during the motion
45°
x
Solution: a) p  m v  2 mv  i  2 mv  j
i
2
2
b)
2
2
pf 
mv  i 
mv  j
2
2
same magnitude
different direction
c)
J  p f  pi   2 m v  j
J  mg t
4
Conical pendulum
Example2: Known: m, L, , v, T. Determine the
impulse of tension and gravity respectively during
one-half period.
Solution: FT  mg / cos 
L 
J G  mgT / 2
J F  m gT / 2 cos 
FT
m
FT is not a constant force!
G
So:

p  2m v
JF 
J  p
2
G
O
2
Direction of these vectors
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Collisions and impulsive force
Concept of momentum and impulse is useful when
dealing with collision problems
Key points:
1) Time interval is usually very short
2) Interaction of collision, or impulsive force is
changing fast and may become very large
The effect is shown by its impulse
3) Impulses of other forces can be ignored
6
Bend your knees
Example3: A 70-kg person jumps from h=3.0m and lands
on firm ground. a) Calculate the impulse experienced;
b) Estimate the average force by the ground if he is stifflegged (s=1.0cm); c) average force with bent legs (s=50cm).
Solution: a) Landing speed v 
2 g h  7 .7 m / s
Impulse J   p  0  m v   5 4 0 N  s
x
1
2
5
b) Fnet  s  0  mv  F n e t   m g h / s   2 .1  1 0 N
2
F n et  F g r  m g  F g r   2 .1  1 0 5 N  6 9 0 N   2 .1  1 0 5 N
c)
F n e t   m g h / s   4 .2  1 0 N
3
impulsive force
F g r   4 .2  1 0 3 N  6 9 0 N   4 .9  1 0 3 N
7
Washing a car
Example4: Water is jetting at rate of R=1.5kg/s with
v=20m/s, and is stopped by a car (no splashing back).
What is the force exerted on the car?
Solution: In a short time interval dt,
How much water hits on the car: dm=Rdt
Change in momentum: dp=vdm=Rvdt
Collision: ignore gravity, only force by car
Fdt  dp  F  R v  30 N
What if the water splashes back?
8
Falling rope
Example5: One end of a hanged rope just touch the table,
then it is released. Prove: Normal force acting on table is
always 3 times of weight of that part already on the table.
Solution: Linear density λ, falling height h
 N  mg  F
Mass of part already on table m = λ h
Falling speed v  2 gh
In time interval dt, dm = λ · vdt
Impulse F·dt = dm·v = λ ·v2dt
F=
λ ·v2 =
λ ·2gh =2mg
 N  m g  F  3m g
dm
v
m
mg F
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Momentum of a system
Consider a system of n interacting particles
P
Total momentum of the system

pi

Change in total momentum dP   Fi  f ij
dt
Internal forces are always in pairs

f ij  0
dP
  Fi
dt
Only external forces can
change the total momentum!

Fi
F1
f1i f
i1
m1
f1n
fn1
mn
mi
fni
fin
Fn
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Conservation of momentum
If the net external force is 0, then
dP
  Fi  0 o r P  c o n s ta n t
dt
This is the law of conservation of momentum:
When the net external forces on a system is zero,
the total momentum remains constant.
Or: The total momentum of an isolated system of
bodies remains constant.
1) isolated
2) system
3) ignoring forces
11
Rifle recoiling
Example6: Calculate the recoil speed of a 5kg rifle
that shoots a 10g bullet at v=700m/s.
Solution: Total momentum is conserved
m RvR  m Bv  0
mB v
 vR  
mR
  1.4 m / s
The rifle moves back and applies a recoil force
12
Billiard ball collision
Example7: A billiard ball with v in the +x direction
strikes an identical resting ball. Directions after
collision are shown in figure, what are the speeds?
Solution: conservation of momentum
mv  mv B  mvY
component form:
x : mv  mv B cos 45   mvY cos 45 
y:
0  mv B sin 45   mvY sin 45 
2
 v B  vY 
v
2
45
45
y
o
x
kinetic energy is also conserved
13
Conservation in component form
If
 F  0 , but  F
x
0
→ x component of momentum is conserved
Example8: Bullet (20g) hits into hanging ball (980g)
with v=400m/s. Determine the speed after collision.
Solution: No net force in horizontal
Horizontal component Px is conserved
m v sin 3 0 o   M  m  v 
mv sin 30 o
 v 
 4 m /s
M m
30 T
v
m
M
(M+m)g
14
Homework
A gun carrier M moves on a frictionless incline, its
speed reduces from v to 0 after shooting a canon-ball
m in the horizontal direction. Is the total momentum
of system (M and m) conserved in this process, and
why? Find out the speed of canon-ball.
v
m v’

15
Distance traveled
Example9: Two objects start from rest, what is the
distance traveled by m2 when m1 reaches the ground?
No friction.
Solution: conservation of momentum in horizontal
m 1 v1 x  m 2 v 2 x  0
b
 m 1  x1  m 2  x 2  0
m1
where  x1   x 2  a  b
m1
S 2   x2 
a  b
m1  m 2

So distance traveled by m2
m2
o
x
a
16
Challenging question
Question: Someone sits on the top of an ice half
sphere on the ground, and then slips down. At what
angle  will he leave the sphere? No friction.
Thinking:
Leave the half sphere: N = 0
m gcos  N  m v 2 / R
m
top
M

?
Relative speed & inertial frame
Conservation of momentum and
mechanical energy
Velocity transformation
17
Elastic collisions
If total kinetic energy is conserved in a collision
1
1
1
1
2
2
2
m1 v1  m 2 v 2  m1 v1  m 2 v 2 2
2
2
2
2
It is called an elastic collision
Ideal model in macro world
head-on collision
Conservation of momentum (1-dimension)
m 1 v 1  m 2 v 2  m 1 v 1  m 2 v 2
 v1  v 2    v1  v 2 
Relative speed: equal but opposite
18
Final speeds
m 1 v 1  m 2 v 2  m 1 v 1  m 2 v 2

1
1
1
1
m1 v12  m 2 v 22  m1 v1 2  m 2 v 2 2
2
2
2
2
v1 
m1  m2
2 m2
v1 
v2
m1  m2
m1  m2
m2  m1
2 m1
v2 
v2 
v1
m1  m2
m1  m2
Special cases:
1) Equal masses v 1  v 2 , v 2  v 1
2) Target m2 at rest
 v 1   v 1 , v 2  0
m1
m2
m1
m 2  v 1  v 1 , v 2  2 v 1
19
Baseball batting
Example10: A baseball with speed 40m/s is hit by a bat
with speed 30m/s in the opposite direction. Determine the
speed of baseball if the collision is elastic and mbat >> mball .
Solution: final speed of ball
 
v ball
m ball  m bat
2 m bat
v ball 
v bat
m bat  m ball
m bat  m ball
  v b a ll  2 v b a t   1 0 0 m / s
Or in the frame of bat:
The ball moves in 70m/s, and rebounds in the same speed
Transform to the frame of ground, it’s 100m/s.
20
Slingshot effect
Example11: Spacecraft Voyager II approaches the Jupiter,
it rounds the planet and departs in the opposite direction.
What is the speed after this slingshot encounter?
( vS =10.4km/s, vJ =-9.6km/s )
Solution: Like an elastic collision
v S  v J    v S  v J 
mJ
m S  v J  v J
speed after slingshot:
v S   v S  2 v J
  2 9 .6 k m /s
21
Inelastic collisions
If kinetic energy is not conserved in a collision
It is called an inelastic collision
Restitution coefficient
e=1:
e<1:
e=0:
v2  v1
e
v1  v2
elastic collision
inelastic collision
completely inelastic collision
How energy transform? heat and sound …
22
Completely inelastic collision
Two objects stick together in collision: v’1 = v’2
If target m2 is initially at rest
Kinetic energy loss Ek 
m2
 Ek 0
m1  m2
Depends on the mass ratio of two objects
m1
m1
m2 :
m2 :
Almost no energy loss
Hammer hit on nail …
Maximum of energy loss
Throw egg against rock …
23
Hit by bullet
Example12: m1=0.49kg and m2= 0.50kg are related by a
massless spring (k=100N/m). The system is initially at rest
on frictionless horizontal plane, then a bullet m3=0.01kg hits
into m1 with v =100m/s. Determine the maximum stretch.
v
m1
m2
Solution: Completely inelastic collision between m1 and m3
m 3 v   m 1  m 3  v   v   2 m /s
Conservation of momentum and mechanical energy
 m 1  m 3  v    m 1  m 2  m 3  v 
1
1
1 2
2
2


 m1  m 3  v   m1  m 2  m 3  v  kx
2
2
2
 x  0 .1 m
24
Collisions in 2-dimensions
Discuss an elastic collision in 2 dimensions
Target initially at rest
Conservation of momentum
y
o
x
m 1 v  m 1 v1 c o s  1  m 2 v 2 c o s  2
1
2
0  m 1 v 1 s in  1  m 2 v 2 s in  2
Conservation of kinetic energy
1
1
1
2
2
m1 v  m1v1  m 2 v 22
2
2
2
It can be solved if one of variables is known
25
Equal-mass case
If it occurs between two objects of equal mass
m1  m 2
Angle between final velocities is always 90°
1   2  90
This is shown to be true
in many experiments
You can try to prove it
(Problem 57, Page 229)
y
o
x
1
2
26
Center of mass
Real body (instead of particle): general motion
One point represents motion of object (system)
As a particle with same mass and same net force
It is called center of mass (CM)
c.
General motion
c
c
= Translational motion of CM
+ Other motion about CM
c.
c
(rotation, vibration…)
Different from center of gravity (CG)
27
CM of several particles
Definition of CM: by weighted average
xC
mx


i
M
1-dimension
i
or
rC
mr


i i
M
more-dimensions
Example13: Determine the CM of stones, where the
mass of black stone is twice as much as white stone.
y
Solution:
25
1  (1  2  2 )  2  ( 2  3  5 )

xC 
1 3  2  3
9
1  ( 3  5  8 )  2  ( 2  2  3 ) 10

yC 
3
1 3  2  3
O
x
28
CM of continuous object
Continuous object: summations become integrals
xC
xdm


or
rdm


rC
M
M
Example14: CM of a thin rod with varying density.
  0 x
Solution: Choose a segment dx
o
dm   dx  0 xdx
The total mass
CM:
1
xC 
M
M   dm 
1
 xdm  M

L
0

L
0
x
dx
L x
 0 xdx   0 L 2 / 2
2
 0 x dx  L
3
2
29
CM of more objects
Example15: CM of uniform thin right triangle.
Solution: Choose a segment dm of the triangle
ax
dm   
bdx
a
 xC


a
0
xdm


a
0

M
Similarly we have
y
abx  bx
dx
a
 ab / 2
b
dx
2
 a /3
O
x
a
x
yC  b / 3
Comparing with Example14
What about other objects?
30
CM and translational motion
rC
mr


i i
M
 M rC 
mr
i i
 M vC 
mv
i i
P
Total momentum of a system is equal to the product of
the total mass M and velocity of CM.
dP

MaC 
dt
F
i
The sum of all the forces acting on the system is equal
to the total mass M times the acceleration of CM.
Or: The center of mass of a system with total mass
M moves like a single particle of mass M acted by
the same net external force.
31
Man on boat
Example16: A man (m) moves from one side of a
boat (M, L) to another side, determine the distance
traveled by the boat. Initially at rest, no resistance.
Solution: No net external force, CM stays at rest
l
l
ml  M
ms  M ( s  )
2 
2
xC 
mM
mM
ml
s 
mM
y
o
s
x
32
Two stage rocket
Example17: A rocket is separated into two parts of
equal mass at its highest point. Part A falls vertically
to Earth from rest, where does part B lands?
Solution: CM still moves along the curve
It lands at x=2d
So part B will land at x=3d
If part A had been
given a kick up or
down?
y
A
d
B
?
x
33
*Rocket propulsion
System of variable mass & ignore external forces
Gas dm is jetting with relative speed vr
Speed of rocket changes from v to v+dv
m v   m  dm   v  dv    v  vr  dm
v
m
vr
dm
dm
 m d v  v r d m  dv  v r
m
v
m
dm
m
dv 
vr
 v  v r ln 0
0
m0
m
m


Tsiolkovsky equation
34