Transcript Systems of Particles

Linear momentum and Collisions
Chapter 9
Center of mass and linear momentum
I. The center of mass
- System of particles /
- Solid body
II. Newton’s Second law for a system of particles
III. Linear Momentum
- System of particles / - Conservation
IV. Collision and impulse
- Single collision / - Series of collisions
V. Momentum and kinetic energy in collisions
VI. Inelastic collisions in 1D
-Completely inelastic collision/
Velocity of COM
VII. Elastic collisions in 1D
VIII. Collisions in 2D
IX. Systems with varying mass
X. External forces and internal energy
changes
I. Center of mass
The center of mass of a body or a system of
bodies is the point that moves as though all the
mass were concentrated there and all external
forces were applied there.
System of particles:
Two particles of masses m1 and m2 separated by a distance d
Origin of reference system
coincides with m1
m2
xcom 
d
m1  m2
System of particles:
Choice of the reference
origin is arbitrary  Shift
of the coordinate system
but center of mass is still
at the same distance
from each particle.
The center of mass lies somewhere between the two particles.
General:
m1 x1  m2 x2 m1 x1  m2 x2
xcom 

m1  m2
M
M = total mass of the system
System of particles:
We can extend this equation to a general situation for n particles
that strung along x-axis. The total mass of the system
M=m1+m2+m3+……+mn The location of center of the mass:
xcom
m1 x1  m2 x2  m3 x3  ........  mn xn
1


M
M
n
m x
i 1
i i
3D:
1 n
xcom   mi xi
M i 1
1 n
ycom   mi yi
M i 1
1 n
zcom 
 mi zi
M i 1
System of particles:
3D: The vector form
Position of the
particle:
Position COM

ri  xiiˆ  yi ˆj  zi kˆ

rcom  xcomiˆ  ycom ˆj  zcomkˆ

1 n 
rcom   mi ri
M i 1
M = mass of the
object
Solid bodies:
Continuous distribution of matter. Particles = dm
(differential mass elements).
1
x dm
3D: xcom 

M
1
ycom 
y dm

M
1
zcom 
z dm

M
M = mass of the object
Assumption: Uniform objects  uniform density
1
xcom   x dV
V
1
ycom   y dV
V
dm M


dV V
1
zcom   z dV
V
The center of mass of an object with a point, line or
plane of symmetry lies on that point, line or plane.
The center of mass of an object does not need to lie
within the object (Examples: doughnut, horseshoe )
Problem solving tactics:
(1) Use object’s symmetry.
(2) If possible, divide object in several parts. Treat each of
these parts as a particle located at its own center of mass.
(3) Chose your axes. Use one particle of the system as origin
of your reference system or let the symmetry lines be your
axis.
II. Newton’s second law for a system of particles
Motion of the center of mass:
Center of the mass of the system moves as a particle
whose mass is equal to the total mass of the system.


Fnet  Macom
Fnet is the net of all external forces that act on the
system. Internal forces (from one part of the system to
another are not included).
The system is closed: no mass enters or leaves the
system during the movement. (M=total system mass).
acom is the acceleration of the system’s center of mass.
Fnet, x  Macom, x
Fnet, y  Macom, y
Fnet, z  Macom, z





Mrcom  m1r1  m2 r2  m3r3  ...  mn rn




drcom
drn
dr1
dr2
Prove:
M
 m1
 m2
 ......  mn
dt
dt
dt
dt





Mvcom  m1v1  m2 v2  m3v3  ...  mn vn
  






Macom  m1a1  m2 a2  m3 a3  ...  mn an  F1  F2  F3  ...  Fn
(*) includes forces that the particles of the system exert on
each other (internal forces) and forces exerted on the
particles from outside the system (external).
Newton’s third law  internal forces from third-law force
pairs cancel out in the sum (*)  Only external forces.
(*)
III. Linear momentum
The linear momentum of a particle is a vector p defined as:


p  mv
Momentum is a vector with magnitude equal mv and have
direction of v .
SI unit of the momentum is kg-meter/second
Newton II law
in terms of momentum:
The time rate of change of the momentum of a particle is
equal to the net force acting on the particle and is in the
direction of the force.

Fnet




dp d (mv )
dv


m
 ma
dt
dt
dt
System of particles:
The total linear moment P is the vector sum of the
individual particle’s linear momentum.
  






P  p1  p2  p3  ....  pn  m1v1  m2v2  m3v3  ...  mnvn


P  Mvcom
The linear momentum of a system of particles is
equal to the product of the total mass M of the
system and the velocity of the center of mass.





dvcom
dP
dP
M
 Macom  Fnet 
dt
dt
dt
Net external force acting on the system.
Conservation:
If no external force acts on a closed, isolated system of
particles, the total linear momentum P of the system cannot
change.

P  const
(Closed , isolated system)




dP
Fnet  0 
 Pf  Pi
dt
Closed: no matter passes through the system boundary in
any direction.
Conservation of Linear Momentum
If no net external force acts on the system of particles the total
linear momentum P of the system cannot change.
Each component of the linear momentum is conserved
separately if the corresponding component of the net external
force is zero.
If the component of the net external force on a closed
system is zero along an axis  component of the linear
momentum along that axis cannot change.
The momentum is constant if no external forces act on a
closed particle system. internal forces can change the linear
momentum of portions of the system, but they cannot
change the total linear momentum of the entire system.
IV. Collision and impulse
Collision: isolated event in which two or more bodies exert
relatively strong forces on each other for a relatively short time.
Impulse: Measures the strength and duration of the collision
force
Third law force pair
FR = - FL
Single collision


p
 dp
 
 t 
F
 dp  F (t )dt   dp  t F (t )dt

dt
p
 t 



J  t F (t )dt  p f  pi  p
f
f
i
i
f
i
Impulse-linear momentum theorem
The change in the linear momentum of a body in a
collision is equal to the impulse that acts on that body.

 

p  p f  pi  J
Units: kg m/s
p fx  pix  p x  J x
p fy  piy  p y  J y
p fz  piz  p z  J z
Favg such that:
Area under F(t) vs Δt curve
= Area under Favg vs t
J  Favg t
Series of collisions
Target fixed in place
n-projectiles  n Δp = Total
change in linear momentum
(projectiles)
J and Δp have opposite
Impulse on the target: J   n  p directions, pf < pi  Δp
left  J to the right.
n/Δt  Rate at which the
J n
n
Favg 

p 
mv projectiles collide with the
t t
t
target.
m  nm in t  Favg
m

v Δm/Δt  Rate at which mass
collides with the target.
t
a) Projectiles stop upon impact: Δv = vf-vi = 0-v = -v
b) Projectiles bounce: Δv = vf-vi = -v-v = -2v
V. Momentum and kinetic energy in collisions
Assumptions: Closed systems (no mass enters or leaves
them)
Isolated systems (no external forces act on
the bodies within the system)
Elastic collision:
If the total kinetic energy of the system of
two
colliding
bodies
is
unchanged
(conserved) by the collision.
Inelastic collision: The kinetic energy of the system is not
conserved  some goes into thermal
energy, sound, etc.
Completely inelastic collision: After the collision the bodies lose
energy and stick together.
Velocity of the center of mass:
In a closed, isolated
system, the velocity of
COM of the system cannot
be changed by a collision.
(No net external force).



P  Mvcom  (m1  m2 )vcom





P conserved  p1i  p2i  p1 f  p2 f






p1i  p2i p1 f  p2 f
P
 vcom 


m1  m2 m1  m2
m1  m2
Completely inelastic
collision  v = vcom
VII. Elastic collisions in 1D
(Total kinetic energy before collision )  (Total kinetic energy after collision )
In an elastic collision, the
kinetic
energy
of
each
colliding body may change,
but the total kinetic energy of
the system does not change.
Stationary target:
Closed, isolated
system 
m1v1i  m1v1 f  m2v2 f
1
1
1
2
2
m1vi1  m1v1 f  m2v22 f
2
2
2
m1 (v1i  v1 f )  m2v2 f
Linear momentum
Kinetic energy
(1)
m1 (v12i  v12f )  m2v22 f  m1 (v1i  v1 f )(v1i  v1 f )
(2)
Stationary target:
Dividing (2) /(1)  v2 f  v1i  v1 f
From (1)  v2 f
m1

(v1i  v1 f )
m2
(1) in (3) v1 f  v2 f
m  m2
v1 f  1
v1i
m1  m2
(3)
m1
 v1i 
(v1i  v1 f )  v1i 
m2
v2 f
2m1

v1i
m1  m2
v2f >0 always
v1f >0 if m1>m2  forward mov.
v1f <0 if m1<m2  bounce back
Equal masses: m1=m2  v1f=0 and v2f = v1i  In head-on
collisions bodies of equal masses simply exchange
velocities.
Massive target: m2>>m1  v1f ≈ -v1i and
v2f ≈
(2m1/m2)v1i  Body 1 bounces back with
approximately same speed. Body 2 moves forward
at low speed.
v1 f 
m1  m2
v1i
m1  m2
v2 f 
2m1
v1i
m1  m2
Massive projectile: m1>>m2  v1f ≈ v1i and
v2f
≈ 2v1i  Body 1 keeps on going barely lowed by the
collision. Body 2 charges ahead at twice the initial
speed of the projectile.
VII. Elastic collisions in 1D
Moving target:
Closed, isolated system 
m1v1i  m2v2i  m1v1 f  m2v2 f
Linear momentum
1
1
1
1
2
2
2
m1vi1  m2 v2i  m1v1 f  m2 v22 f
2
2
2
2
m1 (v1i  v1 f )  m2 (v2i  v2 f )
Kinetic energy
(1)
m1 (v1i  v1 f )(v1i  v1 f )  m2 (v2i  v2 f )(v2i  v2 f )
m1  m2
2m2
Dividing (2) /(1)  v1 f 
v1i 
v2 i
m1  m2
m1  m2
v2 f
2m1
m2  m1

v1i 
v2 i
m1  m2
m1  m2
(2)
VIII. Collisions in 2D
Closed, isolated system 
 


P1i  P2i  P1 f  P2 f
Linear momentum conserved
Elastic collision 
K1i  K 2i  K1 f  K 2 f
Example:
Kinetic energy conserved
x  axis  m1v1i  m1v1 f cos 1  m2v2 f cos  2
y  axis  0  m1v1 f sin 1  m2v2 f sin  2
If the collision is elastic 
1
1
1
2
2
m1v1i  m1v1 f  m2 v22 f
2
2
2
IV. Systems with varying mass
Example: most of the mass of a rocket on its launching is fuel
that gets burned during the travel.
System:rocket + exhaust products
Closed and isolated mass of this
system does not change as the
rocket accelerates.
P=const  Pi=Pf
After dt
dM < 0
Mv  dM U  ( M  dM )  (v  dv)
Linear momentum of
exhaust products
released during the
interval dt
Linear momentum
of rocket at the
end of dt
Velocity of rocket relative to frame = (velocity of rocket relative
to products)+ (velocity of products relative to frame)
(v  dv)  vrel  U  U  (v  dv)  vrel
Mv  dM U  ( M  dM )  (v  dv)
Mv  dM [(v  dv)  vrel ]  ( M  dM )(v  dv)
Mv  vdM  dvdM  vrel dM  Mv  Mdv  vdM  dvdM  Mv  vrel dM  Mdv
dM
dv
Mdv  vrel dM  
vrel  M
dt
dt
R=Rate at which the rocket losses mass= -dM/dt = rate of
fuel consumption
First rocket
dM
dv

vrel  M
 R  vrel  Ma
equation
dt
dt
vf
Mf
i
i

dM
dv
dM
dM

vrel  M  dv  
vrel   dv  vrel 
 vrel ln M f  ln M i
dt
dt
M
v
M M
Mi
v f  vi  vrel ln
Mf
Second rocket equation

Two blocks of masses M and 3M are placed on a horizontal,
frictionless surface. A light spring is attached to one of them, and
the blocks are pushed together with the spring between them. A
cord initially holding the blocks together is burned; after this, the
block of mass 3M moves to the right with a speed of 2.00 m/s.
(a) What is the speed of the block of mass M? (b) Find the
original elastic potential energy in the spring if M = 0.350 kg.
(a) For the system of two blocks
or
p  0
pi  pf
Therefore,
0  M vm   3M
  2.00 m s
Solving gives:
vm  6.00 m s
(b)
(motion toward the left).
1 2 1
1
kx  M vM2   3M  v32M  8.40 J
2
2
2
A 60.0-kg person running at an initial speed of 4.00 m/s jumps onto a 120-kg cart
initially at rest (Figure below). The person slides on the cart’s top surface and finally
comes to rest relative to the cart. The coefficient of kinetic friction between the
person and the cart is 0.400. Friction between the cart and ground can be
neglected. (a) Find the final velocity of the person and cart relative to the ground.
(b) Find the friction force acting on the person while he is sliding across the top
surface of the cart. (c) How long does the friction force act on the person? (d) Find
the change in momentum of the person and the change in momentum of the cart.
(e) Determine the displacement of the person relative to the ground while he is
sliding on the cart. (f) Determine the displacement of the cart relative to the ground
while the person is sliding. (g) Find the change in kinetic energy of the person. (h)
Find the change in kinetic energy of the cart. (i) Explain why the answers to (g) and
(h) differ. (What kind of collision is this, and what accounts for the loss of
mechanical energy?)
(a)
 60.0 kg 4.00 m
s   120  60.0 kgvf
ˆ
v f  1.33 m si
(b)
 Fy  0
n   60.0 kg 9.80 m s2  0
fk  kn  0.400 588 N   235 N
fk  235 N ˆ
i
pi  I pf
(c) For the person,
m vi  Ft m vf
 60.0 kg 4.00 m
s  235 N  t  60.0 kg 1.33 m s
t 0.680 s
(d) person:
cart:
(e)
(f)
ˆ
m v f  m vi  60.0 kg 1.33 4.00 m s  160 N  si
120 kg 1.33 m s  0  160 N  sˆ
i
xf  xi 


1
1
vi  vf t  4.00  1.33 m s 0.680 s  1.81 m
2
2


1
1
xf  xi  vi  vf t  0  1.33 m s 0.680 s  0.454 m
2
2
1 2 1 2 1
2 1
2
m vf  m vi  60.0 kg 1.33 m s  60.0 kg  4.00 m s  427 J
2
2
2
2
(g)
(h)
(i)
1 2 1 2 1
2
m vf  m vi  120.0 kg 1.33 m s  0  107 J
2
2
2
The force exerted by the person on the cart must equal in magnitude and
opposite in direction to the force exerted by the cart on the person. The
changes in momentum of the two objects must be equal in magnitude and
must add to zero. Their changes in kinetic energy are different in
magnitude and do not add to zero. The following represent two ways of
thinking about “way”. The distance the cart moves is different from the
distance moved by the point of applicatio9n of the friction force to the cart.
The total change of mechanical energy for both objects together, -320J,
becomes +320J of additional internal energy in this perfectly inelastic
collision.