“Impulse” and “Momentum”
Last class …
Elastic vs. inelastic collision
Today…
Elastic collision in 1D
Collision
ll
in 2D
Center of mass
Motion of system of particles
(Motion of center of mass)
Elastic Collisions in One Dimension
Both total momentum & total kinetic energy are conserved.
m1v1,i + m2 v2,i = m1v1, f + m2 v2, f
&
1
1
1
1
m1v12i + m2 v22i = m1v12f + m2 v22 f
2
2
2
2
v2i
v1i
Before collision
m1
+x direction
m2
v1f=?
f
v2f=?
v2f
?
After collision
v1, f =
m1 − m2
2m2
v1,i +
v2,i
m1 + m2
m1 + m2
and
v2, f =
2m1
m − m1
v1,i + 2
v2,i
m1 + m2
m1 + m2
(see text for proof)
Example
In 1D elastic collision, if m1=m2, vi1=+10 m/s, v2i=0,
then, after collision v1f= _____ and v2f= _____.
a) 5 m/s; 5 m/s
b) -5 m/s; 5 m/s
c) 0 m/s; 10 m/s
d) 0 m/s; -10 m/s
e) -10 m/s; 0 m/s
V2i=0
V1i=10m/s
Before collision
After collision
v1, f =
m1
m2
v1f=?
m1 − m2
2m2
v1,i +
v2,i
m1 + m2
m1 + m2
v2f=?
and
v2, f =
+x direction
2m1
m − m1
v1,i + 2
v2,i
m1 + m2
m1 + m2
1
Collisions in Two Dimensions
If net external force is zero,
Æ
Or
G
G
Pnet ,i = Pnet , f
, where
net momentum is conserved.
G
G
G
Pnet = m1v1 + m2v2 + ...
y
Pnet ,i , x = Pnet , f , x
V1
Pnet ,i , y = Pnet , f , y
If net kinetic energy is
conserved Æ Elastic collision
not conserved Æ Inelastic collision
v1
m1
If moving together after collision
Æ Perfectly Inelastic collision
(Kinetic energy is not conserved.)
v2 m2
x
V2
Example 2: Collision at an intersection
A 1500 kg car traveling east with a
speed of 25 m/s collides at an
intersection with a 2500 kg van
traveling north at a speed of 20 m/s.
Find the direction and magnitude of
the velocity of the wreckage after
the collision, assuming the vehicles
stick together after the collision.
In this collision,
Is total momentum conserved?
Is total kinetic energy conserved?
How could we analyze the motion of extended objects,
or system of particles?
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Concept of Center of Mass
m1=3 kg
m3=10 kg
m2=6 kg
For a system of particles or an extended object,
“Center of mass” is an “average” position for mass distribution.
Definition of center of mass (com) in 1D
m1
m2
x1
x2
m3
x3
In 1D,
xcom
m x + m2 x2 + ...
= 1 1
=
m1 + m2 + ...
∑m x
∑m
i i
i
i
i
xi
, where
is position of mass
mi
Definition of center of mass (com) in 2D, 3D
∑m x
∑m
my
m y + m y + ... ∑
y =
=
m + m + ...
∑m
mi zi
m z + m2 z2 + ... ∑
zcom = 1 1
= i
m1 + m2 + ...
∑i mi
xcom =
m1 x1 + m2 x2 + ...
=
m1 + m2 + ...
i i
i
i
i
i
1 1
2
1
2
2
i
i
i
G
G
m r + m2 r2 + ...
G
rcom = 1 1
=
m1 + m2 + ...
, where
i
com
( xi , yi , zi )
is the position of
mi
G
∑m r
∑m
i i
i
i
i
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Example 3
Three particles of masses m1 = 1.1 kg, m2 = 2.5 kg, and m3 = 3.4 kg
are located as shown in the figure: m1 is at (0,0), m2 is at (140
m,0), and m3 is at (70 m, 120 m). Find the coordinate of the
center of mass.
Example
A two-section piece, represented by the gray area on the figure, is
cut from a metal plate of uniform thickness. The point that
corresponds to the center of mass of this piece is closest to
(a)
(b)
(c)
(d)
(e)
1
2
3
4
5
Velocity of center of mass (com)
vx ,com =
Δxcom m1vx ,1 + m2 vx ,2 + ...
=
=
Δt
m1 + m2 + ...
∑mv
∑m
i x ,i
i
i
i
(m1 + m2 + ...)vx ,com = m1vx ,1 + m2 vx ,2 + ... = Pnet , x
Similar for y and z components
G
G
m v + m2 v2 + ...
G
=
vcom = 1 1
m1 + m2 + ...
G
∑m v
∑m
i i
i
i
G
G
G
G
Pnet = m1v1 + m2 v2 + ... = Mvcom
i
where
M = m1 + m2 + ...
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Acceleration of center of mass (com)
Δvx ,com
ax ,com =
Δt
=
m1ax ,1 + m2 ax ,2 + ...
m1 + m2 + ...
=
∑m a
∑m
i
x ,i
i
i
i
Similar for y and z components
G
mi ai
G
G
m1a1 + m2 a2 + ... ∑
G
acom =
= i
m1 + m2 + ...
∑ mi
i
G
G
G
G
G
M a CM = m 1a + m 2 a + m 3 a + ...+ m n a
1
2
G
G
G
G
m1a1 = F1, net = F1,ext + F1,int
Newton’s
3rd
Æ
G
M a CM =
3
&
law for internal forces:
G
G
G
F1,int + F2,int + ... + Fn ,int = 0
JG
F
1,ext
+
JG
F
2 ,ext
+
n
G
G
G
G
m2 a2 = F2,net = F2,ext + F2,int
G
G
Ffrom 2 on 1 + Ffrom 1 on 2 = 0
JG
F
3,ext
G
G
Macom = Fnet ,ext
+ ... +
JG
F
n ,ext
Newton’s second law for center of mass
G
Fnet ,ext
G
G
Fnet ,ext = Macom
: Sum of all external forces that act on the system
(Internal forces are not included)
M = m1 + m2 + ...
G
acom
: Total mass of the system
: Acceleration of the center of mass
Internal forces do NOT change the motion of C.O.M.!!
5
Motion of center of mass under gravity force
Newton’s second law for C.O.M.:
Under gravity,
G
G
Fnet ,ext = Macom
G
Fnet ,ext = − m1 gj − m2 gj − ...
= −(m1 + m2 + ...) gj = − Mgj
G
∴− Mgj = Macom
G
∴ acom = − gj
Æ Usual projectile motion
Under uniform gravity force near earth surface,
motion of COM is usual projectile motion
(even though each point may undergo complicated motion)
Center of mass moves like a particle of mass M under the net
external force.
Motion of COM is simple!
Example 1
A 2.0 kg particle has a velocity (2.0 i -3.0 j) m/s, and
a 3.0 kg particle has a velocity (1.0 i + 6.0 j) m/s.
Find (a) velocity of the center of mass and (b) the total momentum
of the system.
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Example
Two objects with unknown mass and velocity collide and stick
together moving at 3 m/s along x direction.
Assuming that net external force on the two objects is zero, what
is the velocity of the center of mass before the collision?
Example 1: 90 degree deflection rule in a game of pool
G
v2, f
m
m
G
v1,i
G
v1,1 f
Assume that the collision is elastic,
and the two balls have the same mass.
Show that the angle between the
outgoing balls is 90 degree.
(No forward, back or side spin is in effect.)
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