Transcript PowerPoint Presentation - Chapter 3 Kinematics in 2d

Chapter 7
Impulse and Momentum
1) Definition
p = mv
2) Impulse
For a constant force,
r
J  Ft
From the 2nd law,
r
r
(mv) p
F

,
t
t
so
r
r
J  Ft  p
If the force is not constant, use the average force
r
r
J  Ft  p
7) Conservation of momentum in 2d
r
r r 
p1  p2  p1  p2
v1
v2
r
r
r
r
m1v1  m2 v2  m1v1  m2 v2
m1v1x  m2 v2 x  m1v1x   m2 v2 x 
m1v1y  m2 v2 y  m1v1y  m2 v2 y
If initial conditions are known, this
gives 2 equations, with 4 unknowns,
so more information is needed.
v1’
v2’
7) Conservation of momentum in 2d
a) Inelastic collision: p1’ = p2’ = p’ reduces unknowns to
px’ and py’
Example: Find v’ if m1 = 1450 kg, m2 = 1750 kg, v1 = 11.5 m/s, v2 = 15.5 m/ss
• conservation of x momentum
m2
v2
m1
v1
v '  vx  2  vy 2  9.95 m/s
tan  
vx 
m1v1x  0  (m1  m2 )vx 
x
vx  

y
vy 
m1v1x  m2 v2 x  m1v1x   m2 v2 x 
   58.4º
v’
m1
v1x  5.21 m/s
m1  m2
• conservation of y momentum
m1v1y  m2 v2 y  m1v1y  m2 v2 y
0  m2 v2 y  (m1  m2 )vy
vy  
m2
v2 y  8.48 m/s
m1  m2
7) Conservation of momentum in 2d
b) Elastic collision: Energy conservation adds 3rd equation:
1
2
mv12  12 mv2 2  12 mv1 2  12 mv2 2
The last condition is determined by the shape & location of impact:
e.g.
F
For a billiard ball collision, the angle of the object ball is
determined by the line through the centres at the point of contact.
7) Conservation of momentum in 2d
b) Elastic collision: Energy conservation adds 3rd equation:
v2’
Example: Cue ball angle:
F
v1
Conservation of momentum
r
r
v1  v1  v2

v2’
v1’
v1
1
• identical masses
• elastic collision
• m2 initially at rest
Find   1   2
2?
Conservation of energy
v12  v1 2  v2 2
v1’
Therefore, by Pythagoras
  90º
8) Centre of Mass
a) Acceleration and force:
The centre-of-mass of a system of particles (or 3d object)
reacts to the sum of the forces like a point particle with a mass
equal to the total mass of the particles.
Ftotal
r
 mtotal aCM
If the total force is zero, the centre-of-mass does not
accelerate. (If there are no external forces, the internal
forces sum to zero by the 3rd law.)
r
r r
r
r
r
r
For 2 masses, Ftotal  F1  F2 , where F1  m1a1 , and F2  m2 a2 are the
forces on the two masses
aCM
r
r
m a  m2 a2
 1 1
m1  m2
8) Centre of Mass
b) Momentum and velocity
Since a = v/t,
r
r
m1 v1  m2 v2
r
vCM 
m1  m2
But, since vCM = 0 if v1 = 0 and v2 = 0, this becomess
vCM
r
r
m1v1  m2 v2

m1  m2
The numerator represents the total momentum, which is conserved
in the absence of external forces, so again, the centre-of-mass
velocity is constant. Note that the CM momentum is simply equal
to the total momentum:
r
r
r
(m1  m2 )vCM  m1v1  m2 v2
8) Centre of Mass
c) Position of the centre of mass
Since v = x/t, the above gives
r
r
m1 x1  m2 x2
r
xCM 
m1  m2
If 2 particles coincide, they also coincide with the CM, so
xCM
r
r
m1 x1  m2 x2

m1  m2
In one dimension,
For m1 = 5.0 kg, m2 = 12 kg, x1 = 2.0 m, and x2 = 6.0 m,
xCM = 4.8 m