Download p = Mv p ≡ mv p = mv

Warm Up:
Which of the following has the greatest
momentum?
a) A 100 g bullet moving at 300 m/s
b) A 5000 kg train car moving at 1.5 m/s
c) A 75 kg motorcyclist moving at 35 m/s
Linear Momentum
What is Linear Momentum?
“Trains are big. They’re hard to stop.”
p ≡ mv
TRAIN
p = Mv
More on Momentum
• Units: kg·m/s
• Momentum is a vector: px=mvx, py=mvy, etc.
• Like KE affected by mass and velocity
BULLET…Also hard to stop
p = mv
Why Study Momentum?
Another powerful way to more
efficiently analyze systems of
interacting objects:
• UNLIKE KE increases linearly with both mass
and velocity
• UNLIKE KE, can be negative or positive!
ESPECIALLY COLLISIONS
and EXPLOSIONS
1
Force and Momentum: Derivation
Recall Newton’s Second Law:
a=
dv
dt
F =m
If mass is
constant, build it
into the derivative.
F=ma
dv
dt
F =
dp
dt
So force and momentum are
related…
F=
dp
dt
What if ΣF=0?
dp
=0
dt
p = constant
Conservation of Linear Momentum
pi = pf
Systems with no net external forces have constant
momentum. (Impulse approximation helps here.)
2-Particle System
p1
m1
F21
Conservation of Linear Momentum
dp1
dt
dp2
F12 =
dt
If there is no net external force, then
Internal Forces
0=
F21 =
F12
p2
m2
system
Newton’s 3rd Law gives:
F12 = −F21
ΣFinternal = 0
Ftot =
Fext +
Fint =
0
0
dp1 dp2
+
dt
dt
d ( p1 + p2 )
dt
ptot = p1 + p2
is conserved.
2
In general
Example two particle system:
ΣFinternal = 0
Ftot =
If:
A bullet with mass 35 g moving at 320 m/s embeds
itself in a 2.40 kg wooden block sitting on a
frictionless surface. How fast do the bullet and
block move together after the collision?
demo
Why?
Fext
Fext =0
Then:
Define system: bullet and block, so
Σpinit = Σpfinal = constant
pi = pf
mbulletvbullet,i= (mbullet+mblock)vf
system
system
vf = 4.6 m/s
.035(320)= (2.4+.035)vf
Conservation of Linear Momentum
Note: this is 3 equations: px, py, pz
Impulse approximation
What if force 0?
dp = F dt
F
Consider a very large force acting over
a very short time.
ex: hitting a hockey puck with a stick.
f
F dt ≡ I
∆p = pf − pi =
i
impulse
Impulse-momentum theorem:
Ι = ∆p
An Impulse is a change in momentum:
- The effect of a force acting over a time
- The agent of momentum transfer.
- Handy for time-varying forces.
t
During the short time of the impulse
from the large force, we can safely
ignore other forces acting on the
puck.
F
F
t
∆t
Ι = F ∆t
Useful to approximate the impulse as due to a
constant force F acting over the same time ∆t.
Applications: collisions & explosions!
demo
3
Impulse Example
Collisions
A 60.0 g tennis ball is initially horizontally east at
50.0 m/s. A player hits the ball with a racket. The
ball is in contact with the racket for 35 ms. After
the hit, the ball is moving west at 40.0 m/s. What
average force did the racket exert on the ball?
Systems of particles interacting via impulsive
forces (large, short time)
Impulse = p=pf- pi =.060 (-40) - .060 (50) =-5.4 kg m/s
But, Impulse also =F t => F(0.035 s)=-5.4 kg m/s
=> F= -154 N
Note: Compare with
force/impulse to stop ball… Negative => Westward
1D collision:
final
m2
v1i
v2i
final
m1
m2
v1i
Choose system.
m2
v1f
v2i
v2f
Two masses
Impulse approximation:
Only consider large contact forces;
Instants just before & after collision
Apply linear momentum conservation.
Types of collisions:
initial
m1
initial
m1
m1
m2
v1f
v2f
Linear momentum conservation:
ΣPi = ΣPf
m1v1i – m2v2i = m1v1f + m2v2f
Given v1i and v2i we have one equation,
but two unknowns.
Elastic
linear momentum is conserved
total kinetic energy is conserved
Inelastic
linear momentum is conserved
total kinetic energy is NOT conserved
Totally inelastic collisions: particles stick
together.
Where does the energy go?
demo
4
1D Perfectly Inelastic Collisions
initial
m1
final
m1+m2
m2
v1i
One more inelastic example:
Explosion!
5.0 kg vi=0
m2
vf
v2i
v1f = v2f = vf
2.0 kg,
v1f =3.6 m/s
Momentum conservation:
m1v1i – m2v2i = m1v1f + m2v2f
m1v1i – m2v2i = (m1+m2)vf
3.0 kg,
v2f =?
pi=pf
0 = 2(-3.6) +3(v2f) => v2f = 2.4 m/s
vf = (m1v1i - m2v2i)/(m1+m2)
Caution with signs!
1D Elastic Collisions
initial
m1
final
m1
m2
v1i
1D Elastic Collisions cont.
v1f
v2i
Some algebra shows:
m2
v1i – v2i = − (v1f − v2f)
v2f
Momentum conservation:
m1v1i – m2v2i = m1v1f + m2v2f
Kinetic Energy is conserved:
½m1v21i + ½m2v22i = ½ m1v21f + ½ m2v22f
Two equations
can solve for two unknowns.
Use these:
v1 f =
m1 − m2
2m2
v1i +
v 2i
m1 + m2
m1 + m2
v2 f =
m − m1
2m1
v1i + 2
v 2i
m1 + m2
m1 + m2
5