Download Ch6 momentum and collision

Ch6 momentum and collision
6.1 Newton’s second law revisited.
The Newton’s second law is,


F  ma
Originally Newton stated his second law a little
differently.


F  ma

v
m
t

(mv )

t
He defined force as the change in an object’s
momentum(He used Latin word “motus”)
divided by the elapsed time.
Here, we define momentum,


p  mv
SI unit of momentum is Kg m/s
Also Kinetic energy can be written as
p2
KE 
2m
From the force –momentum equation, we can
study the characteristics of the momentum.


Ft  p (Impulse-momentum theorem)
6.3 and 6.4 Collisions
Momentum is conserved for any type of
collisions.
Elastic collision.
Both momentum and kinetic energy is
conserved.
We can find a general relationships between the
velocities of two objects
6.2 Newton’s third law revisited.
Newton’s third law is


F12  F21
From this equation, we can discover an
interesting law.

 v
F21   F12


p1
p 2

t
t


p1   p2




mv1 f  mv1i  (mv2 f  mv2i )




mv1i  mv2i  mv1 f  mv2 f
m1v1i  m2 v2i  m1v1 f  m2 v2 f
m1 (v1 f  v1i )   m2 (v2 f  v2i )
1
m1v12i  m2 v22i  m1v12f  m2 v22 f
2
m1v12f  m1v12i  (m2 v22 f  m2 v22i )
m1 (v1 f  v1i )(v1 f  v1i )   m2 (v2 f  v2i )(v2 f  v2i )
From these equations,
v1i  v2i  (v1 f  v2 f )
Inelastic collision
Kinetic energy is not conserved
As you can see the initial momentum equals the perfectly inelastic collision
The change in momentum of an object equals final momentum. The total momentum of an Two object collide and stick together.
the product of the force and the time elapsed. isolated system does not change over time. This
This is the definition of Impulse.
is the law of conservation of momentum.
 
I  Ft
Isolated system: No external net force.
Momentum and impulse are vectors.
We can think the whole universe is an isolated
All these equations are true even if the force is system. So the total momentum of the universe
not constant. Thus area under force vs. time
is conserved.
graphs is the impulse delivered by the force.
6.5 Rocket propulsion
Force is exerted on an object when mass of the
object is changing instead of the velocity of the
object.
M
v f  vi  ve ln( i )
Mf
Instantaneous thrust is
Fthrust  v2
M
t
In a crash test, a car of mass 1.50 x 103 kg collides
with a wall and rebounds. The initial and final
velocities of the car are vi = -15.0m/s vf = 2.60m/s, A rocket has a total mass of 1.00 x 105 kg and a
respectively. If the collision lasts for 0.150s, find
burnout mass of 1.00 x104 kg, including engines,
(a) the impulse delivered to the car due to the
shell and payload. The rocket blasts off from earth
collision and (b) the magnitude and direction of
and exhaust all its fuel in 4.00 min, burning fuel at
the average force exerted on the car.
a steady rate with an exhaust velocity of v =4.50
103
x
m/s. What thrust does the engine
developed at lift off.
A car with mass 1.50 x 103 kg traveling east at a speed
of 25.0 m/s collides at an intersection with a 2.50
x 103 kg van traveling north at a speed of 20.0
m/s. Find the magnitude and direction of the
velocity of the wreckage after the collision,
assuming that the vehicles undergo a perfectly
inelastic collision and assuming that friction
between the vehicle and the road can be
neglected.
e