Download 4.1 The Concepts of Force and Mass

Chapter 7
Impulse and Momentum
• Impulse-Momentum Theorem
• Principle of Conservation of Linear Momentum
• Collisions in One Dimension
• Collisions in Two Dimensions
• Centre of Mass
Impulse and Momentum
Newton’s 2nd Law:
Or


v
F m
t
So,


Ft  mv
F  ma


Ft is the impulse of the force F


Define the momentum: p  mv



Then, Ft  mv  p (impulse
– momentum theorem)
Impulse = change in momentum



Ft  mv  p


If F  0 then p  0
That is, momentum is conserved when the net force
acting on an object is zero.
This applies also to an isolated system of two or more
objects (no external forces) that may be in contact - the
total momentum is conserved.
Compare Newton’s first law:
velocity is constant when the net force is zero.
Alternative formulation of Newton’s second law



Ft  mv  p
or :
 p
F
t
The net force acting on an object is equal to the rate
of change of momentum of the object.
Example:
A 0.14 kg baseball has an initial velocity v0 = -38 m/s
as it approaches a bat. The bat applies an average
force F that is much larger than the weight of the ball.
After being hit by the bat, the ball travels at speed
vf = +58 m/s.
a) The impulse applied to the ball is

 
mv  m ( vf  v0 )
 0.14 kg 58   38 m/s
 13.44 N  s or kg  m/s
b) The bat is in contact with the ball for 1.6 ms.
So the average force of the bat on the ball is:
 mv 13.44 N  s
F

 8400 N
3
t
1.6  10 s
Problem 7.13:
A 0.047 kg golf ball strikes a hard, smooth floor at an
angle of 30°, and rebounds at the same angle.
What is the impulse applied to the golf ball by the
floor?
Conservation of Momentum
Two isolated masses collide. The
initial total momentum is:
  
p  p1  p2


With p1  m1 v01


p2  m2 v02
While the masses are in contact, they
exert equal and opposite forces on each
other (Newton’s third law).


F12  F21
So the impulse acting on m1 is equal in magnitude and opposite in
direction to the impulse acting on m2


Therefore, p1  p2
(change in momentum = impulse)




After the collision: p1  mv f1  p1  p1






p2  mvf2  p2  p2  p2  p1
So, the total momentum after the collision is:
That is, total







p   p1  p2  p1  p1   p2  p1 
 
 p1  p2

p
 
 
momentum is conserved: p1  p2  p1  p2
Problem 7.C13:
An ice boat slides without friction horizontally
across the ice. Someone jumps vertically down from
a bridge onto the boat.
Does the momentum of the boat change?
As the momentum of the person is
downward, not sideways, the horizontal
momentum of the boat is unchanged.
Mv o  m  M vf
v 0M
vf 
m M
As the mass of the boat is increased by
the mass of the person, the boat moves
more slowly, so that the momentum is
unchanged.
Problem 7.16:
A 55-kg swimmer is standing on a stationary 210-kg
floating raft. The swimmer then runs off the raft
horizontally with a velocity of +4.6 m/s relative to the
shore. Find the recoil velocity that the raft would have
if there were no friction and resistance due to the
water.
The freight car on the left catches up with the one on the right
and connects up with it. They travel on with the same speed vf.
Conservation of momentum:
m1v1  m2v 2  m1  m2 vf
m1v1  m2v 2
vf 
m1  m2
m1v1  m2v 2
vf 
m1  m2
Example:
m1  65,000 kg,
m2  92,000 kg
v 01  0.8 m/s,
v 02  1.3 m/s
65000  0.8  92000  1.3
vf 
 1.09 m/s
65000  92000
Kinetic Energy:
1
1
2
2
KE i  m1v 01  m2v 02
 98540 J
2
2
Missing 5274 J
1
KEf  m1  m2 vf2  93266 J
2
What happened to the missing energy?
This was an inelastic collision – some of the kinetic energy was
converted to heat.
Problem 7.28:
A projectile (mp = 0.2 kg) is fired at and embeds itself in
a target (mT = 2.5 kg). The target (with the projectile in
it) flies off after being struck.
What percentage of the projectile’s incident kinetic
energy does the target (with the projectile in it) carry off
after being struck?
The Ballistic Pendulum
How fast was the bullet?
Mass of bullet: m1 = 0.01 kg
Mass of block: m2 = 2.50 kg
• Elastic collision: the total kinetic energy after collision is equal
to the total before collision.
• Inelastic collision: the total kinetic energy is not conserved. If
objects stick together after collision, the collision is “perfectly
inelastic” – no bouncing of one object from the other.
Elastic Collision
Example: A ball of mass m1 = 0.25 kg makes a perfectly elastic
collision with a ball of mass m2 = 0.8 kg.
Initial momentum  m1v1  0
Momentum after impact  m1v f1  m2v f2
Momentum is conserved : m1v1  m1v f1  m2v f2
Conserved Momentum : m1vi1  m1v f1  m2v f2
So,
m1v i1  m1v f1
v f2 
m2
Then use conservation of energy
1
1
1
2
2
m1v i1  m1v f1  m2v f22
2
2
2
Combine the two
equations and after
some algebra:
 m1  m2 

v f1  v i1 
 m1  m2  if m1  m2 ,
 2m1  then v f1  0, v f2  v i1

v f2  v i1 
 m1  m2 
 m1  m2 

v f1  v i1 
 m1  m2 
 2m1 

v f2  v i1 
 m1  m2 
if m1  0.25 kg, m2  0.8 kg ,
and v i1  5 m/s
then
v f1  2.62 m/s , v f2  2.38 m/s
• so, the lighter ball bounces back from the heavier ball
• momentum is conserved
• kinetic energy is conserved
Problem 7.37:
An electron collides elastically with a stationary
hydrogen atom. The mass of the hydrogen atom is
1837 times that of the electron. Assume that all
motion, before and after the collision, occurs along
the same straight line.
What is the ratio of the kinetic energy of the
hydrogen atom after the collision to that of the
electron before the collision?