Linear Momentum and Second Newton’s Law
2nd Newton’s Law:
Definition of acceleration:


F  ma
We can write 2nd Newton’s Law as:

v
F m
t
Definition of momentum:

 v
a
t
Change in momentum:


p  mv


p  mv
1. Second Newton’s Law in terms of momentum:
 p
F
t
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We can get the same change in momentum
with a large force acting for a short time, or
a small force acting for a longer time.
F
same change
in momentum
This is why you should bend your knees when you land; why airbags work;
and why landing on a pillow hurts less than landing on concrete.
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Example: In an egg-tossing contest, two people toss a raw egg back and forth.
After each successful toss, each person takes a step back. Catching the egg
without breaking it becomes harder and harder. Usually the trick is moving your
hand down with the egg when you receive it. This works better because:
A. It decreases the change in momentum
B. It decreases the force on the egg
• If the flying egg has speed v , the change in momentum is:
Δp = 0 – mv = –mv
(independent of how you catch it)
• By moving your hand with the egg, you are increasing the time interval
over which this Δp must take place. So the average force on the egg
Fev  p / t
decreases.
•Catching the egg is harder and harder because its speed becomes larger
(and the required change in momentum, too), so exerting a small force
becomes harder as well.
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2. System of particles. Conservation of momentum


Ptot
Fext 
t
total momentum:
total external force:
 
 Pi  Ptot
 Fi  Fext



Fext  0  Ptot  0  Ptot  const
The total momentum of an isolated system of objects remains constant!
Example: Suppose rain falls vertically into an open cart
rolling along a straight horizontal track with negligible friction.
As a result of the accumulating water, the speed of the cart
1.
2.
3.
Increases
Does not change
Decreases
Mass is increasing
P = mv must be conserved (Fext = 0)
Speed must decrease
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75.0 kg
Example: Two guys of masses m1 =75kg
and m2=90kg pull on both ends of a rope
on an ice rink. After a couple of seconds,
the thin one is moving at 0.2 m/s. What is
the speed of the big one?
No friction, no net vertical force
p1i  p2i  p1 f  p2 f
v2 f
Fext = 0
ptotal is conserved
0  0  m1v1 f  m2v2 f
m1
75 kg
  v1 f  
0.2 m/s  0.17 m/s
m2
90 kg
Compare changes in linear momentum:
p1  m1  v1 f  v1i   (75 kg)(0.2 m/s - 0)  15 kg m/s
p2  m2  v2 f  v2i   (90 kg)(0.17 m/s - 0)  15 kg m/s
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Collisions (and explosions)
A
vA
vB
B
•Momentum is conserved in all collisions.
•Collisions in which kinetic energy is conserved as well are called elastic
collisions, and those in which it is not are called inelastic.
•With inelastic collisions, some of the initial kinetic energy is lost to thermal
or potential energy. It may also be gained during explosions, as there is
the addition of chemical or nuclear energy.
1. Completely inelastic collisions
(The objects stick together after collision, so there is only one final velocity)



vAf  vBf  v




mAv A  mB vB  mAv Af  mB vBf



mAv A  mB vB  (mA  mB )v


 mAv A  mB vB
v
mA  mB
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Example: A ball of mass, m1 = 2 kg has a horizontal velocity, of v1 = 7 m/s.
The ball collides into a cart full of sand, as shown below.
The cart has a mass, m2 = 8 kg and a horizontal velocity v2 = 1 m/s.
The ball and the cart are moving towards each other.
Find the velocity of the cart and the ball after the ball collides with the cart and
gets stuck in the sand.
m1 = 2 kg
v1 = 7 m/s
m2 = 8 kg
v2 = 1 m/s
v - ?
m1
v1
v2
m2
m1v1  m2v2  (m1  m2 )v
m1v1  m2v2
v
(m1  m2 )
2kg  7m / s  8kg 1m / s
v
 0.6m / s
(2kg  8kg )
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Example (Big block, small block): Consider the following two collisions
between two blocks of masses m and M (> m). In both cases, one of the
blocks is initially moving with speed v and the other is at rest. After the
collision, they move together. The final speed of the two objects is larger
when:
A. The big block is initially at rest
B. The small block is initially at rest
C. The speed is the same in both cases
m vm  MvM
vf 
m  M 
mvm  MvM  m  M v f
A.
vm  v
vM  0
v fA
mv

m  M 
B.
vm  0
vM  v
v fB
Mv

 v fA
m  M 
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Example: You want to knock down a large bowling pin by throwing a ball at it.
You can choose between two balls of equal mass and size. One is made of
rubber and bounces back when it hits the pin. The other is made of putty and
sticks to the pin. Which ball do you choose?
A. The rubber ball
B. The putty ball
C. It makes no difference
p
putty
p
p’
p+p’
rubber
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Example: Ballistic Pendulum
A simple device for measuring the speed of projectile
m,v1
M m
v2
M
1. Before collision:
p1=mv1
2
mv
K1  1 ;
2
p1=p2
M m
h
2. Immediately after
collision:
3. At the highest point:
p2=(m+M)v2
U1  0
( m  M )v 2
K2 
; U2  0
2
2
v1 
mv1 = (m+M)v2
(K 2  U 2 )  (K3  U 3 )
p3=0
U 3  (M  m) gh
(m  M )
v2
m
(m  M )v2
 ( M  m) gh
2
2
K 3  0;
(m  M )
v1 
2 gh
m
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2. Elastic Collisions
The kinetic energy of the system is conserved:
after the collision it is the same as that before
2a. Elastic Collisions in 1D
Conservation of momentum:
Conservation of energy:
1
2
m Av A  mB vB  m Av Af  mB vBf
(1)
2
m Av A2  12 mB vB2  12 m Av Af
 12 mB vBf2
(2)
(vA  vB )  vAf  vBf 
Velocity of A relative to
B before the collision
(2a)
Velocity of A relative to
B after the collision
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Example: Carefully place a small rubber ball (mass m) on top of a much
bigger basketball (mass M>>m) and drop these from some height h. What is
the velocity of the smaller ball after the basketball hits the ground, reverses
direction, and then collides with small rubber ball?
Relative
speed is 2v
v
m
M
v
Here too!
V?
3v
v
v
v
Remember that relative speed has to be equal before and after collision!
Relative velocity has opposite direction.
Before the collision, the basketball bounces up with v and the rubber ball is
coming down with v, so their relative velocity is –2v. After the collision, it
therefore has to be +2v!!
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This means that, after collision, the velocity of the smaller ball after is 3v.
Example: A steel ball with mass m and initial speed v0 collides headon with another ball of mass 2m that is initially at rest. What are the
final speeds of the balls?
1) Conservation of momentum:
2a) Relative velocity:
mv0  0  mv1  2mv2
v0  0  v1  v2 
Velocity of 1 relative to
2 before the collision
Adding these equations:
2v0  3v2
Velocity of 1 relative to
2 after the collision
v2  23 v0
v1   13 v0
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Example (Newton’s cradle - a row of adjacent steel-ball )
If two balls on the left are pulled to a certain
height h and released, what happens?
A. One ball rises on the right, but higher than h
pi = pf
2mvi = mvf
vf = 2vi ,
but then kinetic energy would not be conserved:
Ki = 2½mvi2 = mvi2
No!
Kf = ½mvf2 = 2mvi2
B. Two balls rise on the right at height h
pi = pf
2mvi = 2mvf
Ki = 2 ½mvi2 = mvi2
vf = vi.
Kf = 2 ½mvf2 = mvi2
Ok!
C. A ball will rise on each side to height h
Same height means |vi| = |vf| (from conservation of energy), but this
violates conservation of momentum: pi = 2mvi ; pf = mvi – mvi = 0
No!
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3. What happens to the total kinetic energy?
1. K is constant
Elastic collisions
When internal forces are conservative or objects are “hard”.
Examples: Elementary particles, billiard balls.
2. K decreases
Inelastic collisions
Whenever a deformation is involved.
Example: Most macroscopic collisions.
Special case: Perfectly inelastic collisions
When the objects stick together
Example: Pin and putty
Explosions
3. K increases
Superelastic
collisions
One body breaks into a number of parts. The
explosion mechanism provides the extra energy.
Some internal energy is transformed into
kinetic energy because of a collision.
Example: An excited atom hits another atom
and drops to a lower state without radiation.
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