Linear Momentum and Second Newton’s Law
 2nd Newton’s Law for any object:
 Definition of acceleration:


F  ma
We can write 2nd Newton’s Law as:

dv
F m
dt
 Definition of momentum:
 Change in momentum:


dp  mdv


p  mv
Second Newton’s Law in terms of momentum:
 dp
F
dt

 dv
a
dt
or
 d(mv)
F
dt
or


Fdt  dp
Impulse


Fdt  dp
p2
 t2 
 


J   Fdt   dp  p 2  p1  p
t1
p1
 

J  Fev t 2  t1   Fev t  p
The impulse tells us that 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.
Collisions and Impulse
The impulse tells us that 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.
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.
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 impulse
C. 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)
• The impulse is just the same!
J = Δp = - mv
• 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.
By the way: 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.
System of particles. Conservation of momentum


dP1 
 F12  F13  ...
dt



dP2
 F21  F23  ...
dt

dP1 
 F31  F32  ...
dt
................................


dPi
 dt   Fi

 F1

 F2

 F3

F13

F23

F21

F12

F32

F31
 
 Pi  Ptot total momentum!
 Fi  Fext total external force!


F12   F21


F13   F31


F23   F32


dPtot
 Fext
dt



Fext  0  dPtot  0  Ptot  const
The total momentum of an isolated system of objects remains constant
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
ptotal is
conserved
Fext = 0
No net vertical force
p1,i  p2,i  p1,f  p2,f
0  0  mv
1 1,f  m2v2,f
75 kg
m1
  v1,f  
0.2 m/s  0.17 m/s
m2
90 kg
v2,f
Compare changes in linear momentum:

 m v

  (90 kg)(0.17 m/s - 0)  15 kg m/s
p1  m1 v1,f  v1,i  (75 kg)(0.2 m/s - 0)  15 kg m/s
p2
2
2,f
 v2,i
The force produced a momentum transfer.
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.
Completely inelastic collisions
The objects stick together after collision, so there is only one final velocity
'
'
'
'
A
B
A A
B B
A A
B B



v v v




m v m v m v m v



mAvA  mBvB  (mA  mB )v


 m A v A  mB v B
v
m A  mB
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 - ?
v
m1
1
v
2
m2
m1v1  m2 v2  (m1  m2 )v
m1v1  m2 v2
v
(m1  m2 )
2kg  7m / s  8kg 1m / s
v
 0.6m / s
(2kg  8kg)
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.
mvm  Mv M  m  M v f
A.
vm  v
vM  0
v fA
mv

m  M 
mvm  Mv M
vf 
m  M 
B.
vm  0
vM  v
v fB
Mv

 v fA
m  M 
Example: An eagle of mass, mA = 6.0 kg moving with speed vA = 5.0 m/s
is on collision course with a second eagle of mass mB = 4.0 kg moving at
vB = 10.0 m/s in a direction perpendicular to the first. After they collide,
they hold onto one another. At what speed they moving after the
collision?
mA=6.0 kg
vA=5.0 m/s
mB=4.0 kg
vB=10.0 m/s
v-?

vB

pB

vA

pA
m  m A  m A  6.0 kg  4.0 kg  10.0 kg
p A  mA p A  6.0kg  5.0m / s  30.0kg  m / s
pB  mB pB  4.0kg 10.0m / s  40.0kg  m / s
p
v 
m
p A2  pB2

m
 

p  pA  pA
p
p A2  p B2
30.0kg  m / s 2  40.0kg  m / s 2
10kg
 5m / s