Transcript Momentum - Ohio Wesleyan University

Momentum
• Consider systems consisting of more than a single
particle and how these multiple particles interact
with each other
– “Many–body” problems  more complicated, so we need
effective ways of describing “systems” of particles




v mv 
Law: Fnet  ma  m t  t
• Start with Newton’s 2nd

• Quantity m v is called the momentum of a particle:


p  mv
• For a system of particles:
m1
m1v1
m2v2
m2
m3v3
m3

  



Total momentum of system: ptot  p1  p2  p3  mv1  mv2  mv3
Impulse
• Thus Newton’s 2nd


p
Law can be written as: Fnet  t
– This is (essentially) the original way Isaac Newton wrote it!
– Assumes constant force producing constant acceleration
– The shorter the time period over which a given momentum
change occurs, the larger the external force
• Product of the net force and the time
 interval over
which it acts is called the impulse J of the net force:
 

– Assumes constant force
J  Fnet t  Fnet t2  t1 
– If net force is not constant,
impulse is determined by 1 of 2 ways:
Area under F
vs. t curve
Fave  t
Impulse
 –
 Momentum Theorem

p2  p1
– So Fnet 
and thus
t 2  t1
  
J  p2  p1 (Impulse –
Momentum Theorem)
– Same result holds when net force is not constant
• Now consider a system consisting of the Sun and
the Earth:
Sun (m1)

F
Earth (m2)

F
– Sun and Earth exert equal magnitude (but opposite
direction) gravitational forces on each other
– No other external forces are acting on them
• Apply Newton’s 2nd Law to each mass:
 p1
F
t
 p 2
F 
t
Conservation
of
Momentum
 





p tot
 p1  p2 
• Thus p1 p2
0
0

 F F 0
t
t
t
t

 
• Where ptot  p1  p2 = total momentum of the system
• This means that the total momentum of the system
is conserved (or constant) if the net external force
acting on the system is zero
• Note that this also follows from Newton’s 3rd Law
• Objects exchange momentum through mutual
interactions (gravity, collisions, etc.)
• Momentum conservation can hold even when
mechanical energy isn’t conserved
CQ 1: Two 1-kg carts with spring bumpers undergo a
collision on a frictionless track as shown in the before
and after pictures below. The total momentum of the
system is equal to:
A)
B)
C)
D)
0 kg-m/s before the collision and 0 kg-m/s after the collision.
–4 kg-m/s before the collision and 4 kg-m/s after the collision.
–8 kg-m/s before the collision and 8 kg-m/s after the collision.
8 kg-m/s before the collision and 0 kg-m/s after the collision.
CQ2: Interactive Conceptual Example:
Basketball vs. Flower Pot Drop
How does the change in momentum of the
basketball compare with the change in momentum
of the flowerpot after each object hits the table?
A)
The change in momentum of the basketball is the same
as the change in momentum of the flowerpot.
B) The change in momentum of the basketball is twice the
change in momentum of the flowerpot.
C) The change in momentum of the basketball is half of the
change in momentum of the flowerpot.
D) The change in momentum of the flowerpot is twice the
change in momentum of the basketball.
PHYSLET Exercise 8.5.1, Prentice Hall (2001)
Example Problem #6.24
A 730-N man stands in the middle of a frozen pond of
radius 5.0 m. He is unable to get to the other side
because of a lack of friction between his shoes and
the ice. To overcome this difficulty, he throws his
1.2-kg physics textbook horizontally toward the
north shore at a speed of 5.0 m/s. How long does
it take him to reach the south shore?
Solution (details given in class):
62 s
CQ 3: Interactive Example Problem:
Saving an Astronaut
How much time elapses from when the
astronaut removes her oxygen tank until she
arrives back at the ship?
A)
B)
C)
D)
E)
5s
10 s
15 s
20 s
60 s
(ActivPhysics Online Problem #6.6, Pearson/Addison Wesley)
CQ 4: A boy is sliding down a long icy hill on his
sled. In order to decrease his mass and
increase his velocity, he drops his heavy winter
coat and heavy boots from the sled while he is
moving. Will his strategy work?
A) No, because he loses the potential energy
of the objects that he leaves behind.
B) No, because although his kinetic energy
increases, his momentum decreases.
C) Yes, because although his kinetic energy
decreases, his momentum increases.
D) Yes, because although his momentum
decreases, his kinetic energy decreases.
Collisions
• Inelastic collisions
– Example: cars stick together after a collision
BEFORE collision:
v1i
You (m1)
Me (m2) v2i = 0
AFTER collision (cars locked together):
vf
Energy used to do work (destroy fenders)
What is vf?


Conservation of momentum: pbefore  pafter



pbefore  m1v1i  0  m1v1i


pafter  m1  m2 v f
Collisions


m1v1i  m1  m2 v f

 vf 
m1 
v1i
m1  m2
• Note that if m1 = m2 then vf = ½ v1i
• In inelastic collisions, the kinetic energy of system
after collision < kinetic energy of system before
collision (since mechanical energy is typically
expended to do work)
– K.E. before collision = K1 = ½ m1v1i2
– K.E. after collision = K2 = ½ (m1 + m2)vf2 =
½ (m1 + m2)[m1 / (m1 + m2)]v1i2
– So K2 / K1 = m1 / (m1 + m2) < 1 so K2 < K1
• Elastic collisions
– Forces between colliding bodies are conservative
– Kinetic energy is conserved (may be temporarily
converted to elastic potential energy)
– Momentum is conserved
Elastic Collisions
• Head-on collision where one object is at rest before
m2, v2i = 0
the collision:
m1, v1f
m1, v1i
• Conservation of kinetic energy gives:
½ m1v1i2 = ½ m1v1f2 + ½ m2v2f2
• Conservation of momentum gives:
m1v1i = m1v1f + m2v2f
• Combining these 2 equations:
m1  m2
v1 f 
v1i
m1  m2
v2 f
2m1

v1i
m1  m2
m2, v2f
Elastic Collisions
• Interesting cases: (1) m1 = m2
– v1f = 0 and v2f = v1i
• (2) m1 << m2
– v1f  – v1i
– v2f << v1i
– Similar to result of ping-pong ball (m1) striking stationary
bowling ball (m2)
• (3) m1 >> m2
– v1f  v1i
– v2f  2v1i
– Similar to result of bowling ball (m1) striking stationary
ping-pong ball (m2)
• If v2i ≠ 0, then in general: v1i – v2i = –(v1f – v2f)
CQ 5: A block of mass m1 slides across a frictionless
surface with speed v1 and collides with a stationary
block of mass m2. The blocks stick together after the
collision and move away with speed v2. Which of the
following statements is (are) true about the blocks?
I.
m1v1  m1  m2 v2
1
1
2
II.
m1v1  m1  m2 v22
2
2
III. v1  v2
A)
B)
C)
D)
I only
II only
I and II only
I, II and III
Example Problem #6.73
A tennis ball of mass 57.0 g is held just
above a basketball of mass 590 g.
With their centers vertically aligned,
both balls are released from rest at the
same time, to fall through a distance of
1.20 m, as shown at right.
(a) Find the magnitude of the downward
velocity with which the basketball
reaches the ground.
(b) Assume that an elastic collision with
the ground instantaneously reverses
the velocity of the basketball while the
tennis ball is still moving down. Next,
the two balls meet in an elastic
collision. To what height does the
Solution (details given in class):
tennis ball rebound?
(a) 4.85 m/s
(b) 8.41 m
2 – D Collisions
• Can also have a “glancing” collision:
m2, v2i = 0
q
m1, v1f
j
m1, v1i
x
m2, v2f
• Now the final velocities (and hence the momenta)
have horizontal and a vertical components
• Use the component form of conservation of
momentum for the system of both balls:
px,initial = px,final
2-D Collisions Interactive
py,initial = py,final
• True for elastic or inelastic collisions
Example Problem #6.49
A 2000-kg car moving east at 10.0 m/s collides with a 3000-kg car
moving north. The cars stick together and move as a unit after
the collision, at an angle of 40.0° north of east and a speed of
5.22 m/s. Find the speed of the 3000-kg car before the
collision.
BEFORE collision:
AFTER collision: y (N)
y (N)
v1 = 10 m/s
vf = 5.22 m/s
vfy
x (E)
40°
m1 = 2000 kg v
vfx
2
x (E)
m +m
1
m2 = 3000 kg
Solution (details given in class):
5.59 m/s
2
vfx = vf cos40° = 4.00 m/s
vfy = vf sin40° = 3.36 m/s
CQ 6: Ball A moving at 12 m/s collides elastically with
ball B as shown. If both balls have the same mass,
what is the final velocity of ball A?
(Note: sin60° = 0.87; cos60° = 0.5)
A)
B)
C)
D)
3 m/s
6 m/s
9 m/s
12 m/s
Example Problem #6.41
A 12.0-g bullet is fired horizontally into a 100-g
wooden block that is initially at rest on a
frictionless horizontal surface and connected
to a spring having spring constant 150 N/m.
The bullet becomes embedded in the block. If
the bullet–block system compresses the
spring by a maximum of 80.0 cm, what was
the speed of the bullet at impact with the
block?
Solution (details given in class):
273 m/s
Example Problem #6.56
A bullet of mass m and speed v passes completely through a
pendulum bob of mass M as shown above. The bullet
emerges with a speed of v/2. The pendulum bob is
suspended by a stiff rod of length l and negligible mass.
What is the minimum value of v such that the bob will barely
swing through a complete vertical circle?
Solution (details given in class):
4M
v
m
gl
Rocket Propulsion
• Rocket propulsion can be understood in two ways:
1. Conservation of momentum: Spent fuel (gasses)
have momentum directed away from rocket; rocket
has momentum in opposite direction
2. Newton’s Third Law: Rocket exerts force on spent
fuel gasses, expelling them outward; gasses in
response exert force on rocket in opposite direction
• Rocket needs nothing to “push against” in order to
move, otherwise it would not work in the vacuum of
space!