Mar 12 Discussions
Conservation of Momentum
Space Odyssey
• You have been hired to check the technical correctness of
an upcoming made-for-TV murder mystery. The mystery
takes place in the space shuttle. In one scene, an
astronaut's safety line is sabotaged while she is on a space
walk, so she is no longer connected to the space shuttle.
She checks and finds that her thruster pack has also been
damaged and no longer works. She is 200 meters from the
shuttle and moving with it. That is, she is not moving with
respect to the shuttle. There she is drifting in space with
only 4 minutes of air remaining. To get back to the shuttle,
she decides to unstrap her 10-kg tool kit and throw it away
with all her strength, so that it has a speed of 8 m/s. In the
script, she survives, but is this correct? Her mass, including
space suit (but not the tool kit), is 80 kg.
Space Odyssey
D  200m
t max  4 min  240s
need V  D /t max  200/240s  0.83m/s
ptot  0  MV  m v
M  80kg
m  10kg
v  8m/s
V  m v/ M  (10kg)(8m/s)/(80kg) 1.0m/s
She makes it !
• But, what then…?

Superhero physics
• Superman is impervious to human weapons but not the laws
of physics. An evildoer does not realize this and fires a
machine gun at Superman and hits him in the chest with 15
bullets per second. Each bullet has a mass of 20 grams and a
speed of 500 meter/second; the bullets bounce back at the
evildoer at the same speed. What average force does
Superman need to exert on the bullets to achieve this feat?
Hint: Use impulse in this problem. What should one use for Δt?
Superhero physics
• Superman is impervious to human weapons but not the laws of physics.
An evildoer does not realize this and fires a machine gun at Superman and
hits him in the chest with 15 bullets per second. Each bullet has a mass of
20 grams and a speed of 500 meter/second; the bullets bounce back at
the evildoer at the same speed. What average force does Superman need
to exert on the bullets to achieve this feat?
J  Ft  mv  m(v f  v i )
1
s
15
m  0.020kg
v  500m/s
v' 500m/s
t 
F
m(v f  v i ) (0.020kg)(500m/s- (-500m/s))

 300N (about 67 lb)
1 s
t
15
At the golf club
• In testing golf balls, they use a standard golf club,
which strikes the ball being tested, which is
initially at rest, at 120 mi/hr (54 m/s). If the head
of the golf club is 225 grams and the ball has a
mass of 45 grams, what speed would the ball be
traveling after a head-on, perfectly elastic
collision?
• Hint—In addition to conservation of momentum
use the handy formula for head-on elastic
collisions:


v

v


(
v

v
)
AB
AB
At the golf club
ptot  m A v A  m B v B  m A v A  mB v B
v A  v B  (v A  v B )
mA  0.225kg
mB  0.045kg
v A  54 m/s
v B  0 m/s
v A  v B  54 m/s  (v A  v B )  v A  v B  54 m/s
ptot  (0.225kg)(54 m/s)  (0.225kg)v A  (0.045kg)v B
 (0.225kg)(v B  54 m/s) (0.045kg)v B
v B 

2(0.225kg)(54 m/s)
 90 m/s
(0.225kg  0.045kg)
Dramatic rescue
• You are watching a Saturday morning cartoon concerning a jungle hero
called George of the Jungle. George attempts to save his friend, an ape
named Ape, from a stampeding herd of wildebeests. Ape is at the base of
a tall tree which has a vine attached to its top. George is in another tree
holding the other end of the vine. George plans to swing down from the
tree, grab Ape at the bottom of the swing, and continue up to safety on a
ledge which is half of George's initial height in the tree. Assuming that Ape
weighs the same as George, will they successfully make it to the top of the
ledge?
Hints: Assume conservation of George’s energy during the downswing,
reaching a final speed before the collision of vB.
Is this an elastic or inelastic collision? What is the speed of the two of
them immediately after the collision?
What speed will they need to have after the collision to reach a final
height at least half of George’s initial height?
Dramatic rescue?
• Hints: Assume conservation of George’s energy during the downswing,
reaching a final speed before the collision of vB.
Is this an elastic or inelastic collision? What is the speed of the two of
them immediately after the collision?
What speed will they need to have after the collision to reach a final
height at least half of George’s initial height?
ptot  m A v A  m B v B  m A v A  m B v B
Inelast ic: v A  v B
m A  m B /2
vA  0
ptot  m B v B  (m B /2  m B )v B 
2
v B  v B
3
3
m B v B
2
2gh  v
2
vB 2
 h0 
2g
0
B
Assuming conservation of energy on the
way down, and zero final and initial speeds,
v B 2 (2v B /3) 2 4
1
2
2gh f  v B  h f 

 h0  h0
they don’t
 make it!
2g
2g
9
2
George better push off when he leaves the first tree.
The gravitational slingshot effect. The figure shows the planet Saturn moving in
the negative x direction at its orbital speed (with respect to the Sun) of 9.6km/s.
The mass of Saturn is 5.69 x1026 kg. A spacecraft with mass 825 kg approaches
Saturn. When far from Saturn, it moves in the +x direction at 10.4km/s. The
gravitational attraction of Saturn (a conservative force) acting on the spacecraft
causes it to swing around the planet (orbit shown as dashed line) and head off in
the opposite direction. Estimate the final speed of the spacecraft after it is far
enough away to be considered free of Saturn’s gravitational pull.
Hints:
Does the velocity of Saturn
change by an appreciable amount?
Is this an elastic collision?