Download Collisions and Conservation of Energy

Collisions and Conservation of Energy
Consider the above system.
A) What is the momentum for each object?
B) What is the total momentum for the system?
Bellwork Answer
• the 3 kg mass has a positive momentum of 6
kg m/sec
• the 2 kg mass has a negative momentum of 6
kg m/sec
• each mass is moving, but the total momentum
of the system equals 0.
Warm-Up 12/09/09
A net force of 200 N acts on a 100-kg boulder,
and a force of the same magnitude acts on a
130-g pebble. How does the rate of change of
the boulder’s momentum compare to the rate
of change of the pebble’s momentum?
A.
greater than
B.
less than
C.
equal to
D.
Cannot be determined
Solution to Warm-Up
• C – equal to
• The rate of change of momentum is, in
fact, the force. Remember that
F = Dp/Dt. Since the force exerted on
the boulder and the pebble is the same,
then the rate of change of momentum
is the same.
Follow-up
A net force of 200 N acts on a 100-kg
boulder, and a force of the same
magnitude acts on a 130-g pebble. How
does the rate of change of the boulder’s
velocity compare to the rate of change of
the pebble’s velocity?
A.
B.
C.
D.
greater than
less than
equal to
Cannot be determined
To understand the properties of momentum, we
must first re-examine Newton's Laws
Newton's 3rd Law
FAB = - FBA
FAB t = - FBA t
mBvfB - mBvoB = - (mAvfA- mAvoA)
mBvfB + mAvfA = mAvoA + mBvoB
Σpf = Σpo
The Law of Conservation of Momentum states the
sum of the momenta before a collision equals the
sum of the momenta after a collision.
Conservation of Momentum
Σpo = Σpf
• The total linear momentum of a system is conserved
if the net external force on the system is zero.
• Consider the cue ball
that makes a head on
collision with an 8 ball
at rest.
• Define the system.
– Before the collision
– After the collision
8
8
Finding the Sum of Momentum
• Remember: momentum is a vector quantity. You
must use vector addition to determine the total
momentum of a system.
• In a collision, conservation of momentum can be
written as
m1v1 +m2v2 = m1v1o +m2v2o
Practice with Conservation of Momentum: Solve
problems A-F
http://dev.physicslab.org/DocumentPrint.aspx?do
ctype=5&filename=Compilations_CPworkbook_
ConservationMomentum.xml
Example 1: A 50 kg pitching machine (excluding
the baseball) is placed on a frozen pond. The
machine launches a 0.40 kg baseball with a
speed of 35 m/s in the eastward direction.
Describe the velocity of the machine after the
ball is launched.
m1v1 +m2v2 = m1v1o +m2v2o
Conceptual Reasoning for Example 1
• Before the ball is launched, neither the ball
nor the machine is in motion so the total
initial momentum is zero. When the ball is
launched in the eastward direction, it has a
final momentum in the eastward direction.
Therefore, the machine must have a final
momentum in the westward direction to
conserve the total momentum of the system.
Therefore, the velocity of the machine after
the ball is launched is in the westward
direction.
Solution to Example 1
• Given: m1=50 kg (machine) m2=0.40 kg (ball)
v1o = 0
v2o= 0
v1 = ?
v2 = 35 m/s east
Here the “collision” is the launching of the ball.
m1v1 +m2v2 = m1v1o +m2v2o
solve for v1
Example 2
A major league catcher catches a fastball
moving at 95 mi/h and his hand and glove
recoil 10.0 cm in bringing the ball to rest. If it
took 0.00470 seconds to bring the ball (with a
mass of 250 g) to rest in the glove, (a) what is
the magnitude and direction of the change in
momentum of the ball? (b) Find the average
force the ball exerts on the hand and glove.
Solution to Example 2
Given
a) The change in momentum is:
b) The average force is:
Example 3: A 10-gram bullet moving at 250. m/sec
bores through a piece of lucite 2.0 cm thick and
emerges out the other side at 200. m/sec.
• How much did the bullet’s momentum change
as it moved through the lucite?
• If the bullet takes 0.00025 s to travel through
the lucite, what is the average force exerted
on the bullet by the lucite?
• What acceleration did the bullet experience
while traveling through the lucite?
Given:
Solution to Example 3
m= 10 g = 0.010 kg, vf = 200 m/s, v0 = 250 m/s, t = 0.00025 s
Quick Quiz
Amy (150 lbs) and Gwen (50 lbs) are standing on
slippery ice and push off each other. If Amy
slides at 6 m/s, what speed does Gwen have?
1) 2 m/s
2) 6 m/s
3) 9 m/s
4) 12 m/s
5) 18 m/s
150 lbs
50 lbs
Elastic and Inelastic Collisions
• There are two general categories of collisions: elastic
and inelastic.
• Elastic collisions occur when objects bounce off each
other AND kinetic energy is conserved during the
collision. These types of collisions are more "ideal"
in nature and are difficult to observe.
• Inelastic collisions occur when objects collide and
energy is lost during the collision. That is,
ΣKEbefore > ΣKEafter.
• During perfectly inelastic collisions the objects stick
together during the collision and leave as one single
mass.
• Now solve problems G-K
Air Track Collisions
• The degree to which a collision is considered to
be elastic or inelastic is measured by a quantity
called the coefficient of restitution, e, which is
defined as the ratio of the (relative velocities of
recession) /(relative velocities of approach) for
the two objects involved in the collision.
• e = (vf2 - vf1) / (vo1 - vo2)
• When e = 1, the collision is perfectly elastic;
when e = 0 is it perfectly inelastic.
• http://faraday.physics.utoronto.ca/PVB/Harrison
/Flash/ClassMechanics/AirTrack/AirTrack.html
• Example 4: Suppose that a 3-kg mass moving at 5
m/sec strikes a stationary 1-kg mass whereupon they
stick other. What will be the final velocity of the
combined mass, vc, after this inelastic collision?
• Example 5: Suppose that a 3-kg mass moving at 2
m/sec strikes a 1-kg mass moving towards it at 10
m/sec whereupon they stick together. What will be the
final velocity of the combined mass, vc, after this
inelastic collision?
• Example 6: Suppose a 1-kg pistol containing a 10-gram
bullet is resting on a table when it is accidentally
discharged. If the bullet has a muzzle velocity of 150
m/sec, how fast will the pistol recoil?
• Example 7: How do the impulses an object receives
compare when (1) it strikes a wall and sticks to it,
versus (2) it rebounds elastically off of the wall?
Solution to Example 4
• Given: m1 = 3 kg
m2 = 1 kg
v1o = 5 m/s
v2o = 0 m/s
vcombined = ?
• Σmvbefore = Σmvafter
(3kg)(5m/s) + (1kg)(0m/s) = (3kg+1kg)vc
15kgm/s = 4kg vc
vc = 3.75 m/sec
Solution to Example 5
• Given: m1 = 3 kg
m2 = 1 kg
v1o = +2 m/s
v2o = -10 m/s
Σmvbefore = Σmvafter
(3kg)(2m/s) + (1kg)(-10m/s) = (3kg+1kg)vc
-4kgm/s = 4kg vc
vc = - 1 m/sec
• Note that the original velocity of the 3-kg mass is +2
m/sec while the original velocity of the 1-kg mass is 1 m/sec signifying that they are traveling towards
each other in opposite directions.
• Finally, vc being negative tells us that the combined
mass is moving in the original direction of the 1-kg
mass.
Solution to Example 6
• Given: m1 = 1 kg
m2 = 10 g = 0.01 kg
v1o = 0 m/s
v2o = 0 m/s
v1’ = ? m/s
v2’ = 150 m/s
• Σmvbefore = Σmvafter
(1kg +.01kg)(0m/s) = (0.01kg)(150m/s) + (1kg)v1’
0 kgm/s = 1.5 + vf
vf = -1.5 m/sec
• Note that in order to compare their respective momenta,
both the pistol and the bullet have to have their masses
stated in kilograms (10 grams = 0.01 kg). Also note that the
negative sign on the pistol's final velocity signifies that it is
traveling in a direction opposite to the direction of the
bullet. We usually say that the pistol "recoils" with a speed
of 1.5 m/sec.
Solution to Example 7:
the ball that bounces receives an
impulse that is twice as large as
the ball that sticks to the wall
When the ball sticks to When the ball rebounds
the wall,
off of the wall,
its final velocity equals its final velocity equals zero.
vo.
The impulse delivered to The impulse delivered to
the ball
the ball
by the wall equals:
by the wall equals
(net F)t = mvf - mvo
(net F)t = mvf - mvo
(net F)t = m(0) - mvo
(net F)t = m(-vo) - mvo
(net F)t = - mvo
(net F) t = - 2mvo
Homework Pass Challenge: When objects travel along
diagonal paths, they have momentum in each of the x
and y dimensions. Consequently momentum vectors
must be resolved into their x- and y-components when
working two-dimensional collisions. Consider the
following scenario
A 2-kg mass traveling left at 3 m/sec
along the positive x-axis and a 3-kg
mass traveling upwards along the
positive y-axis at 2 m/sec collide with
each other. If the two masses stick
together during the collision, how
fast and in what direction will they
leave the collision?
x-components
y-components
BEFORE
COLLISION
Green ball 2(-3) = -6 kgm/s
Purlple ball 0
0
3(2) = +6 kgm/s
AFTER
COLLISION
Green and
purple balls
stick
together
-(2kg+3kg)vc cosq
+(2kg + 3kg)vc sinq
ANALYSIS
Conservation of Momentum
-6kgm/s + 0 = -(5kg)vc cosq
vc cosq = 6/5
0 + 6kgm/s = +(5kg)vc cosq
vc sinq = 6/5
tan(q) = 1
q = 45o
vc cos(45o) = 6/5
0.707 vc = 1.2
vc = 1.2/0.707
vc = 1.70 m/s