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IB Physics 12
Mr. Jean
September 11th, 2013
The plan:
• Video clip of the day
• Momentum
• Ballistic Pendulums
• Bouncing collisions
Muzzle Speeds:
Ballistic Pendulum
• The ballistic pendulum is a classic example of a dissipative collision in which conservation of momentum can be used for analysis, but conservation of energy during the collision cannot be invoked because the energy goes into inaccessible forms such as internal energy. After the collision, conservation of energy can be used in the swing of the combined masses upward, since the gravitational potential energy is conservative.
Ballistic Pendulum Info:
• Marble = 7.5 grams (approximately)
• Block = 80 grams (approximately)
Ballistic Pendulum Lab:
• Complete the Ballistics Pendulum Lab– Here is some time to work on the lab.
Lab Requirements:
• Data & Observations – Table of data– Graph of data
• All calculations– Calculations from conservation of energy to conservation of
momentum• All formulas
– Conservation of energy– Conservation of momentum
• Conclusion– Three Bullet points at the bottom of the lab.
Due: Monday September 16th, 2013
New material:
• Bouncing Collisions: When two objects collide and bounce off each other, there are two possible energy interactions. Kinetic energy can be conserved or not conserved. – If it is conserved, then we have a perfectly elastic
collision. – The other possibility is that some of the kinetic energy
is transformed into other forms of energy. This is mostly what happens in the real world.
• The type of problem you will be tasked with solving will have two objects bouncing off each other. – You will know three of the four velocities and
solve for the fourth. – You may be asked whether kinetic energy is
conserved and, if it isn’t, how much kinetic energy is lost.
Perfectly Elastic collision with conserved Energy:
• Example: Two balls hit head on as shown, what is the final velocity of the second ball if the first one’s final velocity is –1.50 m/s?
m1m2
v1v2
= 1.50 kg = 1.85 kg
= 2.30 m 's = 1.30 m 's
The general conservation of momentum equation is:
• All we have to do is solve for v2’
1 1 2 2 1 1 2 2' 'm v m v m v m v
1 1 2 2 1 12
2
''
m v m v m vv
m
2
1.50 2.30 1.85 1.30 1.50 1.50' 1.78
1.85
m m mkg kg kg
ms s svkg s
Example #2:• Two balls roll towards each other and collide as
shown. The second’s ball velocity after the collision is 3.15 m/s.
– (a) What is the velocity of the first ball after the collision? – (b) How much kinetic energy is lost during the collision?
m1m2
v1v2
= 2.50 kg = 2.85 kg
=4.75 m 's = 8.60 m 's
A) Velocity:
1 1 2 2 1 1 2 2' 'm v m v m v m v
1 1 2 2 2 21
1
''
m v m v m vv
m
1
2.50 4.75 2.85 8.60 2.85 3.15' 8.64
2.50
m m mkg kg kg
ms s svkg s
(b) Difference in kinetic energy before and after collision
• Question #3: Wow this looks just like a test question.A 2.0 kg block is sliding on a smooth table top. It has a totally inelastic collision with a 3.0 kg block that is at rest. Both blocks, now moving, hit the spring and compress it. Once the blocks come to rest, the spring restores itself and launches the blocks. They slide off the right side of the table. The spring constant is 775 N/m. Find: – (a) the velocity of the two blocks after the collision. – (b) The distance the spring is compressed. – (c) The velocity of the blocks after they leave the spring. – (d) The distance the blocks travel before they hit the deck after the leave
the tabletop. – (e) The kinetic energy of the blocks just before they hit the deck.
1 .2 m
v = 10 .0 m /s
Solutions:
• C) If energy is conserved in the spring, they should have the same speed when they leave as they had going into the spring. So they should be moving at 4.0 m/s to the right.
• D) Distance they fall from table:
2
2
2 1.21 20.495
2 9.8
myy at t s
mas
4.0 0.495 2.0m
x vt s ms
• (e) The kinetic energy of the blocks at the bottom (just before they hit) must equal the potential energy at the top of table plus the kinetic energy the blocks had before they began to fall.
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