10/14 Energy Practice
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10/14 Energy Practice Text: Chapter 6 Energy Lab: “Ballistic Pendulum” Energy example (Like “Ramp Launch”) No HW assigned Exam 2 Thursday, 10/17 5-7 Wit 116 6-8 Wit 114 (only if needed) Please send email if other time needed Also let me know if you are a “6-8” person
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10/14 Energy Practice. Text: Chapter 6 Energy Lab: “Ballistic Pendulum” Energy example (Like “Ramp Launch”) No HW assigned Exam 2 Thursday, 10/17 5-7 Wit 116 6-8 Wit 114 (only if needed) Please send email if other time needed Also let me know if you are a “6-8” person. v. h. v. - PowerPoint PPT Presentation
Transcript of 10/14 Energy Practice
10/14 Energy Practice Text: Chapter 6 Energy Lab: “Ballistic Pendulum” Energy example (Like “Ramp Launch”) No HW assigned Exam 2 Thursday, 10/17
5-7 Wit 116 6-8 Wit 114 (only if needed)Please send email if other time neededAlso let me know if you are a “6-8” person
Your Lab “Ballistic Pendulum”
v
v
v = 0
h
Use conservation of momentum for the collision
Use conservation of energy for the swing
You will find the initial velocity of the ball by measuring the height. (and some math)
You will also find the initial velocity of the ball by measuring its velocity with projectile motion ideas.
Your Lab “Ballistic Pendulum”Here is a quick primer on momentum.
When two blocks collide, they exert equal and opposite forces on each other. (Third Law)Since the contact time is the same for both we can say that:
FA,Bt = -FB,AtWhere the minus sign means direction. We call FA,Bt the “change in momentum” and the changes in momentum for colliding objects are equal and opposite.If we consider the system of both objects, then these third law pairs are “internal forces” and the momentum of the system as a whole does not change. It is conserved.
Your Lab “Ballistic Pendulum”Note that FA,Bt can be seen hiding in the 2nd law, Fnet = ma.
Fnet = ma = mv/t.
So Fnett = mv and for the system, mv is 0 and mv is constant. (conserved)
The symbol for momentum is p and it is a vector. Changes in p are handled just like changes in v.
p = mv and p = mv
Momentum of the system is the vector sum of the individual momenta.
Bullet sticks in block A
A
vB,i = 100 m/smB = 0.05 kg
mA = 2 kg
vA+B,f = ?
A
5 + 0 = (2.05)vA+B,f
vA+B,f = 2.4 m/s
p = m(v) = Ft
pA+B,i = pA+B,f mBvB,i +mAvA,i = (mB + mA)vA+B,f
Is energy conserved? No. Mechanical energy was lost in friction between the bullet and block.
(All directions to the right)For system, Fnet = 0
Block up a ramp with friction: A block is held against a compressed spring then released. Find:The velocity at BThe height at CThe velocity at D
Lost energy = -fR,B x = kNR,B x
NR,B = WY = 4/5WE.B = 4/5mg = 4N
PEg,B = mghB = 15J
PE and h = 0
PEs,A - lost energy = KEB + PEg,B
vB = 8.9m/s
B
AD
C
KE PEg PEs KE PEg PEs
KE PEg PEs
0.5m
5.0m
3.0m
4.0mPEs,A = 1/2kx2 = 40J
= 0.25(4)(5) = 5J
40 5 15J
?
KEB = 20J = 1/2mv2
m = 0.5kg k = 320N/m k = 0.25 g = 10N/kg
WE,B
WY
NR,B
WX
fR,BKE PEg PEs
20J
?
m = 0.5kg k = 320N/m k = 0.25 g = 10N/kg
Block up a ramp with friction: A block is held against a compressed spring then released. Find:The velocity at BThe height at CThe velocity at D
PEC = mghC = 22.2J
PE and h = 0
KEB + PEg,B = KEC + PEg,C
vC = vX,B = 4/5vB = 7.16m/s
Energy conserved
KEC = 1/2mv2 = 12.8J
hC = 4.44m
KE PEg PEs KE PEg PEs
KE PEg PEsKE PEg PEs
0.5m
40J
20J 15J
vB = 8.9m/s
20J 15J 12.8J ?
vC = ?vC = vX,B
Projectile motion!B
AD
C
5.0m
3.0m
4.0m
Block up a ramp with friction: A block is held against a compressed spring then released. Find:The velocity at BThe height at CThe velocity at D
KED = 1/2mv 2 = 35J
PE and h = 0
KEB + PEg,B = KED
vD = 11.8m/s
KE PEg PEs KE PEg PEs
KE PEg PEsKE PEg PEs
0.5m
20J 15J 35J
vB = 8.9m/s
vX,D = 7.16m/s
Also note:
7.16
11.8The angle = Arccos = 53
40J
20J 15J 12.8J 22.2J
35J
B
AD
C
m = 0.5kg k = 320N/m k = 0.25 g = 10N/kg
5.0m
3.0m
4.0m
