Download 10/14 Energy Practice Problem and this weeks Lab

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
Your Lab “Ballistic Pendulum”
Use conservation of energy
for the swing
v=0
v
h
v
Use conservation of
momentum for the collision
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,Bt = -FB,At
Where the minus sign means direction.
We call FA,Bt 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,Bt can be seen hiding in the 2nd law, Fnet = ma.
Fnet = ma = mv/t.
So Fnett = mv and for the system, mv 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 = mv and p = mv
Momentum of the system is the vector sum of the individual
momenta.
p = m(v) = Ft
Bullet sticks in block A
vB,i = 100 m/s
mB = 0.05 kg
mA = 2 kg
A
vA+B,f = ?
A
pA+B,i = pA+B,f
For system, Fnet = 0
mBvB,i +mAvA,i = (mB + mA)vA+B,f
5 + 0 = (2.05)vA+B,f (All directions to the right)
vA+B,f = 2.4 m/s
Is energy conserved?
No. Mechanical energy was lost in
friction between the bullet and block.
Block up a ramp with friction: A block is held against a
compressed spring then released. Find:
The velocity at B m = 0.5kg k = 320N/m k = 0.25 g = 10N/kg
The height at C
fR,B
C
The velocity at D
KE PEg PEs
KE PEg PEs
WX
B
0.5m
A
5.0m
4.0m
?
D
PEs,A - lost energy = KEB + PEg,B
40
5
20J
15J
PEs,A = 1/2kx2 = 40J
PEg,B = mghB = 15J
KE PEg PEs
?
Lost energy = -fR,B x = kNR,B x
NR,B = WY = 4/5WE.B = 4/5mg = 4N
= 0.25(4)(5) = 5J
KEB = 20J = 1/2mv2
WY
WE,B
3.0m
KE PEg PEs
NR,B
vB = 8.9m/s
PE and
h=0
Block up a ramp with friction: A block is held against a
compressed spring then released. Find:
The velocity at B m = 0.5kg k = 320N/m k = 0.25 g = 10N/kg
The height at C
20J 15J
Energy conserved
C
The velocity at D
KE PEg PEs
vB = 8.9m/s
vC = ?
vC = vX,B
B
A
5.0m
4.0m
20J
15J
12.8J
vC = vX,B = 4/5vB = 7.16m/s
0.5m
Projectile motion!
3.0m KE + PE = KE + PE
B
g,B
C
g,C
40J
KE PEg PEs
KE PEg PEs
KEC = 1/2mv2 = 12.8J
PEC = mghC = 22.2J
hC = 4.44m
?
D
KE PEg PEs
PE and
h=0
Block up a ramp with friction: A block is held against a
compressed spring then released. Find:
The velocity at B m = 0.5kg k = 320N/m k = 0.25 g = 10N/kg
The height at C
20J 15J
12.8J 22.2J
C
The velocity at D
KE PEg PEs
vB = 8.9m/s
0.5m
B
3.0m
40J
KE PEg PEs
KE PEg PEs
A
5.0m
4.0m
35J
KEB + PEg,B = KED
20J 15J
35J
KED = 1/2mv 2 = 35J
vD = 11.8m/s
Also note:
vX,D = 7.16m/s
The angle  = Arccos
7.16
= 53
11.8
D
KE PEg PEs
PE and
 h=0