momentum. the “quantity of motion” p = momentum m = mass v = velocity unit: vector direction of...
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Momentum
Momentum
• The “Quantity of Motion”
• p = momentum• m = mass• v = velocity
• Unit:
• Vector• Direction of momentum is determined by the
direction of the velocity
smkg sN or
Momentum
amFNET
t
vvm if
t
p
t
vmvm if
t
pFNET
Newton’s 2nd Law of Motion in terms of momentum
Net External Force = time rate of change of momentum
External to the system
Momentum
t
pF
ptF
Impulse
• I = Impulse
• F= force
• ∆t = change in time
ptFavg
I
p mv f m
v i
Force vs Time Graphs
F
t
Area = Impulse = Δp
Impulse
F
t
Area =
Actual force experienced during a collision
Favg
Impulse - Problems
F (N)
t (s)
Force on a 1 kg object
10
2 4
If v = 0 at t = 0, find vf at t = 4 s
Impulse - Problems
• Find the change in momentum
2 kg
2 kg
5 m/s
3 m/s
+-
If ∆t = 0.020 s, find the average force the wall exerts on the ball
Impulse - Problems
If mass = 2 kg and v = 0 at t = 0, find vf at t = 12 s
F (N)
t (s)
20
4 8 12
Momentum
40°
10 m/s
10 m/s
Find ΔP
m = 2 kg
Momentum
30°
15 m/s
15 m/s
Find ΔP
m = 3 kg
Momentum Example 2 cont’d
pFind
m = 2 kg
10 m/s
6 m/s
m = 2 kg
60°40°
Conservation of Momentum
• If no external unbalanced forces, NETF
• so
if ppp 0
if pp Conservation of momentum for a system
w/ no external imbalance in forces Total initial momentum of a system =
Total final momentum of a system
0
t
pFNET
0
1-D Explosion
5 kg 10 kg
0ivBefore
5 kg 10 kg
?fv After
smv 3
2-D Explosions
6 kg 3 kg
4 kg
6 kg
3 kg
4 kg
?afv
θ=?
30 m/s
50 m/s
Before
2 kg3 kg
10 m/s0biv
After
2 m/s
?afv
3 kg
2 kg
40◦
θ=?
Collisions and Explosions
Examples
3 kg 7 kg
0iv
Before
sm2
3 kg 7 kg?fv
After
sm8.0
Examples
5 kg 7 kg?fv
After
5 kg 7 kg
0iv
Before
3ms
Explosions and Collisions
ma= 950 kg
mb= 1300 kg
vai= 16 m/s
vbi= 21 m/s
= ?vf = ?
Types of Collisions
• Elastic Collisions – Kinetic energy is conserved• KEi = KEf
• Inelastic collision – Some kinetic energy is lost• Typically lost to friction, sound, deformation, etc.• Most collisions in “real life”