Date post: | 01-Apr-2015 |
Category: |
Documents |
Upload: | rebecca-mifflin |
View: | 214 times |
Download: | 0 times |
Linear Momentum
For an individual mass we define the linear momentum to be:
vmp
From the 2nd law we have: am
dt
pd
Fnetdt
pd
This is actually how Newton originally formulated the 2nd law. The “ma” is a special case when m is not changing.
Linear Momentum of a System of Particles
For a system of many particles, we can define the “total linear momentum”:
...21 ppP
Then we can write:
...2211 vmvmP
CMCM VMdt
rdMP
The total linear momentum is a result of the total mass moving with the velocity of the CM
Let us differentiate the total momentum:
exttotalCM FaM
dt
Pd
If there are no external forces, then momentum is conserved:
fPPdt
Pd
00
This is a vector equation. If there are not external forces in say the x-direction, then Pxo = Pxf even if there are external forces in the y-direction.
Elastic Collisions in 1-D
1v 2v
1v 2v
22112211 vmvmvmvm fPP 0
Elastic means KE stays the same:
222
211
222
211 2
1
2
1
2
1
2
1vmvmvmvm
222111 vvmvvm
22222
21
211 vvmvvm
2222211111 vvvvmvvvvm
Dividing two equations gives (upon rearrangement):
1221 vvvv The relative velocity changes direction but keeps same magnitude.
Only true in 1-D, elastic collisions.
Example: Ping pong ball collides with stationary bowling ball.
211 vMvmmv
121 vvv Multiply by M and subtract
11 vMmvMm 1111 vvvv
Mm
Mm
Bounces back with same speed
Example: Bowling ball hits stationary ping pong ball.
211 vmvMMv
121 vvv Multiply by M and add
212 vMmMv
112 22
vvmM
Mv
Barry Bonds uses a light bat
A two dimensional collision
fPP
0
xfx PP 0
1v
1v
2v
coscos 221111 vmvmvm
yfy PP 0
sinsin0 2211 vmvm
Example: a 4 kg mass heading in the – y direction at 12 m/s collides and sticks to a 6 kg mass moving in the + x direction at 10 m/s. Find the magnitude and direction of the final velocity.
xfx PP 0 s
mvv xx 610106
yfy PP 0 s
mvv yy 8.410124
8.
6
8.4tan
22 8.46 v
Impulse
For a constant force, let is define a vector quantity called impulse as the product of the force times the time over which it acts:
tFJ
In one dimension, we need only worry about the sign. If the force is not constant during the time over which the force acts, we define through an integral:
2
1
t
t
dtFJ
Note the analogy to our definition of work. Of course a huge difference is work is a scalar and impulse is a vector. For any component, the impulse will be the area under the Fi vs t graph.
So why bother with impulse?Suppose we focus on the impulse delivered by the net force.
2
1
2
1
t
t
t
t
dtdt
pddtFnetJ
ppppdJ
p
p
12
2
1
curveunderareadtFJt
t
xx 2
1
Impulse – Momentum Theorem
p = pf – pi = F dttf
ti
The impulse of the force F acting on a particle equals the change in the
momentum of the particle.
J = p
Example: A 0.1 kg mass moving with a speed of 10 m/s along the x-axis collides head on with a stationary 0.1kg mass. The magnitude of the force between the two is shown below as a function of time. Find the final speed and direction of each mass.
areaJp
)]12)(10*20(2
11210*30)12)(10*20(
2
1[01. 333 fv
s
mv f 6
For 2nd block:
)]12)(10*20(2
11210*30)12)(10*20(
2
1[101. 333 fv
s
mv f 4
For 1st block:
Impulse is negative by Newton’s 3rd Law