Download Chapter 7: Momentum

Chapter 7
LECTURE NOTES
Linear Momentum
A vector quantity, linear momentum p = mv points in the same direction as the velocity and has
units kg • m/s.
Newton expressed his second law in terms of momentum.
 

m( v f  v i )
p
Since F = ma =
=
or
t
t
Ft = p
An impulse produces a change in momentum.
If the force is not constant but varies with time, we find impulse is the area under the curve of a F
vs t graph. If the impulse is zero, there is no change of momentum -- it is conserved.
Example: A rubber ball mass 0.100kg moving east at 10.0 m/s hits a wall. The ball is in contact
with the wall for 0.100s and then rebounds to the west at 5.00 m/s. What force did the wall exert
on the ball?
Solve F  t = m (vf - vi) for F. Let vf = -5.00 m/s and vi = +10.0 m/s (east is +).
= -15.0 N (the - sign indicates the wall exerts a force to the west).
Then F
Conservation of Momentum
In a collision between two objects, the force object 1 exerts on object 2 is equal in magnitude but
oppositely directed to the force object 2 exerts on object 1 (this is Newton's third law). Each
force acts for the same time of collision  t and changes the momentum of the object such that






p1i  p 2i  p1f  p 2f or p i  p f . If no outside forces act, the total momentum of a system is constant.
Example: A penguin (mass l0.0kg, initially at rest) standing on wet ice (assumed frictionless)
throws a 0.25kg snowball at another penguin to the north at 10.00m/s. Find the recoil velocity of
the first penguin.
Initially, the momentum of penguin + snowball is zero since they are at rest. After the toss, the
final momentum is psnowball + ppenguin
psnowball = msnowball vsnowball = 2.5 kg m/s assuming north is + .
ppenguin = mpenguin vpenguin = 10kg vpenguin.
Thus pi = pf gives 0 = 2.5 m/s + 10 kg vpenguin or vpenguin = - .25 m/s (- signifies south).
Collisions
We deal with three types of collision:
 elastic:
pi = pf and KEi = KEf
 inelastic:
pi = pf
 perfectly or totally inelastic (objects stick together) pi = pf
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Chapter 7
LECTURE NOTES
Example: Totally inelastic head-on collision in one-dimension.
before
m1
m2
v 1i
after
+ east
v 2i
m1 + m2
vf
Mass m1 = 1000kg travels east at 20.0m/s and collides with Mass m2 = 2000kg going west at
14.5m/s. The wreckage sticks together (clue to totally inelastic collision). Find the final velocity
of the wreckage.


Since no external forces act in the east/west directions, we have m1 v1i + m2 v 2i = (m1 + m2) vf


by conservation of momentum. Let v1i = +20.0 m/s and v 2i =-14.5 m/s. We find vf = -3.00 m/s
or 3 m/s, west
Example: Elastic head-on collision in one dimension.
Mass m1 = 1000 kg travels east at 20.m/s and collides with mass m2 = 2000 mg also going east at
10.0 m/s.

If the collision is elastic, find the velocity of each mass after the collision. Take east to
be +.
Conservation of linear momentum



m1 v1i + m2 v 2i = mi v 1f + m2 v 2f
Conservation of kinetic energy
1
1
1
1
m1 v12i + m 2 v 2 i2  m1 v1 f2  m 2 v 2 f2
2
2
2
2
The squares in the kinetic energy expression complicate the solution. We can replace the
condition above by
Conservation of kinetic energy




v1i  v 2i = – ( v1f  v 2f )


(1000kg) (20.0 m/s) + (2000kg) ( 10.0m/s) = (1000 kg) v1f + (2000kg) v 2f


20.0 m/s - 10.0m/s = – v1f + v 2f
or upon dividing the first equation by 1000 kg on each side


40.0m / s  v1f  2v 2f


10.0 m / s   v1f  v 2f
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Chapter 7
LECTURE NOTES


Thus v 2f = + 16.7m/s (east), v1f = 6.7 m/s (east)
The masses were closing at 10 m/s relative speed; now they recede at the same rate.
If this collision was elastic but the wreckage did not stick together, we would only have
the one conservation of momentum equation to find the velocity of the two masses. We
would need more information (like one of the final velocities) to solve the problem.
Example: Perfectly inelastic glancing collision in two-dimension.

+y
vf

m1

v1i
+x

v 2i
m2
Mass m1 = 1000 kg travels 30.0 m/s east and is struck by mass m2 = 2000 kg going north
at 15.0 m/s. The wreckage sticks together. Find its final velocity.
pi = pf is a vector equation equivalent to
x:
y:




p i x  p f x or m1 v1i  (m1  m 2 )v f x




p i y  p f y or m 2 v 2i  (m1  m 2 ) v f y


Solving we find v f x = 10.0 m/s and v f y = 10.0 m/s.
Thus vf = 14.1 m/s at  = 45 N of E.
Center of Mass
We have previously used the idea that forces act on the "center" of a body. To make this
idea definite we define the center of mass of a body to have coordinates:
x cm 
m i x i
m y
and y cm  i i
m i
m i
where the sum is over the i particles comprising the body.
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