Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Chapter 7: Linear Momentum
•Definition of Momentum
•Impulse
•Conservation of Momentum
•Center of Mass
•Motion of the Center of Mass
•Collisions (1d, 2d; elastic, inelastic)
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§7.1 Momentum
Consider two interacting bodies:
F21
F12
Fnet
t
If we know the net force on each body then v  at 
m
The velocity change for each mass will be different if the
masses are different.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Rewrite the previous result for each body as:
m1v1  F21t
m2 v 2  F12t  F21t
Combine the two results:
m1v1  m2 v 2
m1 v1 f  v1i   m2 v 2 f  v 2i 
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The quantity (mv) is called momentum (p).
p = mv and is a vector.
The unit of momentum is kg m/s; there is no derived unit for
momentum.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
From slide (3):
m1v1  m2 v 2
p1  p 2
The change in momentum of the two bodies is “equal and
opposite”. Total momentum is conserved during the
interaction; the momentum lost by one body is gained by
the other.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§7.2 Impulse
Definition of impulse:
p  Ft
One can also define an average impulse when the force
is variable.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text conceptual question 7.2): A force of 30 N is
applied for 5 sec to each of two bodies of different masses.
30 N
Take m1<m2
m1 or m2
(a) Which mass has the greater momentum change?
p  Ft
Since the same force is applied to
each mass for the same interval, p
is the same for both masses.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(b) Which mass has the greatest velocity change?
p
v 
m
Since both masses have the same p,
the smaller mass (mass 1) will have the
larger change in velocity.
(c) Which mass has the greatest acceleration?
v
a
t
Since av the mass with the greater
velocity change will have the greatest
acceleration.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 7.9): An object of mass 3.0 kg is
allowed to fall from rest under the force of gravity for 3.4
seconds. What is the change in momentum? Ignore air
resistance.
Want p =mv.
v  at
v   gt  33.3 m/sec
 p  mv  100 kg m/s (downward)
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 7.10): What average force is
necessary to bring a 50.0-kg sled from rest to 3.0 m/s in a
period of 20.0 seconds? Assume frictionless ice.
p  Fav t
p mv
 Fav 

t
t

50.0 kg 3.0 m/s 
Fav 
 7 .5 N
20.0 s
The force will be
in the direction
of motion.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§7.3 Conservation of Momentum
v1i
m1>m2
v2i
m1
m2
A short time later the
masses collide.
m1
m2
What happens?
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
During the interaction:
y
N1
N2
x
F21
F12
w1
F
F
y
 N1  w1  0
x
  F21  m1a1
w2
F
F
y
 N 2  w2  0
x
 F12  m2 a2
There is no net
external force on
either mass.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The forces F12 and F21 are internal forces. This means that:
p1  p 2
p1 f  p1i  p 2 f  p 2i 
p1i  p 2i  p1 f  p 2 f
In other words, pi = pf. That is, momentum is conserved.
This statement is valid during the interaction only.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 7.18): A rifle has a mass of 4.5 kg and
it fires a bullet of 10.0 grams at a muzzle speed of 820 m/s.
What is the recoil speed of the rifle as the bullet leaves the
barrel?
As long as the rifle is horizontal, there will be no net
external force acting on the rifle-bullet system and
momentum will be conserved.
pi  p f
0  mb vb  mr vr
 0.01 kg 
mb
820 m/s  1.82 m/s
 vr  
vb  
mr
 4.5 kg 
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§7.4 Center of Mass
The center of mass (CM) is the point representing the mean
(average) position of the matter in a body. This point need
not be located within the body.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The center of mass (of a two body system) is found from:
xcm
m1 x1  m2 x2

m1  m2
This is a “weighted” average of the positions of the particles
that compose a body. (A larger mass is more important.)
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
mr

m
i i
In 3-dimensions write
rcm
where mtotal   mi
i
i
i
i
The components of rcm are:
m x

m
xcm
i
i
i
m y

m
i
i i
ycm
i
i
i
m z

m
i i
i
zcm
i
i
i
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 7.27): Particle A is at the origin and has
a mass of 30.0 grams. Particle B has a mass of
10.0 grams. Where must particle B be located so that the
center of mass (marked with a red x) is located at the point
(2.0 cm, 5.0 cm)?
y
xcm
ma xa  mb xb
mb xb


ma  mb
ma  mb
ycm
ma ya  mb yb
mb yb


ma  mb
ma  mb
x
A
x
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
xcm 
10 g xb
10 g  30g
xb  8 cm
 2 cm
ycm 
10 g  yb
30 g  10 g
yb  20 cm
 5 cm
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 7.30): The positions of three particles
are (4.0 m, 0.0 m), (2.0 m, 4.0 m), and (-1.0 m, -2.0 m). The
masses are 4.0 kg, 6.0 kg, and 3.0 kg respectively. What is
the location of the center of mass?
y
2
1
x
3
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
xcm
m1 x1  m2 x2  m3 x3

m1  m2  m3

4 kg 4 m   6 kg 2 m   3 kg  1 m 

4  6  3 kg
 1.92 m
ycm
m1 y1  m2 y2  m3 y3

m1  m2  m3

4 kg 0 m   6 kg 4 m   3 kg  2 m 

4  6  3 kg
 1.38 m
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§7.5 Motion of the Center of Mass
For an extended body, it can be shown that p = mvcm.
From this it follows that Fext = macm.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 7.35): Body A has a mass of 3.0 kg
and vx = +14.0 m/s. Body B has a mass of 4.0 kg and has vy
= -7.0 m/s. What is the velocity of the center of mass of the
two bodies?
Consider a body made up of many different masses
each with a mass mi.
The position of each mass is ri and the displacement of
each mass is ri = vit.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
For the center of mass:
rcm  v cm
 m r
t 
m
i
i
i
i
i
Solving for the velocity of the center of mass:
ri
t
 i

 mi
 mi
v cm
i
m v
m
i
i
i
i
i
Or in component form:
m v

m
i i,x
v cm, x
i
i
i
m v

m
i i, y
v cm, y
i
i
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
Applying the previous formulas to the example,
vcm , x 

vcm , y 
ma va , x  mb vb , x
ma  mb
3 kg 14 m/s   4 kg 0 m/s   6 m/s
3  4 kg
ma va , y  mb vb , y
ma  mb

3 kg 0 m/s   4 kg  7 m/s 

 4 m/s
3  4 kg
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§§ 7.6 & 7.7 Collisions in One and
Two Dimension
When there are no external forces present, the momentum of
a system will remain unchanged. (pi = pf)
If the kinetic energy before and after an interaction is the
same, the “collision” is said to be perfectly elastic. If the
kinetic energy changes, the collision is inelastic.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
If, after a collision, the bodies remain stuck together, the loss
of kinetic energy is a maximum. This type of collision is called
perfectly inelastic.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 7.41): In a railroad freight yard, an
empty freight car of mass m rolls along a straight level track
at 1.0 m/s and collides with an initially stationary, fully
loaded, boxcar of mass 4.0m. The two cars couple together
upon collision.
(a) What is the speed of the two cars after the collision?
pi  p f
p1i  p2i  p1 f  p2 f
m1v1  0  m1v  m2 v  m1  m2 v
 m1 
v1  0.2 m/s
 v  
 m1  m2 
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(b) Suppose instead that both cars are at rest after the
collision. With what speed was the loaded boxcar moving
before the collision if the empty one had v1i = 1.0 m/s.
pi  p f
p1i  p2i  p1 f  p2 f
m1v1i  m2 v2i  0  0
 m1 
 v2i   v1i  0.25 m/s
 m2 
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 7.49): A projectile of 1.0 kg mass
approaches a stationary body of 5.0 kg mass at 10.0 m/s and,
after colliding, rebounds in the reverse direction along the
same line with a speed of 5.0 m/s. What is the speed of the
5.0 kg mass after the collision?
pi  p f
p1i  p2i  p1 f  p2 f
m1v1i  0  m1v1 f  m2 v2 f
v2 f
m1
v1i  v1 f 

m2
1.0 kg
10 m/s   5.0 m/s   3.0 m/s

5.0 kg
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 7.58): Body A of mass M has an
original velocity of 6.0 m/s in the +x-direction toward a
stationary body (B) of the same mass. After the collision,
body A has vx=+1.0 m/s and vy=+2.0 m/s. What is the
magnitude of body B’s velocity after the collision?
Final
Initial
A
A
vAi
B
B
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
y momentum:
x momentum:
pix  p fx
piy  p fy
p1ix  p2ix  p1 fx  p2 fx
p1iy  p2iy  p1 fy  p2 fy
m1v1ix  0  m1v1 fx  m2 v2 fx
Solve for v2fx:
v2 fx 
m1v1ix  m1v1 fx
m2
 v1ix  v1 fx
0  0  m1v1 fy  m2 v2 fy
Solve for v2fy:
v2 fy 
 m1v1 fy
m2
 v1 fy
 2.00 m/s
 5.00 m/s
The mag. of v2 is
v2 f  v 2 2 fy  v 2 2 fx  5.40 m/s
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Summary
•Definition of Momentum
•Impulse
•Center of Mass
•Conservation of Momentum
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