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MOMENTUM AND COLLISIONS
Chapter 5, p. 230-271
In Newton’s time it was known that momentum of objects was
conserved in collisions. Momentum is defined as:


pmv
units of kgm/s 

dp
Newton stated his second law of motion in F 
terms of changing momentum.
dt
vector!!!!
p
F
t

If the force is constant then:


F t  p
Another equation for impulse assumes the
mass remains constant (not always true).


 p  m v

-impulse (Ns)
- change in
momentum
In fact the force acting on an object is often not constant. The
force in the preceding equations then must be thought of as an
average force over the time interval.
In many interactions this net force is very large and this allows
one to ignore small forces like gravity and credit the force to the
main agent causing the impulse.
Take note at this time that the area under a F vs. t graph would
also yield the impulse.
a)What is the impulse given to a 3 kg ball that moves towards a
wall with a velocity of 12 m/s [E] and bounces off the wall with a
velocity of 10 m/s [W]?
b) If the ball was in contact with the wall for 0.0056 s then what
was the average force exerted by the wall on the ball?
a) 66 Ns [W]
b) 11 786 N [W]
LAW OF CONSERVATION OF LINEAR MOMENTUM
If the net force acting on a system of interacting objects is zero,
then the linear momentum of the system before the interaction
equals the linear momentum of the system after the interaction.
This law is more fundamental than the
conservation of energy and is
considered the most important law of
mechanics.
Consider a system where two objects exert equal and opposite
forces on each other (from Newton’s Third Law).


F AB   F BA


F AB t   F BA t


mB  v B   m A  v A


 pB   p A


 p A   pB  0




mA v A1  mB vB1  mA v A2  mB vB 2
The mechanical energy of a system is conserved only when
conservative forces alone act on the system. Momentum is
conserved regardless of the nature of the internal forces.
There is only one Ft within an interaction but there is more than
one possible Fd. Some Fd terms serve to remove kinetic
energy from the system under study. Studying the complete
system would conserve energy but this can be difficult to do.
a) Calculate the recoil velocity of an unconstrained rifle of mass 5
kg after it shoots a 50 g bullet at a speed of 300 m/s. with respect
to Earth.
assume the bullet is shot in the positive direction




mb vb1  mr vr1  mb vb 2  mr vr 2

m
0  0  (0.05kg)( 300 )  (5kg) vr 2
s

m
vr 2  3.00
s
b) A loaded railway car of mass 6000 kg is rolling to the right at 2
m/s when it collides and couples with an empty car of mass 3000
kg , rolling to the left on the same track at 3 m/s. What is the
velocity of the pair of cars after the collision?


right is the positive direction




ma va1  mb vb1  ma va 2  mb vb 2
the cars will have the same final velocity




ma va1  mb vb1  ma v2  mb v2
ma va1  mb vb1 
 v2
ma  mb
m
m
)  (3000kg)( 3 ) 
s
s v
2
6000  3000
(6000kg)( 2

m
v2  0.333
s
c) A bullet with a mass of 2 g is shot into a block of sand with a
mass of 300 g. The bullet lodges in the sand and rises a height of
7.58 cm. What was the speed of the bullet?
m
v  184.1
s
d) A 1 kg ball moving with a velocity of 2 m/s to the right collides
straight-on with a stationary 2 kg ball. After the collision the 2 kg
ball moves off to the right with a velocity of 1.2 m/s. What is the
velocity of the 1 kg ball after the collision? 
right is the positive direction
v A2
m
 0.400
s
Elastic and Inelastic Collisions
An elastic collision takes place when the total kinetic energy after
the collision equals the total kinetic energy before the collision.
An inelastic collision takes place when the total kinetic energy
after the collision is different from the total kinetic energy before
the collision(mostly lose sometimes gain).
A completely inelastic collision occurs when two objects colliding
stick together after contact. This brings about a maximum
decrease of Ek
In any collision there is a loss of Ek during the collision as the
“collision spring” compresses. In an elastic collision all of the Ek
returns as the “collision spring” decompresses. In an inelastic
collision the “collision spring” doesn’t decompress at all.
MOMENTUM IS CONSERVED IN ALL COLLISIONS!!
Analysis of Elastic Collisions


-one-dimensional only


mA v A1  mB vB1  mA v A2  mB vB 2




 (mB vB 2  mB vB1 )  m A v A2  m A v A1




 mB (vB 2  vB1 )  mA (v A2  v A1 )
(1)
m Av A21 mB vB21 m Av A2 2 mB vB2 2



2
2
2
2
2
2
2
2
mAv A1  mB vB1  mAv A2  mB vB 2
2
2
2
2
 (mB vB 2  mB vB1 )  mAv A2  mAv A1
2
2
2
2
 mB (vB 2  vB1 )  mA (v A2  v A1 ) (2)
divide equation (2) by
equation (1)




vB 2  vB1  v A2  v A1
Use the above equation and equation (1) to solve for the final
velocities. Assume the initial velocity of object B is zero, then . . .
v A2
m A  mB
2m A
(
)v A1 vB 2  (
)v A1
m A  mB
m A  mB
a) A 3 kg ball moving right at 5 m/s collides elastically with a
stationary 2 kg ball. What are the final velocities of each ball?
m A  mB
right is +
v (
)v
A2
m A  mB
A1


m
m
 1.00 [ R] v B 2  6.00 [ R]
s
s
v A2
3kg  2kg
m
v A2  (
)( 5 )
3 2
s
b) A 4 kg ball moving to the right at 5 m/s collides with a 2 kg ball
moving left at 4 m/s. If the collision is elastic then what are the
final velocities of the balls.
To use our derived equations the second object’s velocity must be
zero. This can be done by using a moving frame of reference.


m
If v B1  0 then v A1  9 [ R] This is a subtract 4 m/s [L] shift.
v A2
m  mB
( A
)v A1
m A  mB
v A2  (

v A2
4kg  2kg
m
)( 9 )
42
s

m
m
 3.00 [ R] v B 2  12.00 [ R]
s
s
s
These velocities are in the shifted
frame of reference (subtract 4
m/s [L]). To return to the original
frame of reference add 4 m/s [L].

m
v A2  1.00 [ R]
s

vB 2  8.00
m
[ R]
s
This question shows what can be asked if one is unsure if the
collision is elastic or not.
c) A 5 kg ball travelling 4 m/s right collides with a 3 kg ball
travelling 3 m/s left. After the collision the first ball is travelling 1
m/s right.
i) What is the final velocity of the second ball?
ii) Is this collision elastic or inelastic?
iii) If the collision was completely inelastic then what would the
final velocity of the two objects be?
homework
p. 359 1-4
extra p. 370
23,24,25a
ENERGY CHANGES WITHIN AN ELASTIC LINEAR
COLLISION
An elastic linear collision between two objects has three definite
stages. Let’s say object A collides with a stationary object B. While
they are in contact one can say the following.
Initially A is moving closer to B (vA> vB). At some point A moves at
the same velocity as B (vA= vB). This is minimum separation.
Finally A is moving farther away from B (vA < vB).
Since A and B are in contact with each other they must be
undergoing compression and elongation, in other words they are
acting as springs.
a) A 3 kg ball moving 5 m/s [R] collides elastically with a stationary
2 kg ball. The balls have a radius of 10 cm. The balls have a
spring constant of 1250 N/m.
i) What are the final velocities of the balls?
ii) How much kinetic energy is lost at min. sep.?
iii) How close were the centers of the two balls at min. sep.?
Right is +
v A2
m A  mB
2m A
(
)v A1 vB 2  (
)v A1
m A  mB
m A  mB
2(3kg) m
3kg  2kg m
m
)(5 )
v A2  (
)(5 ) v  1.00 vB 2  (
3 2
s A2
3 2
s
s
m
vB 2  6.00
s
At minimum separation the two objects have the same velocity. To
solve for this velocity we use the conservation of momentum
equation. For one side of this equation we can use the initial
velocities given or the final velocities calculated. Both yield the
same answer.




mA v A1  mB vB1  mA v A2  mB vB 2
m
(3kg)( 5 )  0  (3kg  2kg)v2
s

v 2  3.00
m
[ R]
s
The kinetic energy lost at minimum separation turns into elastic
potential energy due to compression of the balls.
mAv A12 mB vB12 kx12 mAv A2 2 mB vB 2 2 kx2 2





2
2
2
2
2
2
m2
m2
m2
(3kg)(5 )
(3kg)(3 ) (2kg)(3 )
2
kx
s 00 
s 
s 
2
2
2
2
kx2
 15 J
2
This is the energy lost at minimum separation.
Solving for the distortion of the balls . . .
N 2
(1250 ) x
m
 15 J
2
x  0.1549m
Since the balls undergo compression the negative answer is
correct and the balls are 4.51 cm apart at minimum separation.
b) A 4 kg cart moving 3 m/s [R] collides elastically with a 2 kg cart
moving 4 m/s [L]. One of the carts has a 30 cm spring with a
spring constant of 2000 N/m in front of it.
i) What are the final velocities of the balls?
ii) How much kinetic energy is lost at min. sep.?
iii) How close are the two carts at min. sep.?
CONSERVATION OF MOMENTUM IN TWO DIMENSIONS
Up to this point we have limited ourselves to one-dimensional situations
for analysis of momentum. Since momentum is a vector two and three
dimensional situations occur (rather plentifully), however our analysis
will extend only to two dimensions.
Two-dimensional collisions are difficult to analyze because of collision
geometry. In one dimension all collisions are head on but this is not the
case in two dimensions. Glancing contact, more direct contact or
elasticity differences lead to very different results. As a result we simply
use the vector equation of conservation of momentum as the basis of
our analysis involving diagrams and geometry.
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