Example: An eagle of mass, mA = 6.0 kg moving with speed vA = 5.0 m/s
is on collision course with a second eagle of mass mB = 4.0 kg moving at
vB = 10.0 m/s in a direction perpendicular to the first. After they collide,
they hold onto one another. At what speed they moving after the
collision?
mA=6.0 kg
vA=5.0 m/s
mB=4.0 kg
vB=10.0 m/s
v-?

vB

pB

vA

pA
m  mA  mA  6.0 kg  4.0 kg  10.0kg
p A  mA p A  6.0kg  5.0m / s  30.0kg  m / s
pB  mB pB  4.0kg 10.0m / s  40.0kg  m / s
p
v 
m
p A2  pB2

m
 

p  pA  pA
p
p A2  p B2
30.0kg  m / s 2  40.0kg  m / s 2
10 kg
 5m / s
Example: Suppose rain falls vertically into an open cart rolling along a
straight horizontal track with negligible friction. As a result of the
accumulating water, the speed of the cart
1.
2.
3.
Increases
Does not change
Decreases
Mass is increasing
P = mv must be conserved (Fext = 0)
Speed must decrease
Ballistic Pendulum
A simple device for measuring the speed of projectile
m,v1
M m
v2
M
1. Before collision:
p1=mv1
2
mv
K1  1 ;
2
p1=p2
M m
U1  0
3. At the highest point:
2. After collision:
p2=(m+M)v2
p3=0
( m  M )v 2
K2 
;
2
U 2  0 K 3  0;
2
v1 
mv1 = (m+M)v2
(K 2  U 2 )  (K3  U 3 )
h
(m  M )
v2
m
(m  M )v2
 ( M  m) gh
2
2
U 3  (M  m) gh
(m  M )
v1 
2 gh
m
Example:
m1
h1
m2
m1v12
m1 gh1 
2
m1v1  (m1  m2 )v
m1  m2 v
2
2
 m1  m2 gh2
h2
v1  2gh1
v
m1
m1
v1 
2gh1
m1  m2
m1  m2
2
 m1 
v
 h1
h2 
 
2 g  m1  m2 
2
Example: You want to knock down a large bowling pin by throwing a ball at it.
You can choose between two balls of equal mass and size. One is made of
rubber and bounces back when it hits the pin. The other is made of putty and
sticks to the pin. Which ball do you choose?
A. The rubber ball.
B. The putty ball.
p
C. It makes no difference.
putty
p
p’
rubber
p+p’
Elastic Collisions (1D)
The kinetic energy of the system is conserved: after the collision it is
the same as that before
Before
Before
v
v
x
A
x
A
B
After
B
After
vA
vA
vB
vB
x
A
B
x
A
B
Elastic Collisions (1D)

mA (v A
'
A
(v A
v
'
A
(1)
mAv A2 mB vB2 mAv'2A mB v'2B



2
2
2
2
(2)
  m v  v' 
 v )  m v  v 
 v )  v  v 
mA v  v'
2
A
2
A
mA v A  mB vB  mA v' A mB v' B
'
A
B
B
B
2
B
2
B
B
'
B

mA  mB v A  2mB vB

mA  mB
'
B

(v A  vB )   v A'  vB'
v
'
B


mB  m A v B  2m A v A

m A  mB
(2a)
Example: A steel ball with mass m1 = 1 kg and initial speed v0 collides
head-on with another ball of mass m2 = 2 kg that is initially at rest.
What are the final speeds of the balls?
1)Conservation of momentum:
v0  0  v1  2v2
2) Conservation of energy:
1
2
v02  0  12 v12  12 v22
We have 2 equations and 2 unknown: v1 and v2.
It is more convenient to use eq. (2a) instead of eq.(2)
2a) Relative velocity:
v0  0  v1  v2 
Velocity of 1 relative to 2
before the collision
Adding eq. 1and 2a:
2v0  3v2
Velocity of 1 relative to 2
after the collision
v2  23 v0
v1   13 v0
Example: Carefully place a small rubber ball (mass m) on top of a much
bigger basketball (mass M>>m) and drop these from some height h.
What is the velocity of the smaller ball after the basketball hits the
ground, reverses direction, and then collides with small rubber ball?
Relative
speed is 2v Here too!
v
m
M
v
V?
3v
v
v
v
Remember that relative velocity has to be equal before and after collision!
Before the collision, the basketball bounces up with v and the rubber ball
is coming down with v, so their relative velocity is –2v. After the
collision, it therefore has to be +2v!!
This means that, after collision, the velocity of the smaller ball after is 3v.
Newton’s craddle (a row of adjacent steel-ball )
If two balls on the left are pulled to a certain
height h and released, what happens?
A. One ball rises on the right, but higher than h
pi = pf
2mvi = mvf
vf = 2vi ,
but then kinetic energy would not be conserved:
KEi = 2½mvi2 = mvi2
KEf = ½mvf2 = 2mvi2
B. Two balls rise on the right at height h
pi = pf
2mvi = 2mvf
KEi = 2 ½mvi2 = mvi2
vf = vi.
KEf = 2 ½mvf2 = mvi2
Ok!
C. A ball will rise on each side to height h
Same height means |vi|= |vf| (from conservation of energy),
but this violates conservation of momentum: pi = 2mvi ; pf = mvi – mvi = 0
Elastic Collisions in 2D and 3D


'
'
mAvA  mBvB  mAvA  mBvB
mAv A2 mB vB2 mAv'2A mB v'2B



2
2
2
2


v A  (v Ax , v Ay , v Az )
2
2
2
v A2  v Ax
 v Ay
 v Az
In 2D and 3D the initial velocities of the two particles dose not determine
the final velocities. You also need to know the direction of one of the final
particles.
Kinetic energy:
m v2 m v
p2
K


2
2m
2m
2
2D Elastic: Nuclear scattering
A particle of unknown mass M is initially at rest. A particle of known mass m is
“shot” against it with initial momentum pi. After the collision, the momentum of
the particle of known mass is measured again, and it is pf. If the collision is
elastic, that’s all we need to determine M and the final momentum of the target, P.
pf
M (at rest)
m
pi
3 unknowns:
3 equations:
M, Px, Py
P
conservation of momentum in the x direction
conservation of momentum in the y direction
conservation of kinetic energy
pi  pf  P
(2 equations)
pf2 P 2
pi2


2m 2m 2M
P  pi  pf
2
  2
2
P  P   pi  pi 
M 2
2

P 
pi  p 2f 
m
M m
( pi  pf )2
( pi2  pf2 )