Chapter 6 Impulse and Momentum Continued

Chapter 6
Impulse and Momentum
Continued
6.3 Collisions in One Dimension
Elastic collisions with arbitrary masses
m1 − m2
v1f =
v1i
m1 + m2
Little mass
hits big mass
m2 >> m1
Initial
v1i
m1 >> m2
Let m1 → ∞
m2
m1
Bounces back
Let m1 → 0
Big mass
hits little mass
2m1
and v2f =
v1i
m1 + m2
v1i
v1f = −v1i
v1f = −v1i
v1i
m1
m2
Small mass ejected
at twice the speed
v2f ≈ 0
Final
m2
m1
Final
Initial
Equal mass
solution is here too
v1f ≈ v1i
and v2f ≈ 0
v1i
m1
v2f ≈ 2v1i
m2
and v2f ≈ 2v1i
6.3 Collisions in One Dimension
Inelastic collisions (with only internal forces affecting the motion)
v1i
Equal masses:
m1
Final
Initial
v2i = 0
m2
masses stick together
(x − components of the velocities)
Momentum conservation: m1v1i = ( m1 + m2 ) vf
Energy NOT conserved:
vf =
vf
m1 m2
m1
v1i
m1 + m2
How much kinetic energy was converted to heat in the collision.
K i = 12 m1v1i2
⎡ m1
⎤
K f = 12 ( m1 + m2 ) v 2f = 12 ( m1 + m2 ) ⎢
v1i ⎥
⎣ m1 + m2 ⎦
m12
m1
⎡⎣ 12 m1v1i2 ⎤⎦
=
v1i2 =
m1 + m2
2 ( m1 + m2 )
=
m1
K
m1 + m2 i
2
ΔK = K f − K i =
m1
K − Ki
m1 + m2 i
⎛ m2 ⎞
=⎜
K i (converted to heat)
⎟
⎝ m1 + m2 ⎠
If m2 >> m1 , ΔK ≈ K i
(all kinetic energy converted to heat)
Clicker Question 6.5
A 9-kg object is at rest. Suddenly, it explodes and breaks
into two pieces. The mass of one piece is 6 kg and the other
is a 3-kg piece. Which one of the following statements
concerning these two pieces is correct?
a) The speed of the 6-kg piece will be one eighth that of the 3-kg piece.
b) The speed of the 3-kg piece will be one fourth that of the 6-kg piece.
c)
The speed of the 6-kg piece will be one forth that of the 3-kg piece.
d)
The speed of the 3-kg piece will be one half that of the 6-kg piece.
e)
The speed of the 6-kg piece will be one half that of the 3-kg piece.
Clicker Question 6.5
A 9-kg object is at rest. Suddenly, it explodes and breaks
into two pieces. The mass of one piece is 6 kg and the other
is a 3-kg piece. Which one of the following statements
concerning these two pieces is correct?
a) The speed of the 6-kg piece will be one eighth that of the 3-kg piece.
b) The speed of the 3-kg piece will be one fourth that of the 6-kg piece.
c)
The speed of the 6-kg piece will be one fourth that of the 3-kg piece.
d)
The speed of the 3-kg piece will be one half that of the 6-kg piece.
e)
The speed of the 6-kg piece will be one half that of the 3-kg piece.
m1 = 6kg, m2 = 3kg


m1v1f + m2 v 2f = 0
m 
m
3kg
m = 6kg

v1 = − 2 v 2 speeds: v1 = 2 v2 =
v2 ⇒ v1 = 12 v2 m1 = 3kg
2
m1
m1
6kg
6.3 Collisions in One Dimension
Example: A Ballistic Pendulum
The mass of the block of wood
is 2.50-kg and the mass of the
bullet is 0.0100-kg. The block
swings to a maximum height of
0.650 m above the initial position.
Find the initial speed of the
bullet.

v1i
System of
two masses
h = 0.650m
Strategy – 1) After the bullet hits the block
the swing will conserve energy. From the


v
,
also
v
for the swing
height determine the starting velocity.
f
0
2) Use this velocity as the final
velocity of the collision. Then momentum
conservation determines the bullet’s initial velocity.
6.3 Collisions in One Dimension
Apply conservation of energy
to the swinging motion:
K + U = K0 + U 0
0 + mgh = 12 mv02 + 0
v0 = 2gh
Apply conservation of momentum
in the collision:



pTotal ,i = p bullet ,i + p block ,i = m1v1i + 0

pTotal ,f = (m1 + m2 )vf


Momentum conservation: pTotal ,f = pTotal ,i
(m1 + m2 )
(m1 + m2 )
v1i =
vf =
2gh
m1
m1
2.50 + .01
=
2(9.81)(0.65) m/s = 896 m/s
.01

v1i
h = 0.650m


v f , also v 0 for the swing
6.3 Collisions in One Dimension
Inelastic collisions (with only internal forces affecting the motion)
v1i
m1
Final
Initial
v2i = 0
m2
(x − components of the velocities)
Momentum conservation: m1v1i = ( m1 + m2 ) vf
Energy NOT conserved:
masses stick together
vf =
m1
v
m1 + m2 1i
vf
m1 m2
6.3 Collisions in One Dimension
Inelastic collisions (with only internal forces affecting the motion)
Final
Initial
v1i
v2i = 0
m2
m1
(x − components of the velocities)
Momentum conservation: m1v1i = ( m1 + m2 ) vf
Energy NOT conserved:
vf
masses stick together
vf =
m1 m2
m1
v
m1 + m2 1i
Final
Masses initially moving
toward each other
v1i
m1
Initial
v2i
m2
masses stick together
(x − components of the velocities)
Momentum conservation: m1v1i + m2 v2i = ( m1 + m2 ) vf
Energy NOT conserved:
vf =
vf
m1 m2
m1v1i + m2 v2i
m1 + m2
For example: v1i = +5.0 m/s, v2i = −10.0 m/s, m1 = m2 = m (same mass)
vf =
m1v1i + m2 v2i
m
=
(v1i + v2i ) = 0.5(+5 − 10)m/s = −2.5 m/s
m1 + m2
2m
Clicker Question 6.6
A mass with a momentum of +10.0 kg ⋅m / s , collides with a mass
twice as big with a momentum of – 6.0 kg ⋅m / s , and they stick together.
What is the momentum of the combined system after the collision?
a) − 2.0 kg ⋅ m/s
b) + 2.0 kg ⋅ m/s
c) + 4.0 kg ⋅ m/s
d) + 6.0 kg ⋅ m/s
e) + 16.0 kg ⋅ m/s
Clicker Question 6.6
A mass with a momentum of +10.0 kg ⋅m / s , collides with a mass
twice as big with a momentum of – 6.0 kg ⋅m / s , and they stick together.
What is the momentum of the combined system after the collision?
a) − 2.0 kg ⋅ m/s
b) + 2.0 kg ⋅ m/s
c) + 4.0 kg ⋅ m/s
d) + 6.0 kg ⋅ m/s
e) + 16.0 kg ⋅ m/s

p(1+2)f
Momentum is conserved


= p1i + p 2i
= ( +10.0 kg ⋅ m/s ) + (−6.0 kg ⋅ m/s)
= +4.0 kg ⋅ m/s
6.4 Collisions in Two Dimensions

v1i = 0.900m/s
m1 = 0.150 kg
Determine the final momentum vector
for mass 1.

v1f
v1f y = 0.12m/s

v 2i = 0.540m/s
m2 = 0.260 kg

v1f
v1f x = 0.630m/s

v 2f = 0.700m/s
6.4 Collisions in Two Dimensions

v1i = 0.900m/s
m1 = 0.150 kg
Determine the final momentum vector
for mass 1.

v1f
v1f y = 0.12m/s

v 2i = 0.540m/s
m2 = 0.260 kg

v1f
v1f x = 0.630m/s

v 2f = 0.700m/s
+2m
v2f2 v=
m
= vm v+ m
+m
v 2 v2i
x-components:m1vm1f1v+1f m
2f 1 1i1 1i 2 2i
v1i = +0.900sin50° m/s , v2i = +0.540 m/s, v2f = +0.700cos35° m/s
6.4 Collisions in Two Dimensions

v1i = 0.900m/s
m1 = 0.150 kg
Determine the final momentum vector
for mass 1.

v1f

v1f
v1f y = 0.12m/s

v 2i = 0.540m/s
m2 = 0.260 kg
v1f x = 0.630m/s

v 2f = 0.700m/s
x-components: m1v1f + m2 v2f = m1v1i + m2 v2i
v1i = +0.900sin50° m/s , v2i = +0.540 m/s, v2f = +0.700cos35° m/s
y-components: m1v1f + m2 v2f = m1v1i + m2 v2i
v1i = −0.900cos50° m/s v2i = 0, v2f = −0.700sin35° m/s
6.4 Collisions in Two Dimensions

v1i = 0.900m/s
m1 = 0.150 kg
Determine the final momentum vector
for mass 1.

v1f

v1f
v1f y = 0.12m/s

v 2i = 0.540m/s
m2 = 0.260 kg
v1f x = 0.630m/s

v 2f = 0.700m/s
x-components: m1v1f + m2 v2f = m1v1i + m2 v2i
v1i = +0.900sin50° m/s , v2i = +0.540 m/s, v2f = +0.700cos35° m/s
y-components: m1v1f + m2 v2f = m1v1i + m2 v2i
v1i = −0.900cos50° m/s v2i = 0, v2f = −0.700sin35° m/s
final x : v1x = +0.63 m/s final y : v1y = +0.12 m/s
(
)
v1 = v1x2 + v1y2 = +0.64 m/s; θ1 = tan −1 v1y v1x = 11°
6.4 Collisions in Two Dimensions
In the elastic collision, m1 is
deflected upward at 90°.

p1i = 35 kg ⋅ m/s
1
m1 = 3.50 kg

p 2i = 0
2
Determine the final momentum vector
for both masses.
p1fy
1
m1 = 3.50 kg
90°
2
m2 = 10.5 kg
Momentum conservation
x-components: p1i = p2fx = 35 kg ⋅ m/s
y-components: p1fy = − p2fy (need this)
2
m2 = 10.5 kg
p2fx
p2fy

p 2f
6.4 Collisions in Two Dimensions
In the elastic collision, m1 is
deflected upward at 90°.

p1i = 35 kg ⋅ m/s

p 2i = 0
1
2
m1 = 3.50 kg
Determine the final momentum vector
for both masses.
p1fy
1
90°
2
m2 = 10.5 kg
Momentum conservation
x-components: p1i = p2fx = 35 kg ⋅ m/s
y-components: p1fy = − p2fy (need this)
Kinetic Energies
p1i2
Ki =
2m1
Kf =
2
p1fy
2m1
+
p 22fx + p 22fy
2m2
m1 = 3.50 kg
2
m2 = 10.5 kg
p2fx
p2fy

p 2f
6.4 Collisions in Two Dimensions
In the elastic collision, m1 is
deflected upward at 90°.

p1i = 35 kg ⋅ m/s

p 2i = 0
1
2
m1 = 3.50 kg
Determine the final momentum vector
for both masses.
p1fy
1
90°
2
m2 = 10.5 kg
Momentum conservation
x-components: p1i = p2fx = 35 kg ⋅ m/s
y-components: p1fy = − p2fy (need this)
Kinetic Energies
p1i2
Ki =
2m1
2
p1fy
p 22fx + p 22fy
Kf =
+
2m1
2m2
p 22fy
2m1
+
p1i2 + p 22fy
2m2
m1 = 3.50 kg
2
m2 = 10.5 kg
p2fx
p2fy

p 2f
6.4 Collisions in Two Dimensions
In the elastic collision, m1 is
deflected upward at 90°.

p1i = 35 kg ⋅ m/s

p 2i = 0
1
2
m1 = 3.50 kg
Determine the final momentum vector
for both masses.
p1fy
1
m1 = 3.50 kg
90°
2
m2 = 10.5 kg
2
p2fy
Momentum conservation
x-components: p1i = p2fx = 35 kg ⋅ m/s
y-components: p1fy = − p2fy (need this)
p
2
2fy
2m1
+
p +p
2
1i
2m2
2
2fy

p 2f
Energy conservation
Kf = Ki
Kinetic Energies
p1i2
Ki =
2m1
2
p1fy
p 22fx + p 22fy
Kf =
+
2m1
2m2
m2 = 10.5 kg
p2fx
p
2
2fy
p2fy = ±
⎛ 1
⎛ 1
1⎞
1⎞
2
⎜ m + m ⎟ = p1i ⎜ m − m ⎟
⎝ 1
⎝ 1
2⎠
2⎠
1
p1i
⇒
p2fy = −24.7 kg ⋅ m/s
2
therefore, p1fy = +24.7 kg ⋅ m/s
6.5 Center of Mass
The center of mass is a point that represents the average location for
the total mass of a system.
m1x1 + m2 x2
xcm =
m1 + m2
6.5 Center of Mass
m1!x1 + m2 !x2
!xcm =
m1 + m2
m1v1 + m2 v2
vcm =
m1 + m2
6.5 Center of Mass
m1v1 + m2 v2
vcm =
m1 + m2
In an isolated system, the total linear momentum does not change,
therefore the velocity of the center of mass does not change.
6.5 Center of Mass
BEFORE
m1v1 + m2 v2
vcm =
=0
m1 + m2
v2f
AFTER
m1v1 + m2 v2
vcm =
m1 + m2
88 kg ) ( −1.5m s ) + (54 kg ) ( +2.5m s )
(
=
88 kg + 54 kg
= 0.00
v1f