Download Chapter 6 Impulse and Momentum Continued

Chapter 6
Impulse and Momentum
Continued
6.4 Collisions in Two Dimensions
Momentum conservation can be used to solve collision
problems if there are no external forces affecting
the motion of the masses.
Energy conservation can be used to solve a collision
problem if it is stated explicity that the collision is ELASTIC.
If two masses in a collision stick together, the collision
must be INELASTIC. Energy conservation cannot be used
unless the energy lost or gained is provided.
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
54 kg ) ( +2.5m s ) + (88 kg ) ( −1.5m s )
(
=
88 kg + 54 kg
= 0.00
v1f
Forces and Newton’s Laws of Motion
Force Magnitudes
Gravitational Force: FG = mg (magnitude), direction downward, weight
Elastic Forces: (stretched spring) FS = kx (magnitude), inward at the ends
Compression: (squeezed spring or object) normal forces, n, ⊥ contact
Frictional: opposes motion, Static - f Max = µ S n, then Kinetic - f K = µ K n

 
Net Force: FNet = F1 + F2 +… on a single object (magnitude and direction)

Newton's 1st Law: speed and direction do not change, if & only if FNet = 0


Newton's 2nd Law: FNet = ma


Newton's 3rd Law for two masses (1,2) in contact: F12 = −F21

Decomposition to components: F = Fx î + Fy ĵ , Fx = F cosθ , Fy = F sin θ
Components to magnitude angle: F = Fx2 + Fy2 , θ = tan −1 (Fy Fx )
Summaries and Examples
Energy and energy conservation
Momentum and momentum conservation
Energy and energy conservation
Work:W = F(cosθ )Δx
Kinetic Energy: K = 12 mv 2
Work changes kinetic energy: W = K − K 0 = 12 mv 2 − 12 mv02
Conservative forces ⇒ Potential Energy
Gravitational Potential Energy:U G = mgy
Ideal Spring Potential Energy:U S = 12 kx
2
Total Energy: E = K + U
Work by non-conservative forces (friction, humans, explosions)
changes total energy: WNC = ( K − K 0 ) + (U − U 0 )
If WNC = 0, there is total energy conservation:
E = E0
⇒ K + U = K0 + U 0
Average power = Work time = (Energy change) time = F v
Momentum and momentum conservation
 


Momentum: p = mv
Impulse: J = Ft
Net average impulse changes momentum:

 


FΔt = p − p0 = mv f − mv 0



Momentum of 2 masses in collision: pTotal = m1v1 + m2 v 2
No net external force active, collision momentum is conserved






pTotal ,f = pTotal.i ⇒ m1v1f + m2 v 2f = m1v1i + m2 v 2i


m1v1 + m2 v 2
=
m1 + m2
m1x1 + m2 x2

Center of mass position: xcm =
, and velocity v cm
m1 + m2

Conservation of momentum ⇒ v cm remains constant
Clicker Question 6.7
A space probe travels with a momentum of +6.0 × 107 kg ⋅m / s .
A retrorocket fires for 10 s, with a force magnitude of 2.0 × 106 N , that
slows the speed of the probe. What is the momentum of the probe
after the rocket ceases to fire?
a) 2.0 × 107 kg ⋅ m/s
b) 4.0 × 107 kg ⋅ m/s
c) 6.0 × 107 kg ⋅ m/s
d) 8.0 × 107 kg ⋅ m/s
e) 10.0 × 107 kg ⋅ m/s
Clicker Question 6.8
Two identical 2.0-kg objects are involved in a collision. The objects
are on a frictionless track. The initial velocity of object A is +12 m/s
and the initial velocity of object B is – 6.0 m/s. What is the final
velocity of the two objects?
a) Object A moves at – 6.0 m/s and B moves at +12 m/s.
b) Object A moves at + 9.0 m/s and B moves at +9.0 m/s.
c) Object A moves at – 6.0 m/s and B moves at +6.0 m/s.
d) Object A moves at – 12 m/s and B moves at +6.0 m/s.
e) This cannot be determined without more information.
Example:
An 11-g bullet is fired horizontally into a 110-g wooden block
that is initially at rest on a frictionless horizontal surface and
connected to a spring having spring constant 183 N/m. The
bullet becomes embedded in the block. The bullet-block
system compresses the spring by a maximum amount 85 cm.
What was the initial velocity of the bullet?
m1
x
 k
v1
m2
Strategy:
1) Momentum conservation bullet + block
2) Energy conservation in spring compression
m1
x
 k
v1
m2
1) Momentum conservation bullet + block
m1v1 = (m1 + m2 )v ⇒ v = ⎡⎣ m1 (m1 + m2 ) ⎤⎦ v1
m1
x
 k
v1
m2
1) Momentum conservation bullet + block
m1v1 = (m1 + m2 )v ⇒ v = ⎡⎣ m1 (m1 + m2 ) ⎤⎦ v1
2) Energy conservation in spring compression
2
Kinetic Energy of block and bullet : K i = 12 (m1 + m2 )v
Potential Energy of compressed spring : U = 12 kx 2
m1
x
 k
v1
m2
1) Momentum conservation bullet + block
m1v1 = (m1 + m2 )v ⇒ v = ⎡⎣ m1 (m1 + m2 ) ⎤⎦ v1
2) Energy conservation in spring compression
2
Kinetic Energy of block and bullet : K i = 12 (m1 + m2 )v
Potential Energy of compressed spring : U = 12 kx 2
Energy Conservation : K + U = K i + U i ; U i = 0, K = 0
m1
x
 k
v1
m2
1) Momentum conservation bullet + block
m1v1 = (m1 + m2 )v ⇒ v = ⎡⎣ m1 (m1 + m2 ) ⎤⎦ v1
2) Energy conservation in spring compression
2
Kinetic Energy of block and bullet : K i = 12 (m1 + m2 )v
Potential Energy of compressed spring : U = 12 kx 2
Energy Conservation : K + U = K i + U i ; U i = 0, K = 0
1
2
2
kx = (m1 + m2 )v = (m1 + m2 ) ⎡⎣ m1 (m1 + m2 ) ⎤⎦ v12
2
1
2
2
1
2
v1 = k(m1 + m2 ) m12 x
= (183N/m)(0.121kg) (0.011kg)2 (0.85 m) = 364 m/s