Transcript No Slide Title
Chapter 7
Linear Momentum
Why study momentum ?
- In an isolated system, net force is zero, momentum is a conserved quantity Applications: - collision problems determination mass or velocity discovery of missing objects or sub-atomic particles (neutrino 中微子 ) http://www.hk-phy.org/articles/neutrino/neutrino.html
Linear Momentum - Like energy, it is conserved Define momentum
p d
p
dt
m v m d v
dt
v dm dt
F
p
Consider the rate of change of momentum
v dm dt
0 Mass of the system does not change in general Newton ’ s Second Law Newton actually stated his Second Law of motion as
Net External force equals
the
change in momentum
of a system
divided by the time over which it changes
.
Collision with a stationary object
Axis-aligning bounding box means that the sides of the box are horizontal and vertical If the stationary boundary is vertical, v If the stationary boundary is horizontal, v f = [v ix , -v iy ] For incoming velocity v i = [v ix , v iy ] f = [-v ix , v iy ] The angle that the ball comes in the angle of incidence (outgoing).
at must equal the angle at which it leaves , that is (incoming) must equal the angle of reflection Example • Suppose you are coding a simple Pong game, and you want to model the collision of the ball with the paddle . If the ball is approaching the paddle with an incoming velocity of [40, 75] when it hits, what should be its resulting velocity be ?
Answer The final velocity is [-40, 75] q q
Example 7.1
(a) Calculate the momentum of a 110Kg football player running at 8 m/s.
(b) Compare the player m=110Kg, v=8 m/s, m fb ’ s momentum with that of a hard-thrown 0.41Kg football that has a speed of 25 m/s.
=0.41 Kg, v fb =25 m/s p/p fb = ? = 85.9
Example 7.2
What is the average force does exert on a 0.14Kg baseball by a bat, given that the ball ’ s initial velocity is 45 m/s and that its final velocity, after a 1.3 ms impact, is 65 m/s in exactly the opposite direction? m=0.14 Kg, v 0 =45 m/s, v f =-65 m/s, in 1.3 ms F = ? D p/ D t = (p f – p i )/ D t = m(v f -v 0 )/0.0013 = 11800 N ~2600 lb
7.2 Impluse
Effect of a force on an object depends on how long it acts .
D p = (net F) (net F) D D t The change in momentum equals the average net external force multiplied by the time this force acts.
t is called the
impulse, I Impulse
is the
same as change in momentum
,
I=
D
p notice I //
D
p
體操
: for a given
D
p change, net F ↓ if
D
t ↑
less impact
I
D
p
t
1
t
2
F
(
t
)
dt
F eff
(
t
2
t
1 ) A piano hammer striking a string would generate a force similar to
F actual
but its impluse might be the same as that of
F eff
.
Example 7.3
Calculate the final speed of a 110Kg football player running at 8 m/s who collides head on with a padded goalpost and experiences a backward force of 17600 N for 0.055 s .
M=110 Kg, v i =8 m/s, net F=17600 N, D t=0.0550 s v f = ?
Use I = (net F) D t = M(v f – v i ) -17600*0.0550 = 110(v f v f – 8) = -0.800 m/s The minus sign indicates the player bounces backward .
7.3 Conservation of momentum
Under what condition is momentum conserved ?
Net external force = 0 consider a larger system
Consider the impulse
D D
p p
1 2 = F 1 = F 2
D D
p p 2 1 +
D
t
D
t
Newton ’ s Third Law
F 2 = (-F 1 )
D
p 2
D
=0 t = -
D
p 1 = -F 1
p 1 P total + p 2 = p 1 = constant + p 2 = p 1 ’ + p 2 ’ Conservation of Momentum (isolated system, net F = 0 )
P total P total
= constant =
P ’ total
- The three dimension in nature is independent - Momentum can be conserved along one direction and not another .
- Momentum is conserved along the X-direction , but not in Y-direction
7.4 Elastic collisions in one dimension
The two collided bodies moving the along the same direction Elastic collision – both momentum and internal kinetic energy conserved Very nearly elastic collision because some KE heat, sound
Example 7.4 An elastic collision Calculate the velocities of two masses following an elastic collision, given that m A = 0.500 Kg, m B = 3.5 Kg, v A = 4.00 m/s, v B = 0.
What are the final velocities of m A and m B ? v A ’ = ? , v B ’ = ?
Since v ½ m B m A =0, A v A v A 2 = m = p A A ½ = p A ’ v A ’ (m A + p + m v A ’ 2 B B ’ v B ’ + m B v B ’ 2 ) Solve the conservation of momentum for v B ’ first , and substitute into conservation of internal KE to eliminate v B ’ , leaving only v A ’ unknown .
There are two solutions for v A ’ , v A ’ = 4.00 m/s or -3.00 m/s First solution same as initial condition discarded 4.00 m/s v B ’ = 1.00 m/s A small mass m, collide with a larger mass M, the larger mass is knocked forward with a lower speed (here m B = 7m A ).
7.5 Inelastic collision in one dimension
Inelastic collision internal kinetic energy is not conserved , some internal KE may be converted into heat or sound energy Two equal masses head toward one another at equal speeds and stick together. KE int = mv 2 , KE ’ int = 0 internal KE not conserved The two masses come to rest after collision momentum conserved Perfectly inelastic, KE ’ int = 0
Example 7.5 Inelastic collision of puck and goalie (a) Find the recoil velocity of a 70 Kg hockey goalie, originally at rest, who catches a 0.15 Kg hockey puck slapped at him at a velocity of 35 m/s. (b) How much KE is lost in the collision? Assume friction between the ice and the puck-goalie system is negligible.
m 1 =0.15 Kg, v 1 (a) v 1 ’ =v 2 ’ =v ’ = =35 m/s, m ?
2 =70.0 Kg, v (b) How much KE is lost in collision ?
2 = 0 (a) m 1 v ’ v 1 + m 2 v 2 = (m 1 = 0.0748 m/s (b) KE ’ int – KE int = ½ + m = -91.7 J 2 ) v ’ (m 1 +m 2 )v ’ 2 solve for v – ½ m 1 v 1 2 ’
Example 7.6 A collision that releases stored energy is inelastic Two carts collide inelastically, and a spring releases its PE and converts it into internal KE. Mass of the cart on the left is m 1 = 0.35Kg, v 1 = 2.00m/s. Cart on the right has m 2 = 0.5Kg, v 2 = -0.5m/s. After collision, the first cart has a recoil velocity v 1 ’ = -4.00m/s.
(a) What is the final vel. of the first cart, v 2 ’ =? (b) how much energy is released by the spring (assume all internal PE is converted into internal KE) (a) m 1 v 1 + m 2 v 2 = m 1 v 1 ’ + m 2 v 2 ’ Solve the conservation of momentum for v 2 ’ (b) KE int KE ’ int KE ’ int = = ½ – ½ (m KE (m int 1 1 v 1 2 + m 2 v 1 ’ 2 + m = 5.46 J v 2 2 )= 0.763 J 2 v 2 ’ 2 ) = 6.22 J first, v 2 ’ =3.7m/s (internal KE increased energy released by the spring
7.6 Collision of point masses in two dimensions
Complication, if object rotation Point masses Neglect rotation Many scattering experiments have a target mass that is stationary in laboratory X component : m 1 v 1 = m 1 v 1 ’ cos Y component : 0 = m 1 v 1 ’ sin q 1 q 1 + m 2 v 2 ’ cos q 2 + m 2 v 2 ’ sin q 2
Example 7.7 Determine the final velocity of an unseen mass from the scattering of another mass A 0.25Kg mass is slid on a frictionless surface into a dark room, where it strikes an object with m 2 =0.4, v 2 =0. The 0.25Kg mass emerges from the room at an angle of 45 with its incoming direction, The speed of the 0.25Kg mass was v 1 =2 m/s, and it is 1.5 m/s after the collision. Calculate the magnitude and direction of the velocity (v ’ 2 and h) of the 0.4Kg mass after the collision.
X : m 1 v 1 = m 1 v 1 ’ cos45 + m 2 v 2 ’ cos q 2 Y : 0 = m 1 v 1 ’ sin45 + m 2 v 2 ’ sin q 2 m 1 = 0.250Kg, m 2 =0.400Kg
v 1 = 2.00m/s, v 2 =0 v 1 ’ =1.5 m/s solve for sin q 2 q 2 = -48.5
o and cos q 2 tan q 2 Given v 2 ’ one can determine mass of the unseen mass m 2
Elastic Collisions of Two Equal Masses
See p.185
- a special case m 1 = m 2 like the case with with some subatomic collisions billiard balls , or the case - assume v 2
½
mv 1 2 = ½
=0, and elastic collisions
m (v
X and Y components of momentum conserve X component : m Y component : 0 = m (X component) (Y component)
½
mv 1
2 1
1 ’
2
2
v
2
1
+ v
= m 1 v : v 1 2 = (v 1 ’ cos q 1 : 0 = (v
= ½ 2
1 ’
’ m v 1 ’ v 2 ’ 2
1
)
v sin 1 1 ’ ’ q cos 1 sin
m (v 1 ’ 2
q q
cos(
q
1
1 1 + m 2 v internal KE 2 ’ cos + m 2 v 2 ’ sin q 2 q 2 + v 2 ’ cos q 2 ) 2 + v 2 ’ sin q 2 ) 2
+ v 2 ’ 2 ) + m v 1 ’ v 2 ’
conserve
(1) v 1 ’ = 0
(2) v 2 ’ = 0
(3) cos(
q
1 cos(
q
1 +
q
2 )=0 head on collision, incoming mass stop +
q
2 ) no collision, incoming mass unaffected +
q
2 ) = 0
angle of separation is 90 o after collision
Particle Accerlator CERN
- European Organization for Nuclear Research 27 Km in diameter
World Class Particle accelerator
• CERN • http://cernenviro.web.cern.ch/CERNenviro/web/main/main .php
• Large Hadron Collider (LHC) • • http://livefromcern.web.cern.ch/livefromcern/antimatter/hi story/historypictures/LHC-drawing-half.jpg
• FermiLab • http://www.cs.cmu.edu/~zollmann/pics/2004_10_chicago/s lides/fermilab.html
• Stanford Linear Accelerator Center (SLAC) http://www.pbs.org/wgbh/nova/einstein/toda-ocon-01.html
A submicroscopic particle scatters straight backward from a target particle. In experiments seeking evidence for quarks, electrons were observed t occasionally scatter straight backward from a proton.