Momentum 2013

Transcript Momentum 2013

nfl football momentum

Momentum is a commonly used term in sports. A team that has the momentum is on the move and is going to take some effort to stop.

Momentum as a physics term; refers to the quantity of motion that an object has. A sports team which is "on the move" has the momentum. If an object is in motion (on the move) then it has momentum.

An object’s momentum will change if its mass and/or velocity changes.

Most common… a change in velocity.

What is a change in velocity called?

Acceleration a = V f -V o t

a According to Newton’s laws, net force causes an object to accelerate , or change its velocity .

IMPULSE

Impulse J acting for a small time interval D t.

is a force F

Impulse:

J = F

Example 1: The face of a golf club exerts an average force of 4000 N for 0.002 s. What is the impulse imparted to the ball?

Impulse: J = F

J = (4000 N)(0.002 s) J = 8.00 N  s D t F The unit for impulse is the

Newton-second

(N·s)

Impulse from a Varying Force

Normally, a force acting for a short interval is not constant. It may be large initially and then play off to zero as shown in the graph.

F In the absence of calculus, we use the average force F avg .



F avg

time, t Unless told otherwise, treat forces as average forces

Impulse Changes Velocity

Consider a mallet hitting a ball: F



;



v f



v o

t F

 

v f



0 D

  

mv f



mv o

Impulse = Change in “mv”

Momentum Defined

Momentum

and velocity,

is defined as the product of mass

Units:

kg m/s

m = 1000 kg p =

ρ = (1000 kg)(16 m/s) ρ = 16,000 kg m/s

= 16 m/s

Impulse & Momentum

Impulse = Change in momentum

F

t = mv

- mv

D t F

A force

for a time acting on a ball D

increases its momentum

Example 2: A 50-g golf ball leaves the face of the club at 20 m/s. If the club is in contact for 0.002 s, what average force acted on the ball?

Given: m = 0.05 kg; v o = 0; D t = 0.002 s; v f = 20 m/s D t F + mv Choose right as positive .

F D t = mv f - mv o 0 F (0.002 s) = (0.05 kg)(20 m/s) Average Force:

F = 500 N

A 1000 kg car moving at 30 m/s (p = 30,000 kg m/s) can be stopped by 30,000 N of force acting for 1.0 s (a crash!) or by 3000 N of force acting for 10.0 s (normal stop)

http://www.youtube.com/watch?v=vtj6 Th6wb8g

http://www.youtube.com/watch?v=Z5BlObat1pQ

Another applications of impulse Contact time is reduced if arm's deceleration is kept as small as possible.

This is done by using "follow through", which means to continue to push during the contact period.

So to summarize… To minimize the effect of the force on an object involved in a collision, the time must be increased.

To maximize the effect of the force on an object involved in a collision, the time must be decreased.

BUT the change in momentum is the SAME either way!

Vector Nature of Momentum

+ v o Consider the change in momentum of a ball that is dropped onto a rigid plate: v f A 2-kg ball strikes the plate with a speed of 20 m/s and rebounds with a speed of 15 m/s. What is the change in momentum?

D p = mv f - mv o = (2 kg)(15 m/s) - (2 kg)(-20 m/s) D p = 30 kg m/s + 40 kg m/s D p = 70 kg m/s

Example 3: A 500-g baseball moves to the left at 20 m/s striking a bat. The bat is in contact with the ball for 0.002 s, and it leaves in the opposite direction at 40 m/s. What was average force on ball?

m 20 m/s + = 0.5 kg D t F + 40 m/s F D t = mv f - mv o v o = -20 m/s; v f = 40 m/s F(0.002 s) = (0.5 kg)(40 m/s) - (0.5 kg)(-20 m/s) Continued . . .

Example Continued:

m + = 0.5 kg 20 m/s F D t F D t = mv f - mv o + 40 m/s F(0.002 s) = (0.5 kg)(40 m/s) - (0.5 kg)(-20 m/s) F(0.002 s) = (20 kg m/s) + (10 kg m/s) F(0.002 s) = 30 kg m/s

F = 15,000 N

Summary of Formulas:

Impulse J = F avg

t Momentum

= mv

Impulse = Change in momentum

t = mv f - mv o

Conservation of Momentum

According to the

law of conservation of linear momentum

, when the vector sum of the external forces that act on a system of bodies equals zero, the total linear momentum of the system remains constant no matter what momentum changes occur within the system

For two objects interacting with one another, the conservation of momentum can be expressed as:

m v

1 1 

m v

2 2

v 1

and

v 2

velocities



m v

1 1

1 ' ' 

m v

2 2 '

' 2

ELASTIC AND INELASTIC COLLISIONS Elastic Collision : A collision in which objects collide and bounce apart with no energy loss.

Inelastic Collision : A collision in which objects collide and some mechanical energy is transformed into heat energy.

The animation below portrays the elastic collision between a 1000-kg car and a 3000-kg truck. The before- and after-collision velocities and momentum are shown in the data tables.

The animation below portrays the elastic collision between a 3000-kg truck and a 1000-kg car. The before- and after-collision velocities and momentum are shown in the data tables.

Before the collision, the momentum of the truck is 60 000 Ns and the momentum of the car is 0 Ns; the total system momentum is 60 000 Ns. After the collision, the momentum of the truck is 30 000 Ns and the momentum of the car is 30 000 Ns; the total system momentum is 60 000 Ns.

The animation below portrays the inelastic together with the same velocity.

collision

between a very massive diesel and a less massive flatcar. The diesel has four times the mass of the freight car. After the collision, both the diesel and the flatcar move

In elastic collisions no permanent deformation occurs; objects elastically rebound from each other.

In head-on elastic collisions between equal masses, velocities are exchanged.

Example 4: A 0.50-kg ball traveling at 6.0 m/s collides head on with a 1.00-kg ball moving in the opposite direction at a velocity of -12.0 m/s. The 0.50-kg ball moves away at -14 m/s after the collision. Find the velocity of the second ball.

M 1 V 1 = 0.50 kg = 6.0 m/s M V 2 2 = 1.00 kg = -12.0 m/s V f1 = -14 m/s m 1 V 1 p before = p after + m 2 V 2 = m 1 V f1 + m 2 V 2f (.5kg)(6m/s) + (1kg)(-12m/s) = (.5kg)(-14m/s) + (1kg)(V 2f ) V 2f = - 2 m/s

Inelastic collisions are characterized by objects sticking together and permanent deformation.

Example 5: A 3000-kg truck moving rightward with a speed of 5 km/hr collides head-on with a 1000-kg car moving leftward with a speed of 10 km/hr. The two vehicles stick together and move with the same velocity after the collision. Determine the post-collision velocity of the car and truck.

M 1 V 1 M 2 V 2 = 3000 kg = 5.0 km/hr = 1000 kg = -10 km/hr p before = p after m 1 V 1 + m 2 V 2



m v

= (m 1 + m 2 )V 1 1

1  

m v m

2 2 2 (3000kg)(5km/hr) + (1000kg)(-10km/hr) (3000kg + 1000kg) V = 1.25 km/hr, right

Explosions

• • • When an object separates suddenly, as in an explosion, all forces are internal.

Momentum is therefore conserved in an explosion.

There is also an increase in kinetic energy in an explosion. This comes from a potential energy decrease due to chemical combustion.

Recoil

• • • Guns and cannons “recoil” when fired.

This means the gun or cannon must move backward as it propels the projectile forward .

The recoil is the result of action-reaction force pairs, and is entirely due to internal forces . As the gases from the gunpowder explosion expand, they push the projectile forwards and the gun or cannon backwards.

Summary of Formulas:

Impulse J = F avg

t Momentum p = mv

Impulse = Change in momentum

t = mv f - mv o

Conservation of Momentum

m v

1 1 

m v

2 2 

m v

1 1 ' 

m v

2 2 '

Example 6 A 7500-kg truck traveling at 5 m/s east collides with a 1500-kg car moving at 20 m/s from a direction of 210



. After the collision, the two vehicles remain tangled together. With what speed and in what direction does the wreckage begin to move?

m 1 m 2

= 7500 kg v

= 1500 kg v 2 = 5 m/s, 0º = 20 m/s, 210º

m 1 v 1 + m 2

v 2 =( m

1 +m 2 )V

m 1 v 1 + m 2

v 2

x-comp

=( m

1 +m 2 )V

7500 kg (5 m/s) 1500kg (20m/s cos 210º) Σx = 11,519 kg m/s

y-comp

0 1500 kg (20m/s sin 210º) Σy = - 15,000 kg m/s

Initial Momentum

 (11519) 2  (15000) 2

= 18,912.7 kg m/s

initial momentum = final momentum

= (m

1 + m 2 ) V



initial m

1 

2  18912.7

= 2.1 m/s

 

tan -1

15000 11519

= 52.5º I quadrant V = (2.1 m/s, 52.5º )

Example 7: Suppose a 5.0-kg projectile launcher shoots a 209 gram projectile at 350 m/s. What is the recoil velocity of the projectile launcher?

y 3 m/s 2 kg 0 m/s 8 kg x Before y 2 m/s 2 kg 50 o x 8 kg v After

• Sample problem Calculate velocity of 8-kg ball after the collision.

2 m/s y y 2 kg 3 m/s 50 o 2 kg 0 m/s 8 kg x 8 kg x v Before After

2D-Collisions

• • • • Momentum in the x-direction is conserved.

▫ S P x (before) = S P x (after) Momentum in the y-direction is conserved.

▫ S P y (before) = S P y (after) Treat x and y coordinates independently.

▫ Ignore x when calculating y ▫ Ignore y when calculating x Let’s look at a simulation: ▫ http://surendranath.tripod.com/Applets.html

Center of Mass

• • • The center of mass of a system is the point at which all the mass can be assumed to reside.

Sometimes the system is an assortment of particles and sometimes it is a solid object.

Mathematically, you can think of the center of mass as a “weighted average”.

Center of Mass

–

x cm y cm z cm

  

x m i i



M y m i i

 

M z m i i M

system of points

• Analogous equations exist for velocity and acceleration, for example

 

v m i M

2 • Determine the Center of Mass.

y 2 kg 0 x 3 kg 1 kg -2 -2 0 2 4

Certain problems involve the use of conservation of energy and the conservation of momentum These problems require a special kind of Physics aptitude and are only attempted by the most savvy of Physics students…. You are those students.

The Ballistic Pendulum is used in the lab setting to find the velocity of fast moving projectiles

How fast was the bullet traveling before it struck the catcher? Is mechanical energy conserved before and after the collision?

821.3 m/s

Conservation of Momentum and Energy

Example 2 A 4 kg tennis racket moving with a velocity of 21 m/s strikes a 50 g tennis ball initially at rest. Find the velocity of the tennis ball and tennis racket after the collision.