Impulse and Momentum What Is Momentum? Mass in Motion

Chapter 7

Impulse and Momentum What is momentum? Mass in Motion. p mv

How hard something is to stop. hard to stop small mass, big velocity

p = m v hard to stop, big mass, small velocity P = M v Hard to Stop! MV STOP IT!

Dp = F Dt = = F Dt What is Impulse? Impulse changes momentum. Dv F ma  m Dt FD t  m D v Dmv I F D t  D p Stop it! Hay stack or Concrete Wall?

Dp = F Dt = = F Dt Same Impulse & Change in Momentum Different Force, Different Time

Dp = F Dt = = F Dt Stop It!

Same Impulse, different Force and Time. 7.1 The Impulse-Momentum Theorem

There are many situations when the force on an object is not constant.

DEFINITION OF IMPULSE

The impulse of a force is the product of the average force and the time interval during which the force acts:

  J  FDt

Impulse is a vector quantity and has the same direction as the average force.

newton seconds (N s)

  J  FDt

DEFINITION OF LINEAR MOMENTUM

The linear momentum of an object is the product of the object’s mass times its velocity:   p  mv

Linear momentum is a vector quantity and has the same direction as the velocity.

kilogrammeter/second (kg m/s)

   mv  mv F  f o  Dt

    FDt  mvf  mvo

Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. What is the average Force exerted on a 0.140 kg baseball by a bat given that the ball’s initial velocity is 45.0m/s and its final velocity, after 1.3ms impact, is 65.0m/s in the opposite direction?

pmv00 (.14 kg )(  45 ms / )   6.3 kgms  /

pff mv (.14 kg )(65 ms / )  9.1 kgms  /

Dpp f  p0 9.1 kgms  /  (  6.3 kgms  / ) Dp 15.4 kg  m / s  I

Dp (15.4kg m / s ) 4 F  1.18xN 10 Dt 1.3xs 103 + Bug Splat Compared to the FORCE that acts on the bug, how much force acts on the bus? More Same Less

Newton’s 3rd Law:

FFbus bug bug bus Bug Splat Compared to the IMPULSE on the bug, how much impulse is on the bus? More Same Less

FFbus bug bug bus

Fbus bugD t   F bug bus D t Bug Splat Compared to the IMPULSE on the bug, how much impulse is on the bus? More Same Less

FFbus bug bug bus

Impulsebug -Impulse bus Bug Splat Compared to the Change in Momentum of the bug, the momentum of the bus changes? More Same Less

Impulse Dp

Impulsebug -Impulse bus

Dp-bug  Dpbus Bug Splat Which undergoes the greater acceleration?

F Bug Same Bus a  m

Which suffers the greatest damage?

Bug Same Bus Bounce/ No Bounce Bullets You are at a war protest when the cops start shooting pellets at the rowdy anarchists. Which would you rather be hit with, rubber bullets that bounce or don’t bounce? Both have the same mass & initial speed.

Bounce: Dp  m( vf  (  v0 ))

m( v0  v 0 )  2 mv 0

No Bounce: Dp 0   mv00  mv Newton’s Third Law

FFaction reaction Newton’s Third Law

FFaction reaction DDpp ball cannon DDtt

Dppball  D cannon

()()mb v bf m b v b00   m c v cf  m c v c

mb v bf m c v cf  m b v b00  m c v c pp Conservation of total,, before total after Momentum! Conservation of Momentum If there are no external forces acting on a system of objects then the total momentum before an event is equal to the total momentum after an event.

ppbefore after The system

100kg 100kg  pbefore  mv 0  mv

THE EVENT v p 2 m mv  after 2 Conservation Depends on The System

100kg 100kg  pbefore  mv

THE EVENT

External v Force! pm  after 2 Momentum is NOT conserved! Colliding Systems For momentum to be conserved you must define The System so all the forces are internal to it and there are no net external forces. Include everything that gets hit. Conservation of Momentum If there are no external forces acting on a system of objects then the total momentum before an event is equal to the total momentum after an event.

ppbefore after System: You define the system so that momentum is conserved. •No external forces acting on the system •Internal forces cancel out and don’t change momentum •Include all the objects that collide or ‘go bump’ •Always apply conservation laws to a system

When you have defined the system so that the net external forces acting is zero, THEN & ONLY THEN can you invoke Conservation of momentum! ONLY AFTER you have defined the system so that the net external forces acting on it is zero, THEN & ONLY THEN can you invoke Conservation of momentum and apply:

ppbefore after Conservation of Momentum

Momentum is conserved for which system?

(a) (b) (c)

INCLUDE EVERYTHING THAT GOES BUMP!

NOTICE HOW THE INTERNAL FORCES CANCEL! The recoil momentum of a gun that kicks is more than less than the same as the momentum of the bullet it fires.

Momentum is conserved for the system including: gun alone gun + man gun + man + bullet

Frictionless surface Charlie and his gun, m = 80kg, shoots off a bullet, m =.01kg, straight ahead at 300 m/s. What is Charlie’s (+gun) recoil velocity?

1. Define the System so net external forces is zero.

2. THEN momentum is conserved

ppinitial final pcharlie,,,, i p bullet i  p charlie f  p bullet f

0 ppcharlie,, f gun f

ppcharlie,, f bullet f

mc v cf m b v bf .01kg mb  300ms / Frictionless surface vvcf bf 80kg mc .04ms / Other Exploding Systems

ppinitial final Rocket Thrust

ppinitial final

0 Mvrocket mv gas Dprocket Mv

Mvrocket mv gas

Dpprocket  D gas Dpgas mV

Rocket Pushes Gas Out. Gas Pushes Back on Rocket. Collision Systems: Total Momentum is Conserved

Once the System is properly defined Momentum is conserved for all Collisions. Total momentum before = Total momentum after. Elastic Collisions: Bounce Momentum is transferred!

Total momentum before = Total momentum after Inelastic Collisions: Stick

Total momentum before = Total momentum after Collision Decision

Both trucks have the same speed and mass. After they collide they will: a) Bounce off each other b) Stick and move to the left c) Stick and move to the right d) Stick and stop dead in the center e) Vaporize Inelastic Collisions: Stick

Was Momentum Conserved in this case? Depends on the System! Elastic Collisions: Bounce What is the System so momentum is conserved? If the mass is the same what is her final velocity?

1 2 ppinitial final

mv1i mv 2 i  mv 1 f  mv 2 f

v1i v 2 i  v 1 f  v 2 f

2m/s v2f v 1 i  v 2 i  v 1 f

v2 f 2 m / s  (  1 m / s )  (  1 m / s ) 1 2

v2 f  2/ m s Momentum is Exchanged! Elastic Collisions

What if you didn’t know the final velocity of car 1? Could you still find the final velocities? 1 2

Use Conservation of Momentum AND Kinetic Energy! Solve 2 equations for two unknowns: the velocities! pp 1 2 f 0

KEf  KE0 Elastic Collisions

ppf  0

mv1ff mv 2  mv 10  mv 20

1 2 KEf  KE0

12 1 2 1 2 1 2 mv1ff mv 2  mv 10  mv 20 2 2 2 2

Plus lots of Algebra!!

Can you use this method to 1 2 solve for Inelastic Collisions?

Kinetic Energy is NOT Conserved! Inelastic Collisions

Granny (m=50kg) whizzes around the rink at 3m/s and snatches up Andy (m=25kg). What is their final velocity? (Ignore friction)

Before After Granny: m=50kg, v =3m/s Inelastic Collisions 0 Andy: m=25kg, v0=0

What is the system?

ppinitial final

Before After Granny: m=50kg, v =3m/s Inelastic Collisions 0 Andy: m=25kg, v0=0

ppinitial final

m1 v 1if() m 1 m 2 v

mv11i v f  ()mm12 50kg 3 m / s  1 (50kg 25 kg ) 2  2/ms

Before After Collisions Momentum is conserved for all Collisions. Momentum is transferred during Collisions.

Elastic: The total Kinetic Energy of the system is conserved. Think Bounce!

Inelastic: The total Kinetic Energy of the system is not conserved. Think Stick!

Why is momentum conserved during Inelastic Collisions but kinetic energy is not? Inelastic Collisions

Kinetic Energy is ‘lost’ to heat & other kinds of energy. There is only one kind of momentum. Therefore it is conserved. 7.2 The Principle of Conservation of Linear Momentum

Example 6 Ice Skaters

Starting from rest, two skaters push off against each other on ice where friction is negligible.

One is a 54-kg woman and one is a 88-kg man. The woman moves away with a speed of +2.5 m/s. Find the recoil velocity of the man.

m1v f 1  m2v f 2  0

m1v f 1 v f 2   m2

54 kg 2.5m s v    1.5m s f 2 88 kg

A baseball is thrown to you at 15m/s. Now a medicine ball, with a mass 10 times that of the baseball is thrown to you. Rank the following choices from slowest to fastest for the medicine ball:

3 a) the same speed as the baseball vmb 15 m / s

mvbb bb 1 b) the same momentum as the baseball vmb  1.5ms / mmb 2 c) the same kinetic energy as the baseball

1122 mbb 1 mbb v bb m mb v mb vvmb bb  15m / s  4.7 m / s 22 mmb 10 Momentum is a Vector p mv Inelastic Collisions A B C D E

Equal Masses Equal Speeds Inelastic Collisions A B C D E

Unequal Masses Equal Speeds Inelastic Collisions A B C D E

Equal Masses Unequal Speeds Inelastic Collisions A B C D E

Equal Masses Unequal Speeds Inelastic Collisions A B C ? D E

Unequal Masses Unequal Speeds A car and a truck, velocities and masses shown, collide and lock bumpers. Find the final velocity (magnitude and direction) of the pair. v v = 15 m/s f c  370 Mc = 2000 kg

vt = 5 m/s

Mt = 8000 kg Center of Mass The geometric ‘center’ or average location of the mass. Center of Mass: Stability

If the Center of Mass is above the base of support the object will be stable. If not, it topples over. Balance and Stability

This dancer balances en pointe by having her center of mass directly over her toes, her base of support.

https://www.youtube.com/watch?v=o9rFZ_SzGPI Slide 12-88 Center of Mass The geometric ‘center’ or average location of the mass.

System of Particles:

mxii i xCM  Mtotal 7.5 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 Balancing Act

A meter stick has a mass of 75.0 grams and has two masses attached to it: 50.0 grams at the 20.0cm mark and 100.0 grams at the 75.0 cm mark. Find the center of mass of the system - that is, at what mark on the meter stick should the fulcrum be placed so that the system balances? Rotational & Translational Motion Objects rotate about their Center of Mass.

The Center of Mass Translates as if it were a point particle. Center of Mass

The Center of Mass Translates as if it were a point particle and, if no external forces act on the system, momentum is then conserved. This means: EVEN if the bat EXPLODED into a thousand pieces, all the pieces would move so that the momentum of the CM is conserved – that is, the CM continues in the parabolic trajectory!!!! THIS IS VERY VERY IMPORTANT! System of Particles Center of Mass

• A projectile is fired into the air and suddenly explodes • With no explosion, the projectile would follow the dotted line • After the explosion, the center of mass of the fragments still follows the dotted line, the same If no external forces parabolic path the act on the system, projectile would have then the velocity of followed with no the CM doesn’t explosion! change!! 7.5 Center of Mass

If no external forces act on the system, then the velocity of the CM doesn’t change!!

m1Dx1  m2Dx2 m1v1  m2v2 Dxcm  vcm  m1  m2 m1  m2 7.5 Center of Mass

BEFORE

m1v1  m2v2 vcm   0 m1  m2

AFTER

88 kg1.5m s 54 kg 2.5m s v   0.002  0 cm 88 kg  54 kg Using Both Conservation of Energy and Momentum A bullet of mass 0.01 kg is shot into a wooden block of mass 1.00 kg. They rise to a final height of 0.650 m as shown. What was the initial speed of the bullet before it hit the wood block? 7.3 Collisions in One Dimension

Example 8 A Ballistic Pendulim

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. 7.3 Collisions in One Dimension

Apply conservation of momentum to the collision:

m1v f 1  m2v f 2  m1vo1  m2vo2

m1  m2 v f  m1vo1

m1  m2 v f vo1  m1 7.3 Collisions in One Dimension

Applying conservation of energy to the swinging motion:

1 2 mgh  2 mv

1 2 m1  m2 ghf  2 m1  m2 v f

1 2 ghf  2 v f

2 v f  2ghf  29.80m s 0.650 m 7.3 Collisions in One Dimension

2 v f  29.80m s 0.650 m

m1  m2 v f vo1  m1

 0.0100 kg  2.50 kg  2 vo1    29.80m s 0.650 m  896m s  0.0100 kg  Two cars collide on a horizontal frictionless surface. Before the collision, car A is moving due west at 5 m/s and car B is traveling due north at 10 m/s. Car B weighs twice as much as Car A. After the collision, the two bodies are stuck together and move off. What is the speed of the composite body (A + B) after the collision? Ignore Friction. Show a vector diagram. Give the magnitude and direction of the final velocity