Momentum & Impulse. A moving object has momentum Inertia in motion = momentum! Linear momentum = p = (mass) X (velocity)

Ch. 9 Momentum Momentum & Impulse A moving object has momentum  Inertia in motion = momentum!  Linear momentum = p = (mass) X (velocity)  p = m...
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Ch. 9 Momentum

Momentum & Impulse A

moving object has momentum  Inertia in motion = momentum!  Linear momentum = p = (mass) X (velocity)  p = m·v  Where p and v are both vector quantities and are in same direction.  Units: kg·m/s or N·s

    



A massive truck & small car both traveling at same velocity… Which will be more difficult to stop? Which can do more damage upon impact? If they both travel toward each other… Which will experience greater impact force upon colliding? Which experiences more acceleration?

The force upon impact will be... The same for both the truck and the car.  Ftruck = Fcar (Newton’s 3rd Law)  mt·at = mc·ac (Newton’s 2nd Law) 

mt·a 

t

=

ac

mc·

Force is the same, but because mass is different, the accelerations experienced by each vehicle will be different.

 Force

is required to change the momentum of an object.  The rate of change of momentum of a body is proportional to the net force applied to it…  ΣF = Δp/Δt = m Δv /Δt = ma  Going back to our example… During impact, which vehicle had a greater change in momentum?  It’s the same!

What’s required to change the momentum of an object?

What’s required to change the momentum of anobject?

Impulse!





   

In order to change the momentum of an object, an impulse is required… A net force acting for some time will cause an object to change its momentum. ΣF = Δp/Δt ΣF Δt = Δp = mΔv Impulse=change in momentum We assume that the net force is constant throughout the duration of changing momentum. (Usually use average net force.)

Impulse

The force of the foot on the ball is an impulsive force.

Graphical Interpretation of Impulse J = Impulse = area under the force curve  Favg t

The Impulse-Momentum Theorem: Impulse causes a change in momentum:

Example Problem: Two 1-kg stationary cue balls are struck by cue sticks. The cues exert the forces shown. Which ball has the greater final speed?

A. B. C.

Ball 1 Ball 2 Both balls have the same final speed

Answer

Two 1-kg stationary cue balls are struck by cue sticks. The cues exert the forces shown. Which ball has the greater final speed?

A. Ball 1 B. Ball 2 C. Both balls have the same final speed

Example Problem

A 0.5 kg hockey puck slides to the right at 10 m/s. It is hit with a hockey stick that exerts the force shown. What is its approximate final speed?

Answer…  ΣF avg  ΣF avg

Δt = m Δv Δt = m (vf – vi)  (25N)(0.020s) = (0.5kg) (vf – 10m/s)  0.50 Ns = (0.5kg) (vf – 10m/s)  1 m/s = vf – 10m/s  vf = 1 m/s + 10 m/s

vf

= 11 m/s

1.

Impulse is A. a force that is applied at a random time. B. a force that is applied very suddenly. C. the area under the force curve in a force-versus-time graph. D. the interval of time that a force lasts.

Slide 9-5

Answer 1.

Impulse is A. a force that is applied at a random time. B. a force that is applied very suddenly. C. the area under the force curve in a force-versus-time graph. D. the interval of time that a force lasts.

Law of Conservation of Momentum 

The sum of the momentums before a collision equal the sum of the momentums after the collision in an isolated system.

Law of Conservation of Momentum: The total momentum of an isolated system of bodies remains constant.  (Isolated system: meaning that all forces acting on the bodies are included… and the sum of the external forces applied to the system is zero. External forces like Ff or Fg.)  Momentum before = Momentum after  m1v1 + m2v2 = m1v'1 + m2v'2 (Elastic Collision)  m1v1 + m2v2 = (m1 + m2)v‘ (Inelastic Collision)  v = velocity before collision  v' = velocity after collision 

Elastic Collisions!

Elastic Collision

Elastic Collisions: Two or more objects collide, bounce (don’t stick together), and kinetic energy is conserved.  An ideal situation that is often never quite reached… billiard ball collisions are often used as an example of elastic collisions. 

 Kinetic

energy is conserved:  KE1 + KE2 = KE'1 + KE'2  ½m1v12+½m2v22=½m1v'12+½m2v'22  Momentum is conserved:  m1v1 + m2v2 = m1v'1 + m2v'2

Inelastic Collision Inelastic Collision: two or more objects collide and do not bounce off each other, but stick together. Or, an explosion where one object starts w/one momentum and then separates into two or more objects w/separate final momentums.  Kinetic energy is not conserved.  KEbefore = KEafter + heat + sound + etc.  The kinetic energy “lost” is transformed into other types of energy, but…  Total energy is always conserved!  m1v1 + m2v2 = (m1 + m2)v' 

Elastic vs. Inelastic Collisions (a) A hard steel ball would rebound to its original height after striking a hard marble surface if the collision were elastic.  (b) A partially deflated basketball has little bounce on a soft asphalt surface.  (c) A deflated basketball has no bounce at all. 



Show “happy balls”



A ball of mass 0.250 kg and velocity +5.00 m/s collides head on with a second ball of mass 0.800 kg that is initially at rest. No external forces act on the balls. If the balls collide and bounce off one another, and the second ball moves with a velocity of +2.38 m/s, determine the velocity of the first ball after the collision, including direction.

Did you get…?  vf1

= -2.62 m/s (ball 1 rebounds)

Lets prove if energy is conserved….. Remember from a previous slide that for Elastic Collisions: Two or more objects collide, bounce (don’t stick together), and kinetic energy is conserved.  KE1 + KE2 = KE'1 + KE'2  ½m1v12+½m2v22=½m1v'12+½m2v'22  Apply this equation to the last problem and see if it is true. 

Energy is conserved!

2. The total momentum of a system is conserved A. always. B. if no external forces act on the system. C. if no internal forces act on the system. D. never; momentum is only approximately conserved.

Answer

2. The total momentum of a system is conserved A. always. B. if no external forces act on the system. C. if no internal forces act on the system. D. never; momentum is only approximately conserved.

3.In an inelastic collision, A. impulse is conserved. B. momentum is conserved. C. force is conserved. D. Kinetic energy is conserved. E. elasticity is conserved.

Slide 9-9

Answer

3.In an inelastic collision, A. impulse is conserved. B. momentum is conserved. C. force is conserved. D. Kinetic energy is conserved. E. elasticity is conserved.

Slide 9-10

Slide 9-19

Forces During a Collision

Slide 9-20

The Law of Conservation of Momentum

In terms of the initial and final total momenta:

In terms of components:

Example Problem

A curling stone, with a mass of 20.0 kg, slides across the ice at 1.50 m/s. It collides head on with a stationary 0.160-kg hockey puck. After the collision, the puck’s speed is 2.50 m/s. What is the stone’s final velocity?

Slide 9-23



Answer: 1.48 m/s

Rocket propulsion is an example of conservation of momentum:

The rocket doesn’t push on the environment. The rocket pushes the exhaust gas in one direction (backward), and the exhaust gas pushes the rocket in the opposite direction (forward). Newton’s third law, the force and time acting on the rocket and the gas (as a whole) are equal and opposite. The momentum is conserved. The momentum before is zero and the momentum after is a total of zero. Positive momentum of Slide 9-24 the rocket = Negative momentum of the gas.

Inelastic Collisions: For now, we’ll consider perfectly

inelastic collisions:

A perfectly inelastic collision results whenever the two objects move off at a common final velocity.

Slide 9-25

Example Problem

Jack stands at rest on a skateboard. The mass of Jack and the skateboard together is 75 kg. Ryan throws a 3.0 kg ball horizontally to the right at 4.0 m/s to Jack, who catches it. What is the final speed of Jack and the skateboard?

Answer:

0.154 m/s

Recall from a previous slide…

Inelastic Collision: two or more objects collide and do not bounce off each other, but stick together. Or, an explosion where one object starts w/one momentum and then separates into two or more objects w/separate final momentums.  Kinetic energy is not conserved.  KEbefore = KEafter + heat + sound + etc.  The kinetic energy “lost” is transformed into other types of energy, but…  Total energy is always conserved!  m1v1 + m2v2 = (m1 + m2)v'



Prove if the last problem conserved energy…. KE1 + KE2 = KE1+2  ½m1v12+½m2v22=½m1+2v'2  Apply this equation to the last problem and see if it is true. 

Energy is not conserved!



A 20.0 g ball of clay traveling east at 2.00 m/s collides with a 30.0 g ball of clay traveling 30.0o south of west at 1.00 m/s. The two pieces stick together and become one. What are the speed and direction of the final piece of clay?

Momentum is a vector… including direction. Hint: Draw your vectors tip to tail and draw the resultant momentum vector (p final). Resolve all vectors into x and y-components. Determine the sum of the x-components and the sum of the y-components and draw your final resultant vectors making a right triangle. Solve for p final and angle. Solve for v final.