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MOMENTUM & COLLISIONS

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**INTRODUCTION MOMENTUM: p = mv Inertia in motion**

Linear momentum of an object equals the product of its mass and velocity Moving objects have momentum Vector quantity The momentum vector points in the same direction as the velocity vector Proportional to mass and velocity p = mv p = momentum (kg * m/s) m = mass (kg) v = velocity (m/s)

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**collisions Collisions:**

Momentum- Useful concept when applied to collisions In a collision, two or more objects exert forces on each other for a brief instant of time, and these forces are significantly greater than any other forces they may experience during the collision

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**TAXI PROBLEM What is the taxi cab’s momentum?**

* Mass of the taxi = 0.14 kg * Velocity of the taxi = 1.2 m/s Answer: p = mv p = (0.14 kg)(1.2 m/s) p = 0.17 kg * m/s to the left p = 0.17 kg * m/s v = 1.2 m/s

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**Momentum & Newton’s 2nd Law**

Newton’s 2nd Law & the definition of acceleration ΣF = ma = m(Δv/Δt) Momentum p = mv ** Here, you see a car accelerating − or you could say you see a car whose momentum is changing over time. Either statement leads to the conclusion that a net force is acting on the car. In fact, what we call Newton’s second law was expressed by him in terms of a change in momentum, not in terms of acceleration. ΣF = Δp/Δt

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**Momentum-relationship to Newton’s 2nd Law**

Definition: Newton’s 2nd Law: F = ma or, since M=V so, t F= p x v so, v t If you substitute m = p v The result: The change in momentum over time equals the force!

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**Impulse: Change in momentum**

ΣF = Δp/Δt **Net force equals the change in momentum per unit time Rearranging this equation Δp = ΣFΔt Impulse (J) The change in momentum is called the impulse of the force (Impulse- momentum theorem) Vector quantity Units: kg * m/s J = Δp = FavgΔt p = Momentum J = Impulse Favg = Average force Δt = Elapsed time The greater the net force, or the longer the interval of time it is applied, the more the object’s momentum changes the same as saying the impulse increases

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**Changing Momentum: Scenario 1 **

if you want to decrease a large momentum, you can have the force applied for a longer time. If the change in momentum occurs over a long time, the force of impact is small. Examples: Air bags in cars. Crash test video FDt

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**FDt Changing Momentum: Scenario 2**

If the change in momentum occurs over a short time, the force of impact is large. Karate link Boxing video FDt

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Impulsive force Baseball player swings a bat and hits the ball, the duration of the collision can be as short as 1/1000th of a second and the force averages in the thousands of newtons The brief but large force the bat exerts on the ball = Impulsive force

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Impulse problem A long jumper's speed just before landing is 7.8 m/s. What is the impulse of her landing? (mass = 68 kg) J = pf – pi J = mvf – mvi J = 0 – (68kg)97.8m/s) J = -530 kg * m/s *Negative sign indicates that the direction of the impulse is opposite to her direction of motion

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Kinetic Books View Kinetic books section 8.4- Physics at play: Hitting a baseball BASEBALL EXAMPLE The ball arrives at 40 m/s and leaves at 49 m/s in the opposite direction. The contact time is 5.0×10−4 s. What is the average force on the ball? J = Favg Δt J = Δp = mΔv FavgΔt = mΔv Favg = mΔv/Δt Favg = (0.14kg)(49 – (-40)m/s)/5.0×10−4 s Favg = 2.5×104 N

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**Review of equations Momentum**

p = mv Newton’s 2nd Law & the definition of acceleration ΣF = ma = m(Δv/Δt) Different way to express Newton’s 2nd Law Net force is equal to the change in momentum over time ΣF = Δp/Δt

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**Review of equations Impulse = Change in momentum Change in momentum**

J = Δp = FavgΔt Change in momentum Δp = mΔv ** In conclusion, there are 2 different equations for impulse J = Favg Δt J = Δp = mΔv Favg Δt = mΔv Favg = mΔv/Δt

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**Conservation of momentum**

The total momentum of an isolated system is constant No net external force acting on the system Momentum before = Momentum after Playing pool example: Kinetic books 8.6

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**Conservation of momentum**

p = mv Conservation of momentum Momentum before = Momentum after pi1 + pi2 +…+ pin = pf1 + pf2 +…+ pfn pi1, pi2, …, pin = initial momenta pf1, pf2, …, pfn = final momenta m1vi1 + m2vi2 = m1vf1 + m2vf2 m1, m2 = masses of objects vi1, vi2 = initial velocities vf1, vf2 = final velocities The total momentum of a system with no net external forces acting on it stays the same. The equation you see here states this fact mathematically. The vector sum of the momenta of all the objects at an initial time equals the total momentum at any later (final) time.

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**Conservation of momentum**

The law of conservation of momentum can be derived from Newton’s 2nd and 3rd laws Newton’s 2nd law F = ma Newton’s 3rd law Forces are equal but opposite * Refer to Kinetic Books- 8.7 For step-by-step derivation

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**Conservation of momentum**

A 55.0 kg astronaut is stationary in the spaceship’s reference frame. She wants to move at m/s to the left. She is holding a 4.00 kg bag of dehydrated astronaut chow. At what velocity must she throw the bag to achieve her desired velocity? (Assume the positive direction is to the right.)

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**solution VARIABLES: Mass of astronaut ma = 55 kg Mass of bag mb = 4 kg**

Initial velocity of astronaut via = 0 m/s Initial velocity of bag vib =0 m/s Final velocity of astronaut vfa = -0.5 m/s Final velocity of bag vfb = ? EQUATION: m1vi1 + m2vi2 = m1vf1 + m2vf2 mavia + mbvib = mavfa + mbvfb 0 = mavfa + mbvfb Vfb = - (mavfa / mb) Vfb = - ((55kg)(-0.5m/s))/(4kg) = m/s

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**collisions Elastic collision Inelastic collision**

Kinetic energy is conserved KE before = KE after KE = 1/2mv2 Inelastic collision Kinetic energy is NOT conserved KE before ≠ KE after Momentum is conserved in any collision Elastic or inelastic Elastic collision- Game of pool provides a good example of nearly elastic collisions. The collisions between balls are almost completely elastic and little kinetic energy is lost when they collide. Inelastic collision- For example, a car crash often results in dents. This means some kinetic energy compresses the car permanently; other KE becomes thermal energy, sound energy and so on. This means that an inelastic collision reduces the total amount of KE.

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**Collisions Momentum is always conserved in a collision Collision video**

Classification of collisions: Newton’s cradle: Demonstration ELASTIC Both energy & momentum are conserved INELASTIC Momentum conserved, not energy Perfectly inelastic -> objects stick Lost energy goes to heat

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** Examples of Perfectly Inelastic Collisions**

Catching a baseball: Video Football tackle Cars colliding and sticking Bat eating an insect Examples of Perfectly Elastic Collisions Superball bouncing Electron scattering Bouncing ball video

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**Conservation of Momentum**

Mv(initial) = Mv (final) An astronaut of mass 80 kg pushes away from a space station by throwing a 0.75-kg wrench which moves with a velocity of 24 m/s relative to the original frame of the astronaut. What is the astronaut’s recoil speed? 0.225 m/s

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**Perfectly Inelastic collision in 1-dimension**

Final velocities are the same

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Example 6.6 A 5879-lb (2665 kg) Cadillac Escalade going 35 mph =smashes into a 2342-lb (1061 kg) Honda Civic also moving at 35 mph=15.64 m/s in the opposite direction.The cars collide and stick. a) What is the final velocity of the two vehicles? b) What are the equivalent “brick-wall” speeds for each vehicle? a) 6.73 m/s = 15.1 mph b) 19.9 mph for Cadillac, 50.1 mph for Civic

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**Elastic collision in 1-dimension**

Conservation of Energy: Conservation of Momentum: Rearrange both equations and divide:

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Example 6.7 A proton (mp=1.67x10-27 kg) elastically collides with a target proton which then moves straight forward. If the initial velocity of the projectile proton is 3.0x106 m/s, and the target proton bounces forward, what are a) the final velocity of the projectile proton? b) the final velocity of the target proton? 0.0 3.0x106 m/s

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**Elastic collision in 1-dimension**

Final equations for head-on elastic collision: Relative velocity changes sign Equivalent to Conservation of Energy

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Example 6.8 An proton (mp=1.67x10-27 kg) elastically collides with a target deuteron (mD=2mp) which then moves straight forward. If the initial velocity of the projectile proton is 3.0x106 m/s, and the target deuteron bounces forward, what are a) the final velocity of the projectile proton? b) the final velocity of the target deuteron? vp =-1.0x106 m/s vd = 2.0x106 m/s Head-on collisions with heavier objects always lead to reflections

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Center of Mass Video Balancing Activity: video demo

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