MOMENTUM & COLLISIONS

1. MOMENTUM & COLLISIONS

2. INTRODUCTION • MOMENTUM: • 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)

3. 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

4. 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

5. 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

6. 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!

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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

14. Review of equations • Impulse = 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

15. Conservation of momentum • 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

16. Conservation of momentum • 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

17. 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

18. Conservation of momentum A 55.0 kg astronaut is stationary in the spaceship’s reference frame. She wants to move at 0.500 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.)

19. 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) = 6.875 m/s

20. collisions • Elastic 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

21. 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

22. 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

23. 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

24. Perfectly Inelastic collision in 1-dimension • Final velocities are the same

25. 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

26. Elastic collision in 1-dimension • Conservation of Energy: • Conservation of Momentum: • Rearrange both equations and divide:

27. 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

28. Elastic collision in 1-dimension • Final equations for head-on elastic collision: • Relative velocity changes sign • Equivalent to Conservation of Energy

29. 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

30. Center of Mass • Video • Balancing Activity: video demo