2. Conservation of Momentum

3. Conservation of momentum: Split into components: If the collision is elastic, we can also use conservation of energy.

4. Two Methods Method 1 - Graphical Method – Ruler and protractor. 1.Calculate the momentum of each object. 2.Choose an appropriate scale.  Draw each for the momenta before the interaction  Arrange them head to tail  Draw the resultant, the total momentum before 3.Draw each for the momenta after the interaction  Arrange them head to tail  Draw the resultant, the total momentum after 4. Compare the magnitude and direction of the total momentum vectors before and after. If they are identical momentum is conserved.

5. Method 2: Trigonometric Method 1.Calculate the momentum of each from p = mv. 2.Use trig calculate the X and Y components of each object before. 3.Find the x and y components after the collision.  Total the X components before the collision  Total the X components after the collision  Set them equal to each other  Total the Y components before the collision  Total the Y components after the collision  Set them equal to each other in magnitude and direction 4.Solve the individual component equations for the unknowns. 5.Make a chart and fill out as in the following example

6. p xi p yi p xf p yf Mass 1 Mass 2 Total 1.Fill out the chart 2. Solve the component equations

7. Two types of 2-D collisions - One object hits another stationary Type 1 – Glancing collision - One object hits another stationary object they fly off in separate directions.

8. p xi p yi p xf p yf Mass 1 Mass 2 Total Ex: The masses of the green ball and red ball are 2 kg and 3 kg. If the initial velocity of the green ball is 10m/s. Construct the component equations, don’t solve them.

9. Before the collision After the collision

10. Type 2 – Right angle collisions

12. 1.Objects approach at 90 0, one with momentum in the x-direction the other in the y-direction only. 2.They fly off together at an angle , with some final velocity Type 2 – Right angle collisions

13. A 95 kg Physics student running south the hall with a speed of 3m/s tackles a 90 kg Chemistry student moving west with a speed of 5 m/s. If the collision is perfectly inelastic, calculate: 1. The velocity of the students just after the collision 2. The energy lost as a result of the collision. Right angle collisions

14. m 2 =90kg V 2i =5m/s m 1 =95kg V 1i =3m/s Use Pythagorean Theorem and trig N S W E 

15. An eastern kingbird and a bee approach each other at right angles. The bird has a mass of.13 kg and a speed of.6 m/s, and the bee has a mass of 5x10 -3 kg and a speed of 15 m/s. If the bird catches the bee, what’s the new speed of the bird? Right angle collisions

16. Accident investigation. Two cars of equal mass approach an intersection. One is traveling east at 29 mi/h (13.0 m/s) the other north with unknown speed. The vehicles collide and stick together, leaving skid marks at an angle of 55º north of east. The second driver claims he was driving below the speed limit of 35 mi/h (15.6 m/s). 13.0 m/s ?? m/s 1.Is he telling the truth? 2.What is the speed of the“combined vehicles” after the collision? 3. How long are the skid marks (u k = 0.5) Right angle collisions

17. A cue ball traveling at 4m/s makes a glancing, elastic collision with a target ball of equal mass initially at rest. It is found that the cue ball is deflected such that it makes an angle of 30  with respect to its original direction of travel. Set up the conservation of momentum equations in the x and y directions, don’t solve them. Glancing Collision

18. A 2 kg ball is initially at rest on a horizontal frictionless surface. A 4 kg ball moving horizontally with a velocity of 10 m/s hits it. After the collision the 4 kg ball has a velocity of 6 m/s at an angle of 37  to the positive x-axis. 1.Determine the speed and direction of the 2 kg ball after the collision. 2.How much K.E. was lost in the collision? Glancing Collision

20. Two skaters collide and embrace, in a completely inelastic collision. Thus, they stick together after impact, where the origin is placed at the point of collision. Alfred, whose mass m A is 83 kg, is originally moving east with speed v A = 6.2 km/h. Barbara, whose mass m B is 55 kg, is originally moving north with speed v B = 7.8 km/h.