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2. Physics 140 – Fall 2007 lecture #13 : 18 Oct A 50 minute class period lasts roughly one micro-century. – Enrico Fermi Ch 8 topics: elastic and inelastic encounters collisions in two dimensions Announcements: check scores on SAMS send score corrections within two weeks of occurrence Source: National Archives

3. Homework set #6 closes in <15 hours. Half of you need to get going… Set #7 opens today, closes next TUESDAY.

4. The baby cannon on the lecture bench is aimed to shoot at some angle  above the horizontal. Since it is on wheels, it will recoil freely when it fires the tennis ball. Viewed from your seat, what angle above the horizontal does the tennis ball’s trajectory make as it emerges from the cannon’s muzzle? 1. An angle larger than  2. An angle smaller than  3. The same angle 

5. The kinetic energy of an object of mass m and velocity v, K = mv 2 / 2, can be rewritten in terms of momentum, p = mv, becoming K = p 2 / 2m. At fixed momentum, lighter objects have more kinetic energy than massive ones. Kinetic energy and momentum During nuclear fission of Uranium, neutrons (blue dots) carry away more kinetic energy than the `daughter’ products (Barium and Krypton). Source: wikimedia commons

6. Unlike the total momentum, the total kinetic energy of a colliding pair of bodies (1 and 2) may change during the encounter. The total kinetic energy, K=K 1 +K 2, cannot increase (unless some other force, such as a spring or an explosion, assists the encounter), but K can either decrease or remain constant. These outcomes lead to different categories of collisions - elastic (hard bounce): K 1 +K 2 remains constant inelastic (soft bounce): K 1 +K 2 decreases completely inelastic: K 1 +K 2 decreases and v 1f = v 2f (1+2 are stuck together) For inelastic encounters, the “lost” kinetic energy is converted into internal energy (heat/stress/spin) of the bodies involved. Kinetic energy in collisions

7. 1) The collision is completely inelastic. 2) The collision is elastic and M 1 < M 2. 3) The collision is elastic and M 1 > M 2. 4) Either 1 or 2. 5) Either 1 or 3. 6) M 2 will always rise above the initial height of M 1. A ball of mass M 1 is released at rest from an initial angle  from the vertical and collides a suspended ball of mass M 2 hanging at rest. Under what circumstances will the maximum height that M 2 rises lie below the initial height of M 1 ? M1M1 M2M2

8. http://hyperphysics.phy-astr.gsu.edu/HBASE/doubal.html Analysis of Double Ball Drop

9. http://www.spacedaily.com/reports/Jupiter_Encounter_Begins_For_New_Horizons_Spacecraft_On_Route_To_Pluto_999.html http://www.nasa.gov/mission_pages/newhorizons/main/index.html New horizon spacecraft article removed removed Source: NASA

10. A collision of two objects (1 and 2) occurs in a plane defined by the directions of two initial momentums p 1i and p 2i. When no external forces act on the pair, then each component of the total momentum is conserved (m 1 v 1x + m 2 v 2x ) f = (m 1 v 1x + m 2 v 2x ) i (m 1 v 1y + m 2 v 2y ) f = (m 1 v 1y + m 2 v 2y ) i Since there are four unknowns and only two equations, more constraints are needed. As in the 1D case: elastic encounters conserve total kinetic energy (1 added eqn) (K 1 + K 2 ) f = (K 1 + K 2 ) i completely inelastic encounters stick together (2 added eqn’s) v 1f = v 2f = (p 1i +p 2i )/(m 1 +m 2 ) [= v com ] Colliding bodies in 2 dimensions

11. 1) ball 2 is moving faster than ball 1. 2) ball 2 is moving slower than ball 1. 3) the balls have equal speeds. 4) more information is needed. A cue ball (yellow) collides with an equal mass nine (blue) ball at rest on the table. After the collision, the nine ball moves at a 35 degree angle with respect to the incoming cue ball’s direction. What can you say about the final speeds of the two balls?

12. You are in a boat resting on a lake on a perfectly calm day. You fall asleep and lose your oars overboard. Spying two large but unequal size rocks lying in your boat (for ballast? for hitting carp?), you get the brilliant idea of throwing the rocks from the boat away from shore so that your boat will recoil and drift into port. If you throw the rocks with some fixed velocity v rel (relative to you), what scheme will end up giving you and your boat the fastest trip back to shore? 1. throw both rocks simultaneously. 2. throw the lighter, then the heavier. 3. throw the heavier, then the lighter. 4. it doesn’t matter how you throw them. CC: BY-NC-SA Dave Patten (flickr) http://creativecommons.org/licenses/by-nc-sa/2.0/deed.en