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Bellwork 01/21/09

Bellwork 01/21/09. A proton of mass m moving with a speed of 3.0 x 10 6 m/s undergoes a head-on elastic collision with an alpha particle of mass 4m which is initially at rest. What are the velocities of the two particles after the collision? 

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Bellwork 01/21/09

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  1. Bellwork 01/21/09 • A proton of mass m moving with a speed of 3.0 x 106 m/s undergoes a head-on elastic collision with an alpha particle of mass 4m which is initially at rest. What are the velocities of the two particles after the collision?  • **Hint: Assign mass of a proton to be 1 amu (just like in chemistry)

  2. Center of Mass The center of mass is the point where all of the mass of the object is concentrated.

  3. Definition of the center of mass: • The center of mass is the point at which all of the mass of an object or system may be considered to be concentrated, for the purposes of linear or translational motion only. • We can then use Newton’s second law for the motion of the center of mass:

  4. Center of Mass • The momentum of the center of mass does not change if there are no external forces on the system. • The location of the center of mass can be found: • Where x and y are the coordinates of the particle and M = ∑m is the total mass of the system. • This calculation is straightforward for a system of point particles, but for an extended object calculus is necessary.

  5. Example 1: Find the location of the center of mass of the following three-mass system. Mass m1 = 1.0 kg m2 = 2.0 kg m3 = 3.0 kg Location (0,0 (1.0 m, 1.0 m) (2.0 m, -2.0m)

  6. Solution to Example 1 • The location of the center of mass is at • (XCM, YCM) = (1.3m, -0.67m)

  7. Example 2:   For the collection of masses shown below, where is the center of mass? A) B) C)

  8. Solution to Example 2 • xcm = [M(0) + 3M(L) + M(2L)] / 5M      = [0 + 3ML + 2ML] / 5M      = 5 ML / 5 ML      = L • xcm = [M(0) + 3M(L) + M(2L)] / 5M      = [0 + 3ML + 2ML] / 5M      = 5 ML / 5 ML      = L •   xcm = 1.3 L ycm = 1.4 L

  9. The center of mass of a flat object can be found by suspension.

  10. The center of mass may be located outside a solid object. • the cantilevered balconys of Frank Lloyd Wright's Fallingwater, • the leaning tower of Pisa , • a truck parked on a hillside, • a race car moving through a banked curve, • an acrobatic troop's pyramid act in the circus, • a professional unicyclist sensing how to lean as he accommodates the various dynamically-changing weight vectors in the system to keep his center of mass above the wheel's base of support, or simply • a person bending over to pick up an object from the floor.

  11. Independent Practice • p. 214 – 215; 104, 105, 108, 112, 114, 116, 119 • Online Alternate: do practice problems 1-10 instead of 104, 105, 108, 112, and 114 in the book. Complete 116 and 119 on paper. • Finish Conservation of Momentum Practice Problems • Study for Test on Thursday

  12. Bellwork 01/23/09 • Find the center of mass of a system composed of three spherical objects with masses of 3.0 kg, 2.0 kg, and 4.0 kg and centers located at (-6.0, 0), (1.0 m, 0), and (3.0 m, 0), respectively. • A 3.0-kg rod of length 5.0 m has at opposite ends point masses of 4.0 kg and 6.0 kg. (a) Will the center of mass of this system be (1) nearer to the 4.0-kg mass, (2) nearer to the 6.0-kg mass, or (3) at the center of the rod? Why? (b) Where is the center of mass of the system?   (-0.44m, 0m)

  13. (a) The center of mass of the system will be be 2) nearer to the 6.0 kg mass. The center of mass will “weighted” to the more massive side. • (b) Choose the 4.0-kg mass as the origin (x = 0). The center of mass of the rod (alone) is then at 2.5 m.

  14. Bellwork 01/26/09 • A 20.0-g bullet traveling at 300 m/s passes completely through a wooden block, initially at rest on a smooth table. The block has a mass of 1000 g. The bullet emerges traveling in the same direction, but at 50 m/s. • What is the speed of the block afterward? • What fraction (or percentage) of the total initial kinetic energy is lost in this process?  

  15. Solution to Bellwork (#116) (a) po = p’ m1 v1o + m2 v2o = m1 v1 + m2 v2 (0.0200 kg)(300 m/s) + (1.000 kg)(0) = (0.0200 kg)(50.0 m/s) + (1.000 kg) v2. So v2 = 5.00 m/s Ko= ½(0.0200kg)(300m/s)2+1/2 (1.000 kg)(0)2= 900 J. K’ = ½ (0.0200kg)(50.0m/s)2+1/2(1.000kg)(5.00m/s)2 = 37.5 J. So the fraction of the total initial kinetic energy lost is (900J-37.5J)/900J = 95.8%

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