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Physics 221 Chapter 9

Physics 221 Chapter 9. On a Collision Course . Linear Momentum = mass x velocity p = mv Conservation of Momentum : In a collision, momentum BEFORE the collision equals the momentum AFTER the collision. Problem 1 . . . The Odd Couple I.

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Physics 221 Chapter 9

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  1. Physics 221Chapter 9

  2. On a Collision Course ... • Linear Momentum = mass x velocity p = mv • Conservation of Momentum: In a collision, momentum BEFORE the collision equals the momentum AFTER the collision

  3. Problem 1 . . . The Odd Couple I • A model railroad car (mass = 100 g) collides and locks onto a similar stationary car. The coupled cars move as a unit on a straight and frictionless track. The speed of the moving cars is most nearly • A. 1 m/s • B. 2 m/s • C. 3 m/s • D. 4 m/s

  4. Solution 1 . . . The Odd Couple I • Momentum BEFORE = Momentum AFTER • 100 x 4 +0 = 200 x v • v = 2 m/s

  5. Problem 2 . . . The Odd Couple II • Is the K.E. BEFORE = K.E. AFTER? In other words, is the K.E. conserved in this collision?

  6. Solution 2 . . . The Odd Couple II • K.E. BEFORE the collision = (1/2)(0.1)(16) +0 = 0.8J • K.E. AFTER the collision = (1/2)(0.2)(4) +0 = 0.4J • Half of the K.E. has mysteriously disappeared! • BUT this does not mean that ENERGY is not conserved! K.E. was transformed (converted) into other forms of energy.

  7. Problem 3 . . . Big Bang! • A shell explodes into two unequal fragments. One fragment has twice the mass of the other. The smaller fragment moves in the NE direction with a speed of 6 m/s. The velocity (speed and direction) of the other fragment is most nearly • A. 3 m/s SW • B. 3 m/s NE • C. 4 m/s NE • D. 4 m/s SW

  8. Solution 3 . . . Big Bang! • M x 6 = 2 M x V • V = 3 m/s in the SW direction.

  9. Close collisions of the second kind ... • There are TWO types of collisions: • I. Momentum conserved and K.E. also conserved. This type is called an ELASTIC collision. Elastic collisions are “non-sticky” as in billiard balls or steel ball-bearings. • II. Momentum conserved BUT K.E. NOT conserved. This type is called an INELASTIC collision. Inelastic collisions are “sticky” as in coupled railroad cars and putty.

  10. Problem 4 . . . Elastic I • Two hockey pucks collide elastically on ice. P2 is at rest and P1 strikes it “head-on” with a speed of 3 m/s. • A. P1 stops and and P2 moves forward at 3 m/s • B. P1 and P2 move forward at 1.5 m/s • C. P1 and P2 move in opposite directions at 1.5 m/s • D. P1 moves forward at 1 m/s and P2 moves forward at 2 m/s

  11. Solution 4 . . . Elastic I • A. P1 stops and P2 moves forward at 3 m/s. • Please verify that: • 1. Momentum is conserved • 2. K.E. is also conserved (elastic)

  12. Problem5 . . . Elastic II • Two hockey pucks collide elastically on ice. P2 has twice the mass of P1 and is at rest and P1 strikes it “head-on” with a speed of 3 m/s. Calculate their velocities after the collision.

  13. Solution 5 . . . Elastic II • P1 moves backward at 1 m/s and P2 moves forward at 2 m/s. • HINT: • 1. Take 5 sheets of scrap paper ( Trust me!!!) • 2. Write the equation for momentum conservation • 3. Write the equation for K.E. conservation • 4. Solve two equations for two unknowns.

  14. Tell me more about momentum! • O.K. good boys and girls. Here is everything you always wanted to know about momentum but were afraid to ask! • In the beginning there was F = ma • So F = m(vf - vi) / t • If F = 0 then : mvf = mvi • Another interesting observation: If F is NOT zero then momentum WILL change and change in momentum equals F x t. This is called IMPULSE

  15. Problem 6 . . . Tiger in the woods • A golf ball has a mass of 48 g. The force exerted by the club vs. time is a sharp spike that peaks at 180 N for 0.01 s and then drops back to zero as the ball leaves the club head at high speed. It is estimated that the average force is 90 N over a short time (t) of 0.04 s. The speed of the ball is • A. 35 m/s • B. 75 m/s • C. 95 m/s • D. 135 m/s

  16. Solution 6 . . . Impulse to ride the tiger • Impulse = F t = 90 x 0.04 = 3.6 Ns • Impulse = change in momentum • 3.6 = mv -0 • 3.6 = 0.048 x v • v = 75 m/s

  17. Problem 7 . . . Teeter- Totter • FB weighs 180 pounds and sits at the 20 cm. mark. Where should SP (120 pounds) sit in order to balance the teeter-totter at the playground? 0 20 50 100 180 120

  18. Solution 7 . . . See-Saw • 30 x 180 = what x 120 • what = 45 or 95 cm. mark 0 20 50 100 180 120

  19. 0 20 50 80 100 ? 100 300 Problem 8 . . . Center of Mass • The C.M. is a weighted average position of a distribution of masses where the system can be balanced. • Where is the C.M. ?

  20. x 0 20 50 80 100 100 300 Solution 8 . . . Center of Mass (100)( x -20) = (300)(80 - x) x = (100) (20) + (300)( 80) / (100 + 300) x = 65

  21. C.M.. 0 X1 50 X2 100 M 1 M 2 A formula for C.M. C.M. = M1 * X1 + M2 * X2 M1 + M2

  22. 0 20 50 80 100 ? 100 300 Problem 9 . . . more C. M. • Include the mass of the meter stick (60 g). Where is the new C.M. ?

  23. Solution 9 . . . more C. M. C.M. = M1 * X1 + M2 * X2 + M3 * X3 M1 + M2 + M3 C.M. = (100)(20) + (60)(50) +(300)(80) 100 + 60 + 300 C.M. = 63

  24. (R , 0) Problem 10 . . . The Hole • A circular hole of radius R/2 is drilled in a disc of mass M and radius R as shown in the figure. The center of mass is closest to • A. (R/6 , 0) • B. (R/4 , 0) • C. (R/3 , 0) • D. (R/2 , 0)

  25. Circular plate with hole • If the mass of the missing piece is 100 then the mass of the remaining piece is 300 for a total of 400. In other words, if M is the mass of the original disc (no hole) then the mass of the disc with the hole is 3M/4 and the mass of the circular piece removed is M/4. • NOTE: Non-Linear thinking! Disc with HALF the radius has only a quarter of the mass!

  26. 100 300 Solution 10A model for the “hole” problem • (100)(R/2) = (300)(x) • x = R/6 x

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