hw8a' - HW#8a Note: numbers used in solution steps can be...

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HW#8a Page 1 of 7 Note: numbers used in solution steps can be different from the question part. You can practice the methods in the solution and verify with the numbers and answers given in the question part. Or you should practice the methods in the solution and verify your calculation with numbers in your webassign. Problem 1, On a horizontal frictionless table, masses A and B (2 kg, 3 kg) slide to the right and left, respectively. They have speeds of 3 m/s and 1 m/s, respectively. The two masses collide, and bounce off each other. After the collision, they travel in opposite directions at speeds of 1.5 m/s and 2 m/s, respectively. (Note:In this problem, please use + dir = right, - dir = left, to indicate direction.) a.) Calculate A's momentum: before collision: kg·m/s , after collision: kg·m/s Explain why it is NOT conserved. Solution: Speeds were given, assign directions +/- appropriately: v Aix = + 3 m/s v Bix = -1 m/s v Afx = -1.5 m/s v Bfx = +2 m/s p A is not conserved. It goes from +6 kg m/s before the collision to -3 kg m/s after the collision. The reason it is not conserved is that p A is only conserved when the net force on object A equals zero. But A does feel net force during collision (from B). b.) Calculate B's momentum: before collision: kg·m/s after collision: kg·m/s Explain why it is NOT conserved. p B is not conserved. Similar to part A, p B is not conserved since net force on object B is not zero. B feels a force during collision (from A). c.) Calculate the A+B system's momentum: before collision: kg·m/s after collision: kg·m/s Explain why it IS conserved. p of A+B system is conserved, since F net sys = 0. The total net force on the system is zero. The force between A and B are internal forces which do not change the total moments of the system. Use p conservation for A+B, since no net external forces on the system: ± N and mg cancel ± Frictionless ± Collision force are internal to system
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HW#8a Page 2 of 7 Calculate the kinetic energy of the A+B system: before collision: J after collision: J What kind of collision is this? somewhat inelastic KE decreased. Mechanical energy is lost to heat/deformation. This is a somewhat inelastic collision. Mechanical energy is lost to heat/deformation (but don’t stick together, which would be perfectly inelastic). Somewhat inelastic collision does not stick together. Don’t assume elastic and conservation of KE, when object do not stick together. Only if there is no any lose to deformation, heat, sound, etc. you have KE unchanged and that’s called elastic collision. That’s an ideal case for extremely hard materials. Most collisions in reality are in between----- somewhat inelastic collision.
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This note was uploaded on 08/25/2010 for the course PHYS 111 taught by Professor Jamesm.lockhart during the Spring '09 term at S.F. State.

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hw8a' - HW#8a Note: numbers used in solution steps can be...

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