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Lab 7: 2-D inelastic collisions Only 5 more to go!!

Lab 7: 2-D inelastic collisions Only 5 more to go!!. Last week we learned that momentum is conserved in collisions. This led us to the law of conservation of momentum:.

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Lab 7: 2-D inelastic collisions Only 5 more to go!!

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  1. Lab 7: 2-D inelastic collisions Only 5 more to go!! Last week we learned that momentum is conserved in collisions. This led us to the law of conservation of momentum: Also remember last week that the collision happened all in one dimension so we didn’t have to worry much about the fact that momentum is a vector quantity. But this week the collision will occur in two dimensions so you will need to write the conservation of momentum in the x and y directions:

  2. y How would we write the momentum of this object? V What is pX and pY ?  x mass, m Consider this collision. What is the speed of mass 1 and mass 2 after they hit? v1f m1 v1i  m1 m2  V2i= 0 m2 v2f

  3. To solve this we first write the conservation of momentum in the x and y directions: Since V2i = 0 the equations for conservation of momentum become: Now substitute in for the velocity components:

  4. From the equation in the y-direction we can get: Now we can solve for V2f !!!! Now substitute this equation for V2f into the equation in the x-direction: Solve for V1f and you get this big equation:

  5. Now that you have V1f you can substitute back in and find V2f

  6. Let’s look at another example: What type of collision is this? BEFORE COLLISION Cart #1 Cart #2 +x mass = m1 velocity = v1i mass = m2 velocity = v2i=0 AFTER COLLISION +x Cart #2 Cart #1 mass = m1+m2 velocity = v(1+2)f What’s the velocity of the combination after the collision? We can use cons. Of momentum: m1v1 + m2v2 = (m1 + m2) v(1+2)f ; but v2 = 0 so this becomes: m1v1 = (m1 + m2) v(1+2)f

  7. What if I wanted to know how much Kinetic Energy was lost? All you need to do is calculate the initial KE of the system then calculate the final KE of the system and take the difference. Initial KE of system: KEi = ½ m1v12 + ½ m2v22 = ½ m1v12 + 0 = ½ m1v12 Final KE of system: KEf = ½ (m1 + m2) (v(1+2)f)2 KE lost: KEf- KE1 = ½ (m1 + m2) (v(1+2)f)2 - ½ m1v12 What are we going to do today? We are going to look at a 2-d collision and see if momentum is conserved and whether of not KE is conserved

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