1D Collisions

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					1D Collisions

Unit A Momentum
                 Objectives
• You will be able to state and apply the Law of
  Conservation of Momentum to linear
  collisions.
• You will be able to define isolated system, and
  be able to determine if a system is or is not
  isolated.
         Collisions and systems
• A collision is an interaction between two
  objects where a force acts on each object for a
  period of time.
• A group of two or more objects that interact is
  known as a system.
• A system where the system’s mass is constant,
  and no external net force acts on the system is
  an isolated system.
      Momentum of a system…
• … is defined as the sum of the momenta of all
  objects in the system.



       psys  p A  pB  pC  . . .
The belief in the fact that momentum is conserved in all collisions
has led to the Law of Conservation of Momentum.


                                      
      total p before collision  total p after collision
                                     
1)    mAv A  mBvB  mAv A        mBvB  

     Use when objects do not stick together.
               
2)    mAv A  mBvB  mA  mB  v

     Use when objects stick together.
3)   mAv A  mBvB
     Use for explosions (momentum total is zero before).
Note: These formulas are not on your formula sheets and you
      can use the first formula exclusively if you like.
                                                                      *
                   Example:
• Fred sits in a stationary, frictionless wheelchair
  and throws a 3.0 kg ball outward at 8.0 m/s. If
  the mass of Fred and his chair is 75 kg, what
  will be the velocity of Fred and the chair
  immediately after the ball is thrown?
                        Solution
• Before:                           • After

   mB  3.0kg                         mB  3.0kg
  vB  zero                           vB f  8.0m / s
   mF  75kg                          mF  75kg
  vF  zero                           vFf  ??
   psys  zero
Because both velocities are zero.
                   Solution
psysi  psys f
                               mB vB f
psys f  0            vF f 
                               mF
0  pF f  pB f              3.0kg  8.0m / s
                      vF f 
 pF f  pB f                    75m / s
mF vFf  mBvB f            0.32m / s [ R ]
                   Example
• A 50 g bullet is fired into a 650 g block of
  wood, which sits on a frictionless surface.
  After impact the bullet and block move right
  at 30 m/s. Find the bullet’s velocity just before
  impact.
               Solution
• Before:           • After
mB  0.050kg     mBW  mB  mW  0.700kg
vBi  ?          vBW f  30m / s
mW  0.650kg     Because wood and
vWi  zero       bullet move as a single
                 object.
 pWi  zero
                    Solution
                               mBW vBW f
psysi  psys f         vBi 
                                  mB
pBi  pWi  pBW f
                             0.700kg  30m / s
mvBi  0  mBW vBW f   vBi 
                                 0.050kg
                            420m / s [ R ]
                           0.42km / s [ R ]
Using “Conservation of Momentum”
 Colliding Objects Stick Together

    Example : A 1.50 g pellet is fired into a 12.3 g wood block. The
               block and imbedded pellet fly off at 2.78 m/s E. What
               was the pellet’s velocity before impact?
                                 
                        mAv A  mBvB  mA  mB  v

                    0.00150 kg  v A 
                                  
                                                                        
                                          0  0.00150 kg  0.0123 kg  2.78 m
                                                                             s
                                                                                 
                    0.00150 kg  v A 
                                  
                                          0  0.038364 kg  m
                                                            s
                                          
                                          v p  25.6 m E
                                                     s

                                                                Answer is
                                                                “+”

                                                                                     *
Example 2: A 980 kg Toyota going at 52.8 km/h N collides with and
            becomes entangled with a 738 kg Honda going
            79.3 km/h S. What is the velocity of the wreckage
            immediately after impact?

Note: It does not matter what units of mass or velocity are used -
     just be consistent!
                         
                mAv A  mBvB  mA  mB  v

980 kg     52.8 km 738 kg  79.3 km  980 kg  738 kg v'
                     h                     h
                       51744   58523    1718 v'
                                                       
                                           
                                           v '  3.95 km S
                                                       h



                                           Answer was “–”

                                                                        *
Examples of “Explosion” Conservation of Momentum Problems

Example : When a 960 g plate is dropped, the first 370 g piece
           flies off at 2.63 m/s S. What is the velocity of the
           second piece?
                                     
                            mv  mv
                                             
               370 g  2.63 m   590 g  v
                              s
                                
                               v  1.65 m N
                                         s

Note: In order for the total final momentum to be “0”, if one
      piece flies north, the other must fly south.




                                                                  *
Example 2: What is the ratio of the velocities of the 3.46 g piece
           of an exploded firecracker to the other 5.96 g piece?
                             
                        mv  mv
                                        
                3.46 g  v1  5.96 g  v2
                          
                          v1 5.96
                           
                          v2 3.46
                           
                           v1  1.72 x

Note: The smaller piece has a velocity 1.72 times the velocity of
      the larger piece!




                                                                     *

				
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