One Dimensional Inelastic Collisions

 

Red m  kg  v = m/s  v' = m/s
Blue m  kg  v = m/s  v' = m/s
Red  1/2mv2  J 1/2mv'2 =  J
Blue  1/2mv2  J 1/2mv'2 =  J
Initial X Position:  x  m  Initial X Position:  x  m 

 

To change the default values for m and v, press Reset before entering your values for mass and velocity. The velocities
can be either positive or negative. The default x-positions are -1.00 m for the red ball and +1.00 m for the blue ball.
To change the default x-positions, press Reset before entering the values. For instructional purposes, you can verify the
velocities or determine that Δp = 0 or ΔKE ≠ 0 for the collision.
 
Inelastic collisions are those collisions that involve a change in kinetic energy. Generally, the kinetic energy decreases
because some of the kinetic energy is converted to sound, light, heat, and/or a deformation of the colliding objects. It is
important to note that energy is not lost but rather transformed. This animation deals with the special case which is a
completely inelastic collision where the objects stick together after the collision. Examples of such collisions are
railroad cars and football players (i.e. a running back and a linebacker) colliding, sticking together, and moving with the
same velocity.
 
For the system described above, the kinetic energy is not conserved. Mathematically this is expressed as:
 
ΔKE ≠ 0
KEi ≠ KEf
1/2m1v12 + 1/2m2v22 ≠ 1/2m1v1'2 + 1/2m2v2'2
 
It is extremely important to keep in mind that in both elastic and inelastic collisions, momentum is conserved.
Mathematically this is expressed as:
 
Δp = 0
pi = pf
m1v1 + m2v2 = m1v1' + m2v2'
 
Also, check out One Dimensional Elastic Collisions.
 
1) A 2.0 kg block moving 3.5 m/s to the right collides with a 6.0 kg block moving 2.0 m/s to the left. Ignoring
    friction, determine the:
    (a) final velocity if the two blocks stick together.
    (b) amount of heat produced.
    (c) final velocity if the collision was completely elastic.
 
2) A 10.0 g bullet is moving with a horizontal velocity of 40.0 m/s into a 8.0 kg block of wood which is at rest. If the
    bullet becomes embedded in the wood what is the final velocity of the wooden block?

 

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