9.4. Elastic and Inelastic Collisions http://www.ck12.org
Solution:Since the collision is elastic, both momentum and KE are conserved. We use the conservation of momen-
tum and conservation of KE equations.
Conservation of momentum:m 1 v 1 +m 2 v 2 =m 1 v 1 ′+m 2 v 2 ′
Conservation of KE:^12 m 1 v 12 +^12 m 2 v^22 =^12 m 1 v 1 ′^2 =^12 m 2 v 2 ′^2
Sincem 1 ,m 2 ,v 1 , andv 2 are known, onlyv 1 ′andv 2 ′are unknown. When the known values are plugged into these
two equations, we will have two equations with two unknowns. Such systems can be solved with algebra.
( 12. 0 kg)( 2. 40 m/s)+( 36. 0 kg)( 0 m/s) = ( 12. 0 kg)(v 1 ′)+( 36. 0 kg)(v 2 ′)
- 8 = 12. 0 v 1 ′+ 36. 0 v 2 ′
Solving this equation forv 1 ′yieldsv 1 ′= 2. 4 − 3 v 2 ′
1
2 (^12.^0 )(^2.^40 )
(^2) + 1
2 (^36.^0 )(^0 )
(^2) = 1
2 (^12.^0 )(v^1
′) (^2) + 1
2 (^36.^0 )(v^2
′) 2
1 = 12. 0 v 1 ′^2 + 36. 0 v 2 ′^2
76 =v 1 ′^2 + 3 v 2 ′^2
Substituting the equation forv 1 ′into this equation yields
76 = ( 2. 4 − 3 v 2 ′)^2 + 3 v 2 ′^2
76 = 5. 76 − 14. 4 v 2 ′+ 9 v 2 ′^2 + 3 v 2 ′^2
12 v 2 ′^2 − 14. 4 v 2 ′= 0
12 v 2 ′= 14. 4
v 2 ′= 1. 2 m/s
Substituting this result back intov 1 ′= 2. 4 − 3 v 2 ′, we getv 1 ′=− 1. 2 m/s.
So, the heavier car is moving in the original direction at 1.2 m/s and the lighter car is moving backward at 1.2 m/s.
Summary
- Elastic collisions are those in which both momentum and kinetic energy are conserved.
- Inelastic collisions are those in which either momentum or kinetic energy is not conserved.
Practice
Questions
The following video is a demonstration of elastic and inelastic collisions. Use this resource to answer the questions
that follow.
http://www.teachertube.com/viewVideo.php?video_id=30870
MEDIA
Click image to the left for use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/82509- Explain what happened in the first demonstration on elastic collisions.
- Explain what happened in the second demonstration on inelastic collisions.