Elastic vs Inelastic Equation?
Ball A of mass 3.10-kg, moving to the right at a velocity of +3.40 m/s on a frictionless table, collides head-on with a stationary ball B of mass 8.10-kg. Find the final velocity of ball A if the collision is elastic. Give your answer in SI units.
What is the difference in the equations between elastic and inelastic equations? Can you give the equations?
- Anonymous8 years agoFavorite Answer
for elastic collision, coefficient of restitution is
e = - (V' - v')/(V - v) = 1
- (V' - v')/(0 - 3.4) = 1
V' = v' + 17/5
conservation of momentum law,
ΔP = 0
m v - M V' - m v' = 0
(3.1)(3.4) - (8.1)( v' + 17/5) - 3.1 v' = 0
v' = -1.51786 m/s
negative means in opposite with respect to initial direction...
and for inelastic collision,
ΔP = 0
m v - (M + m) v' = 0
(3.1)(3.4) - (8.1 + 3.1) v' = 0
v' = 0.941071 m/s
difference = |v' elastic - v' inelastic|
difference = |1.51786 - 0.941071|
difference = 0.576789 m/s
- lindmanLv 43 years ago
the answer to the first question is bound, notwithstanding a lot momentum is lost from motor vehicle a million is the same because the momentum benefit in motor vehicle 2 assuming that's a splendidly elastic collision. assuming that's a splendidly inelastic collision for problem 2, then the momentum of motor vehicle a million previously the outcome is now equivalent to the momentum of motor vehicle a million+ motor vehicle 2 after the outcome. in any problem the position the collision is both not completely inelastic or elastic, then some a lot more suitable complicated math per many variables comes into play, purely hardship about completely elastic and inelastic.
- RLv 58 years ago
Elastic means kinetic energy is conserved.
Inelastic means it is not, some energy is lost in the noise, crunch, etc.
Totally inelastic is a special case of inelastic, where you have the extra information that the two objects stick together after the collision.
Totally inelastic is the easiest to solve, because there is only one v after the collision.
Elastic is harder, because you have two equations for two velocities: conservation of momentum, and conservation of energy. In fact, a lot of people find the quadratic equation that results from substitution to be a pain, so there is actually a shortcut that works ONLY for elastic, head-on (i.e. one dimensional) collisions:
v1 + v1' = v2 + v2'
So if you combine that with
m1v1 +m2v2 = m1v1' +m2v2'
You get two easier equations in two unknowns.
Good luck!Source(s): I tutor physics.