And explain why.
Also..formula for inelastic collision is..
M(a)V(a)+M(b)V(b)=M(a)V(a2)+M(b)V(b2)?
And elastic is..
M(1)V=M(1)V(1)+M(2)V(2)
?
2006-12-09 10:21:05 · 4 answers · asked by pugfug90 2 in Science & Mathematics ➔ Physics
INELASTIC
MOMENTUM is conserved but
ENERGY is lost
ELASTIC collison ==> (m1u1) + (m2u2) = (m1v1) + (m2v2)
INELASTIC collision ==> (m1u1) + (m2u2) does not equal (m1v1) + (m2v2)
2006-12-09 10:24:54 · answer #1 · answered by Anonymous · 0⤊ 0⤋
It can be treated as elastic. Although no collision in real life if perfectly elastic, bowling is close -- there is negligible energy lost. Because all the momentum is transferred into the pins and remaining motion of the ball.
2006-12-09 18:27:28 · answer #2 · answered by Julian A 4 · 0⤊ 0⤋
inelastic.
totally elastic collisions only occur on the quantum level (theoretically)
(if the collision between the ball and pins was elastic then how could you hear it? ;-)
2006-12-09 18:29:28 · answer #3 · answered by wynnr 2 · 0⤊ 0⤋
It is inelastic the pins and the ball do not deform.
2006-12-09 18:25:51 · answer #4 · answered by shadouse 6 · 0⤊ 0⤋