Bat Hitting Ball: Find Speed After Collision

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In the discussion about a ball and bat collision, participants focus on solving for the speed of the ball after an elastic collision, given both objects approach each other at 2.30 m/s. The conservation of momentum and kinetic energy equations are emphasized as essential for the solution. Users express confusion about how to apply these equations, particularly in relating the mass of the bat to the ball. The conversation highlights the assumption that the bat's mass is significantly larger, allowing simplifications in the equations. Overall, the thread revolves around clarifying the application of physics principles to find the post-collision speed of the ball.
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Homework Statement



A ball and bat, approach one another each with the same speed of 2.30 m/s, collide. Find the speed of the ball after the collision. (Assume the mass of the bat is much much larger than the mass of the ball and that this is an elastic collision with no rotational motion).

Homework Equations



m1v1i + m2v2i = m1v1f + m2v2f


The Attempt at a Solution



Im stuck at beggining :/
 
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hi alfredo24pr! :smile:

elastic collision means that you use conservation of energy also :wink:
 
tiny-tim said:
hi alfredo24pr! :smile:

elastic collision means that you use conservation of energy also :wink:

hi tiny tim..so that means I have to relate kinetic energy?
 
(just got up :zzz: …)
alfredo24pr said:
hi tiny tim..so that means I have to relate kinetic energy?

yup! :biggrin:

what do you get? :smile:
 
tiny-tim said:
(just got up :zzz: …)


yup! :biggrin:

what do you get? :smile:


umm.. I do not understand. I know that I have to relate to v used in momentum to the v in kinetic energy :confused:
 
write out the conservation of momentum equation and the conservation of energy equation :smile:
 
tiny-tim said:
write out the conservation of momentum equation and the conservation of energy equation :smile:

m1v1i + m2v2i = m1v1f + m2v2f

and

1/2mvi2 + mghi = 1/2mvf2 + mghf
 
hi alfredo24pr! :smile:

ok, now use 2.30, and assume that m2/m1 is very much less than 1 :wink:
 
tiny-tim said:
hi alfredo24pr! :smile:

ok, now use 2.30, and assume that m2/m1 is very much less than 1 :wink:

so m2 / m1 is negligible therefore I can write it out as 0?

I still do not understand how to work with both equations.
 
  • #10
start :smile:
 
  • #11
tiny-tim said:
start :smile:

uhh..

m1v1 = m1(2.3)
m2v2 = m2(2.3)

1/2mvi2 = 1/2m(2.3)2

and then i substitute

m1(2.3) + m2(2.3) = m1vif + m2vif

1/2mvi2 + 0 = 1/2mvf2??
 
  • #12
alfredo24pr said:
1/2mvi2 + 0 = 1/2mvf2??

what is this supposed to be? :confused:

(in particular, what is the "0" ?)
 
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