Bat Hitting Ball: Find Speed After Collision

[alfredo24pr]

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

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

The Attempt at a Solution

Im stuck at beginning :/

 

[tiny-tim]

hi alfredo24pr!

elastic collision means that you use conservation of energy also

 

[alfredo24pr]

hi alfredo24pr!

elastic collision means that you use conservation of energy also

hi tiny tim..so that means I have to relate kinetic energy?

 

[tiny-tim]

(just got up :zzz: …)

hi tiny tim..so that means I have to relate kinetic energy?

yup! …

what do you get? ​

 

[alfredo24pr]

(just got up :zzz: …)


yup! …

what do you get? ​

umm.. I do not understand. I know that I have to relate to v used in momentum to the v in kinetic energy

 

[tiny-tim]

write out the conservation of momentum equation and the conservation of energy equation

 

[alfredo24pr]

write out the conservation of momentum equation and the conservation of energy equation

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

and

1/2mv_i² + mgh_i = 1/2mv_f² + mgh_f

 

[tiny-tim]

hi alfredo24pr!

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

 

[alfredo24pr]

hi alfredo24pr!

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

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

I still do not understand how to work with both equations.

 

[tiny-tim]

start

 

[alfredo24pr]

start

uhh..

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

1/2mv_i² = 1/2m(2.3)²

and then i substitute

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

1/2mv_i² + 0 = 1/2mvf2??

 

[tiny-tim]

1/2mv_i² + 0 = 1/2mvf2??

what is this supposed to be?

(in particular, what is the "0" ?)