What is the speed of the ball after colliding with a bat at 1.7 m/s?

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In a perfectly elastic collision between a ball and a bat, both initially moving at 1.7 m/s, the speed of the ball after the collision can be determined using conservation of momentum and kinetic energy principles. Given the bat's much larger mass, it can be treated as stationary, allowing the ball's speed to effectively double to 3.4 m/s in the bat's reference frame. When converting back to the original frame, the ball's speed remains 3.4 m/s after the collision. Participants discuss the implications of treating the bat as a wall and the confusion surrounding the calculations. Understanding these concepts is crucial for accurately determining the ball's final speed post-collision.
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A ball and bat, approaching one another each with the same speed of 1.7 m/s, collide. Find the speed of the ball after the collision. (Assume the mass of the bat is very much larger than the mass of the ball, a perfectly elastic collision, and no rotational motion).


so i am going to use of energy, and since it is an elatic collison, it will be kenitic energy

.5m1v1i^2+.5m2v2i^2=.5m1v1f^2+.5m2v2f^2

since the mass of the bat is much larger than that of the ball, i am going going to use

v1i+v2i=v1f+v2f

so, if the bat and ball are approching each other at the same speed i am going to take the ball approching the bat to be negitive, so i have

1.7-1.7=v1f+v2f

the answer isn't 0...so this is where i am stuck
 
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Rasine said:
since the mass of the bat is much larger than that of the ball, i am going going to use

v1i+v2i=v1f+v2f

Could you explain this assumption please?
 
i really don't know! i was just trying to go somewhere...can you help me?
 
Rasine said:
(Assume the mass of the bat is very much larger than the mass of the ball, a perfectly elastic collision, and no rotational motion).

Does this mean that the collision can be treated like a 'ball vs. wall' collision?
 
i don't know...but i was also thinking that the bat would be taken as 0 and the velociyt of the ball could be 2(1.7)
 
congrats, radou. Seems like everyone's turning gold around here! :approve: (I know your thinking; you just want more medals! :-p)

(sorry for the OT comment Rasine)
 
Rasine said:
i don't know...but i was also thinking that the bat would be taken as 0 and the velociyt of the ball could be 2(1.7)
Actually, you are on the right track with this thinking. More precisely: In a frame in which the bat is at rest, the ball moves with speed 2x(1.7). In that frame, treating the bat as hugely massive, what's the rebound velocity of the ball? Then convert back to the original frame to find the ball's speed with respect to the ground.
 
Rasine said:
i don't know...but i was also thinking that the bat would be taken as 0 and the velociyt of the ball could be 2(1.7)

Just to add, if you assume m2 >> m1, (where m2 is the mass of the bat), you can easily verify your result by simply using conservation of momentum, unless I'm missing something here.

cristo said:
(I know your thinking; you just want more medals! :-p)

(sorry for the OT comment Rasine)

Bingo. :biggrin:
 
so if i do that ussing conseravation of energy i will have

0+m2(2*1.7)=0+m2(2*1.7)...right?
 
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i am so confused
 
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