What is the maximum height reached by the other ball after collision?

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    Balls Collison
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Homework Help Overview

The problem involves two identical steel balls in a collision scenario, where one ball is released from a height and collides elastically with the other. The objective is to determine how high the second ball rises after the collision.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the first ball's velocity just before the collision using energy conservation principles. There is a question about the accuracy of the height calculation and the use of trigonometric functions to find the vertical component of the height.

Discussion Status

Some participants are exploring different methods to calculate the height reached by the second ball, while others are questioning the assumptions made in the calculations. Guidance has been suggested regarding the need for a diagram to clarify the geometric relationships involved.

Contextual Notes

There is an emphasis on understanding the geometry of the situation, particularly in relation to the angles involved and the height calculations based on trigonometric functions.

Rasine
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Two identical steel balls, each of mass 1.4 kg, are suspended from strings of length 36 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle q = 36° with the vertical and let it go. It collides elastically with the other ball. How high will the other ball rise?

ok so what i am using is pi=pf and ei=ef

first i calcuate the velocity of of th first ball when it is just about to hit the second which is
mgh=1/2mv^2
(1.4)(9.8)36cos36=1/2(1.4)v^2

then v=23.89

so pi=pf of the system so m1v1+m2v2=m1v1f+m2v2f

which is (1.4)(23.89)+0=0+1.4v2f

so v2f=23.89 and if that is true, wouldn't ball 2 go as high as ball 1?
 
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Rasine said:
first i calcuate the velocity of of th first ball when it is just about to hit the second which is
mgh=1/2mv^2
(1.4)(9.8)36cos36=1/2(1.4)v^2

Reconsider the red step.
 
would it be 36sin56 instead?
 
Rasine said:
would it be 36sin56 instead?

I suggest you draw a diagram on this one.
 
well i want to find the length in the y direction and i see 2 ight triangles i could get y from 36cos36 or take 90-36=56 and do 36sin56
 

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