Solving Ball Roll Down Ramp: 200 m/s Speed for Steel Ball

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To determine the height from which a 110 g marble must be released to give a 290 g steel ball a speed of 200 m/s after a collision, the discussion emphasizes using conservation of energy and the principles of elastic collisions. The final kinetic energy of the steel ball is calculated using KE = 0.5 * m * v^2, resulting in 5800 J. This value is then used to find the velocity of the marble just before impact, which is calculated to be 324.7 cm/s. The discussion also highlights the importance of using the incline angle (39 degrees) to calculate the marble's acceleration down the ramp. Ultimately, the relationship between kinetic energy and potential energy is crucial for finding the required height for the marble's release.
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A 110 g marble rolls down a 39.0 deg. incline. At the bottom, just after it exits onto a horizontal table, it collides with a 290 g steel ball at rest.

How high above the table should the marble be released to give the steel ball a speed of 200 m/s?

well first i tried to use the formula for perfectly elastic collisions
(Vf)steel= (2m(marble)/ m(marble)+m(steel))*v_f(ball1)

and i solved for v_f and then i plugged that into mgh(initial)= 0.5m*v^2

and none that didnt work so now I am confused again
 
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I just want to be sure. Do you mean final velocity of 20 m/s? Or do you really mean 200 m/s?

In an elastic collision you can assume that the Kinetic Energy of the balls will be the same before and after.

KE = 1/2 mv^2

1) Find the final KE of the steel ball after impact (v=20 or v=200).
2) Use this number to calculate the velocity of the smaller ball just prior to impact.
3) Calculate the initial height necessary in order to accelerate the ball to the velocity found in step 2.

Furthermore if I'm not mistaken, use the sin(39) to help calculate the acceleration of the ball down the ramp. After all if the ball were on a table, the angle would be zero and sin(0) = 0. If the ball were up against a wall the angle would be 90 (straight down), and sin (90) = 1.
 
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o whoops.. i meant cm/s
 
ok this is what i got so far

0.5*m*v^2= 0.5*.290kg*200cm/s^2= 5800

5800= (0.5)(.110kg)(v^2)
i solved for v and i got 324.7cm/s^2

now i know this is the velocity at the bottom of the ramp but i don't know hwo to get the height from here.
 
Use conservation of energy.

gain in KE = loss in PE.
 
what is the unit of energy you got?
 
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