# Ball rolls down ramp

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 im 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.

Last edited:
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 dont know hwo to get the height from here.

Fermat
Homework Helper
Use conservation of energy.

gain in KE = loss in PE.

mukundpa
Homework Helper
what is the unit of energy you got???