Solving for Translational Speed of a Bowling Ball on a Vertical Rise

[kpangrace]

Help! Only Have Until 11:30

A bowling ball encounters a 0.76 m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. If the translational speed of the ball is 4.10 m/s at the bottom of the rise, find the translational speed at the top



aahh! finals are next week and i don't know what I'm doing

 

[BobG]

You have a certain amount of kinetic energy at the bottom of the ramp and no potential energy (using the bottom of the ramp as your base).

At the top of the ramp, you have both kinetic and potential energy, since you gained some height. Your total energy (kinetic plus potential) has to stay constant.

In other words, the kinetic energy at the bottom of the ramp is equal to the sum of the kinetic energy and potential energy at the top of the ramp.

And no, the mass won't matter. If you take your equation for kinetic energy and potential energy, you'll notice you can pull the mass outside your parenthesis via the distributive property. The mass will cancel out since it's outside the parenthesis on both sides of the equation.

 

[kpangrace]

ok i put (1/2)v^2= 1/2 v^2 + gh now is that right? i plug everything in and get .957... which doesn't show up right... you know what I'm doing wrong? thanks in advance

 

[BobG]

Yeah, you stopped too soon.

You solved $$1/2 v_i^2-gh$$
The other side of your equation still has $$1/2 v_f^2$$

You need to multiply your answer by 2 to get rid of the 1/2.
Then you need to take the square root to get rid of power of 2.