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Sample solution

Giancoli 7th Edition, Chapter 5, Problem 12

(4:49)

Recent questions and answers

Giancoli 7th "Global" Edition, Chapter 12, Problem 35

By maia.fehr.oth23 on Thu, 12/01/2016 - 21:44

Why 331 and not 343?

Giancoli 7th Edition, Chapter 10, Problem 40

By odelay.chewy on Fri, 11/25/2016 - 00:47

I am struggling understanding the funky algebra used in this step. You kind of gloss over it. Could you show it step for step?

By Mr. Dychko on Sat, 11/26/2016 - 11:39

Hi odelay.chewy,

Thanks for the question. I'm away from my recording equipment at the moment, but I'll try to give some highlights here. We start with saying the total forces directed up equal the total forces directed down, then I make a substitution for each up force with the archimedes principle that the buoyant force is the weight of fluid displaced, which is $F = \rho V g$, and the down forces are the usual weight formula $F = mg$. Then, in green, the volume of the wood and lead is expressed in terms of their density and mass. This is in turn substituted (using red) for each volume factor in the up force terms. The specific gravity of wood replaces the ratio of the density of water to wood. Then comes a confusing line where the mass of the wood is shown moving to the left, but actually ends up on the right. The thing to notice with that line is that we're collecting like terms. We're placing the wood terms together on one side, and the lead terms on the other side. The mass of lead and mass of wood are factored out on their respective sides, leaving us with some bracket multiplied by a mass on each side. We're interested in knowing the mass of lead, so we divide both sides by the bracket by which the lead mass is multiplied. Then we plug and chug. Hopefully this video commentary is a bit helpful...

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 13, Problem 14

By saie.joshi on Thu, 11/24/2016 - 08:42

Why do you subtract the values instead of adding them together? Don't both the water and the glass reduce volume?

By saie.joshi on Thu, 11/24/2016 - 08:43

never mind I got it!

Giancoli 7th Edition, Chapter 8, Problem 22

By taylorreilley on Mon, 11/14/2016 - 08:32

Why did you plug the Wf value in for b when you wrote Wi

By Mr. Dychko on Wed, 11/23/2016 - 00:32

Hi taylerreilley,

Thanks for the question. "initial" and "final" is a bit confusing in this questions since there are more than one time periods to consider, each with their own "initial" and "final" moment. For part a), the "final" moment is when the car has reached $55 \textrm{ km/h}$, making $\omega_f = 38.194 \textrm{ rad/s}$. However, for part b), the initial moment is when the car is at $55 \textrm{ km/h}$, so for part b) $\omega_i = 38.194 \textrm{ rad/s}$. The initial moment of part b) is the final moment of part a).

I can see where you're coming from, and technically it would be more correct for me to put a subscript on the variables to assign them to parts of the question, such as for the angular speed for part a) written as $\omega_{iA}$ and $\omega_{fA}$, and then say $\omega_{fA} = \omega_{iB}$, and so on, but I think that also creates it's own source of confusion by presenting so many more subscripts.

All the best,
Mr. Dyckho

Giancoli 7th Edition, Chapter 7, Problem 41

By sheumangutman on Fri, 11/11/2016 - 16:12

Why is this collision deemed "inelastic" in terms of the treatment of kinetic energy conservation. Seems based on the diagram, the objects are not sticking together, which would suggest its inelastic.

By Mr. Dychko on Sat, 11/12/2016 - 13:16

Hi sheumangutman,

Thanks for the question. The term "inelastic" means there is less total kinetic energy after a collision compared with before a collision. Since that's the case here, we can call it inelastic. It is not "completely inelastic", that's true. A completely inelastic collision involves the two particles sticking together, and this type of collision results in the maximum loss in kinetic energy that's possible while still conserving momentum. Seeing two particles bounce apart means either the collision is "inelastic" (but not "completely inelastic"), or maybe the collision is "perfectly elastic", in which case the total kinetic energy after collision is the same as it was before. In this case, the collision is a regular inelastic collision since some kinetic energy was lost, but not the maximum loss that would have otherwise been possible had they stuck together.

Best,
Mr. Dychko

Giancoli 7th Edition, Chapter 6, Problem 9

By sheumangutman on Thu, 11/10/2016 - 08:28

Could you also approach the problem by first solving for final velocity by using V=a*t, and then plugging this velocity into v^2=vo^2+2AD to solve for displacement?

Thanks.

By Mr. Dychko on Sat, 11/12/2016 - 13:03

Hi sheumangutman,

Yes you can! Substituting $v=at$ info $v^2 = v_o^2 + 2aD$ gives $a^2t^2 = v_o^2 + 2aD$, which, with $v_o = 0$ rearranges to $D = \dfrac{1}{2}at^2$ after you divide both sides by $2a$. $D = \dfrac{1}{2}at^2$ is the formula I used in the video. Good work!

All the best,
Mr. Dychko

Giancoli 7th Edition, Chapter 4, Problem 52

By schamorg on Wed, 11/09/2016 - 07:09

Thank you so much. This was very helpful actually.

By Mr. Dychko on Wed, 11/09/2016 - 07:11

Super! I'm glad, and thanks for the feedback. :)

Giancoli 7th Edition, Chapter 3, Problem 37

By kmoons25 on Mon, 11/07/2016 - 20:00

why do you do these things?

By Mr. Dychko on Tue, 11/08/2016 - 00:54

Hi kmoons25, Well, I know physics can be confusing. When you have a more specific question, just let me know.

All the best,
Mr. Dychko