Problem 8.79 - Combining Conservation Laws

A 5.00-kg chunk of ice is sliding at 12.0 m/s on the floor of an ice-covered valley when it collides with and sticks to another 5.00-kg chunk of ice that is initially at rest. Since the valley is icy, there is no friction. After the collision, how high above the valley floor will the combined chunks go?

Figure 8.79

My Attempt / Work

$$m_a=[5.00]kg$$ $$m_b=[5.00]kg$$ $$m=m_a=m_b$$ $$v_a=[12]\frac{m}{s}$$ $$v_b=[0]\frac{m}{s}$$ $$v'=v'_a=v'_b$$ $$P=m_av_a+0=[(5.00)(12)]\frac{kg}{s}m$$ $$P=[60]\frac{kg}{s}m$$ $$[60]\frac{kg}{s}m=m_av_a+0=[(5.00)(12)]\frac{kg}{s}m$$ $$P=m(v'_a+v'_b)=2mv'$$ $$[60]\frac{kg}{s}m=m(v'_a+v'_b)=2mv'$$ $$[\frac{60}{2}]\frac{kg}{s}m=mv'$$ $$[30]\frac{kg}{s}m=mv'$$ $$\frac{30}{m}=v'$$ $$v'=[6]\frac{m}{s}$$ $$\Delta K=K'-K=[-180]\frac{m}{s}$$

Since both masses are the same ma and mb are both equal. Initially object A had an initial velocity and the Block B as shown in the figure is at rest. In addition, both the objects were stuck together when they collided, I was able to use conservation of momentum (No external forces). Which led final velocity v' being equal to 6 m/s both the final velocities for both object A and B have to be equal because they got stuck together. I am assuming here that it was a perfect inelastic collision, so by subtracting the final kinetic energy with the initial kinetic energy, I was able to find that 180 joules of energy was lost; hence the minus sign.

My Question

Well I know what the problem is asking but I seem to be stuck on what to do next. I do not even know if the information I already have and found is enough and is even correct. On top of that I am not sure if this is a completely inelastic collision; I think so because they both stuck together and 180 joules of energy has been lost. Do I need more information before answer the problem? What can I do, I already know the answer by looking at the back of the book but the answer does not mean anything to me. So please be mindful that I am still learning and anything to help will be great.

  • $\begingroup$ Three remarks: 1. Never put only a list of equation after eachother. Without explanation, they do not mean anything to me. Explain which concept or conservation law you are using where, then we'll get your line of thought. (e.g. what is $v'$? No clue. 2. Never fill in number in equations, only at the ultimate final line, if desired. 3. If you ultimately plug in numbers, always use units! Can you update your question according to the above three remarks? It'll be easier to spot your mistake, or get your misunderstanding. $\endgroup$ – Bernhard Feb 21 '13 at 21:52
  • $\begingroup$ @Bernhard Well I agree with you on the units. I didn't include them because I am no Latex expert and on top of that I thought it was self explanatory if you want me to explain I will; again I am sorry if I left out on a piece of information. $\endgroup$ – Daniel Lopez Feb 21 '13 at 21:55
  • $\begingroup$ Best is to leave them out altogether. Well, you're asking for guidance, so I won't be giving you the answer, but try to carefully explain the steps you took. $\endgroup$ – Bernhard Feb 21 '13 at 21:57
  • $\begingroup$ @Bernhard Check my edit to see if there is anything missing, or wrong. $\endgroup$ – Daniel Lopez Feb 21 '13 at 22:13

I feel stupid for forgetting about potential energy, but any way I figured it out. What I what I had to do was equate the final kinetic energy K with gravitational potential energy U. I actually attempted that before but the problem was I did it a little wrong. I didn't include it in my work here on StackExchange because I wanted only the part where I knew what to do and stop where I didn't know where to go next. But since I understood where I had to go next I answered my own question which is not what I was expecting.

Solution: $$g=[9.8]\frac{m}{s^2}$$ $$0.5(m+m)(v')^2 = U$$ $$m(v')^2=(m + m)gh$$ $$m(v')^2=2mgh$$ $$(v')^2=2gh$$ $$\frac{(v')^2}{2g}=h$$ $$\frac{36}{g}=[1.8]m$$

P.S. Thanks for anybody who gave me tips. I appreciate it.

| cite | improve this answer | |
  • 1
    $\begingroup$ It is always best to find the answer yourself, as you won't forget in anymore. Don't forget to accept this answer as the correct one. $\endgroup$ – Bernhard Feb 22 '13 at 6:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.