Homework Hint: Problem 3/189
- For part (a) of this problem, treat A and B together as a single system. With this choice of system, the coupling force during impact is an internal force. As a result, linear momentum in the horizontal direction is conserved.
- For part (b), you are asked to find the loss of energy. This is simply the change of kinetic energy from before to after impact.
- Question: What happens to the energy lost in the system during impact? (This question becomes very important to us on Friday when we discuss impacts. During that lecture, we will need to address the fact that energy is almost never conserved during impact.)
I am having problems coming up with the right energy lost. how do you equate the kinetic energies? the only equation I could come up with is (.5m(a+b)v(f)^2)-(.5m(a)v(a)^2+.5m(b)v(b)^2)=delta(E). Is that the correct equation but I am messing up the conversions between mi/hr and ft-lb or did I mess up somewhere else?
ReplyDeleteremember that lbs in this book refer to weight, not mass. You also have to convert speed to ft/s for the kinetic energy equation to work out to ft-lb.
ReplyDeletehmm...I converted all units to ft/s and lb mass but I am still get an answer that is 192 lbs off. The equation I am using is:
ReplyDeletedelta E = (0.5(mA + mB)vf^2) - [ (0.5(mA)(vA)^2) + (0.5(mB)(vB)^2) ]
any ideas what I am doing wrong.
sri-
ReplyDeleteI had the exact same problem. It's just caused by rounding the vf term before plugging it into your equation.
I agree with Panda Watch in that the problem likely lies in rounding. When I carry through with the accuracy of my calculator and using g = 32.2 ft/sec^2, I agree with the solution in the book.
ReplyDeleteSince your answer is off by 192 ft-lb, your answer is accurate to only 1 significant digit. This means that at least some of your calculations were made using at most 2 significant digits.
Look back on the rounding that you used in using v_2 or in your number of digits in g.
ah, right! Thanks!
ReplyDelete