* echo

## Results

The results of your exam are displayed below. Depending on what you've set the SHOW level to be you'll see various amounts of grading information. In the grading report (if it exists!) question numbers in bold are questions you missed. For SHOW in the range 5-7 questions marked "not sure" but correctly answered will be reported, but not in bold. Additional "helpful" information may also be displayed.

* noecho 66 b Work = F·s. The Work-Energy Theorem says that the work done by a force is minus the change in PE. In this case the force of gravity F is down whereas the displacement s is partially up. Thus the dot product of these two vectors is negative; only B is negative! Additionally gravitational PE is:
PE = mgh
thus the increase in PE of mgh means a negative work of -mgh. 67 a Use conservation of energy! The initial PE of mgh will be converted to KE of ½ mv2. Thus:
mgh = ½ mv2
2gh = v2
so the answer is A 68 d Use conservation of momentum! Initially all the momentum is in the red sphere: mvbefore. Finally a total mass of m+5m=6m is moving at some speed vafter. Thus:
mvbefore = 6mvafter
(1/6) vbefore = vafter
so the answer is D. 69 c Use your experience or simultaneous conservation of energy and momentum or the fact that the relative velocity is reversed. In the inelastic collision mechanical energy is converted to heat. Thus with an elastic collision there will finally be more total kinetic energy than in an inelastic collision. So something must be moving faster with an elastic collision! In an equal mass elastic collision, the target ends up moving with the initial speed of the projectile and the projectile ends at rest. Thus the target moves faster than in an inelastic case and the projectile moves slower. In an elastic collision where the projectile is much lighter than the target, the projectile rebounds with an almost undiminished speed. The target must then end up with about twice the initial momentum of the projectile, so that total momentum is conserved. Thus both the target and the projectile end up moving faster than in an inelastic collision. Hence see that B and C are the only options, and that since our 5-to-1 mass ratio is most similar to the latter case, suspect C is correct. Here are the simultaneous equations solved:

mvbefore = mv1 + 5mv2   (conservation of momentum)
vbefore = -v1 + v2   (reversed relative velocity)

vbefore = v1 + 5v2 (1)
vbefore = -v1 + v2 (2)

2vbefore = 6v2   [(1)+(2)]
4vbefore = -6v1   [5×(2)-(1)]

(1/3)vbefore = v2   [(1)+(2)]
-(2/3)vbefore = v1   [5×(2)-(1)]

Both of these are faster than the inelastic case. 70 c Galileo commented on this almost 400 years ago! Pendulum period depends mostly on L and g; for "small" swings it is independent of amplitude. Thus same L and g, small swings in both cases, same time C. 71 d The speed and hence the momentum of the projectile changes as it falls. Momentum is not conserved since there are external forces. During the inelastic collision mechanical energy is converted to heat and sound. 72 d Use conservation of energy! Conservation of momentum tells us that the speed after the collision is only (1/6) that before. Conservation of energy says that the maximum height is related to the square of the speed:
½ v2 = gh
Thus with (1/6) the speed, we get (1/36) the height. 73 d Combine conservation of momentum and conservation of energy! Conservation of energy on the down swing says:
½ (vbefore)2 = ghinitial
Conservation of momentum at the collision says:
mredv0 = (mred+mblue)v1
(2ghinitial)½ = vbefore = (1+mblue/mred) v1 = (1+mblue/mred) (2ghfinal)½
SO: hinitial = (1+mblue/mred)2 hfinal
(1+mblue/mrede)-2 hinitial = hfinal
AND doubling hinitial doubles hfinal whereas doubling mred increases hfinal by (12/7)2. Same result for doubling mblue. 74 a In the horizontal direction we have an approximation to isochronous simple harmonic motion. The jump in velocity is the collision. B shows motion where the post-collision period is longer than the pre-collision period. C shows the velocity at the time of release is non-zero. D shows constant acceleration motion. A is correct. 75 a When are the forces large? Near equilibrium (i.e., near collision) the forces are zero. The largest acceleration occurs at the largest amplitude. * echo

### Summary Results

Out of the * write nquest i3 questions you answered * let ntry=nquest-nblank * write ntry i3 questions and got * write nright i3 correct -- * if (ntry.gt.0) then that's * write nint(nright/ntry*100) i3 percent correct overall. * endif * if (ntry.eq.0) then if you want to get some correct, you really should try to answer some! * endif * if (n_quest.gt.0) then * if (ntry.gt.n_quest) then

Of the * write n_quest i3 questions you attempted and marked "not sure", * write nint(n_right/n_quest*100) i3 percent were in fact correct. Your answers to questions not marked "not sure" were correct * write nint((nright-n_right)/(ntry-n_quest)*100) i3 percent of the time. * endif * endif

On the real exam, you should be sure to guess an answer for each question. However, on this web exam I give you the expected ¼ of a correct answer per blank answer. Thus if you have no idea what the correct answer is leaving the web answer form blank does no harm on average. Your expected score (including blanks) is: * let score=nright+nblank*.25 * write score f5.1 * log score '6Hscore=,f5.1'

*show