* 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