*echo
*set show 1
This was a **timed test** you spent
*let timet=secnds-time0
*if (timet.lt.0) then
*let timet=timet+24*3600
*endif
*write timet/60 f5.1
minutes on it.

## Question 1

*let guess=ran
*if (guess.lt..2) then
*if (answer(1).eq.66) then
Correct! The force of gravity is partially in the opposite direction as the
displacement, so Work = **F·s** is negative.
*else
Your answer:
*write answer(1) a1
is incorrect; B is correct.
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*.
*endif
*elseif (guess.lt..4) then
*if (answer(2).eq.65) then
Correct! The lifting force is more-or-less in the same direction as the
displacement, so Work = **F·s** is positive.
*else
Your answer:
*write answer(2) a1
is incorrect; A is correct.
Work = **F·s**. The Work-Energy Theorem says that the work done by a
non-conservative force is the change in total energy (which includes the PEs
of all the conservative forces). In this case the KE is zero before and after the lift
and the gravitational PE has been increased by *mgh*. The the work done in
lifting is equal to the change in total energy is *mgh*. Note that in this case
the lifting force is more-or-less in the same direction as the
displacement, so Work = **F·s** is positive.
*endif
*elseif (guess.lt..6) then
*if (answer(3).eq.65) then
Correct! Conservation of energy assures us that the initial
PE: *mgh* equals the final KE (½ *mv*^{2})
*else
Your answer:
*write answer(3) a1
is incorrect; A is correct.
Conservation of energy assures us that the initial
PE: *mgh* equals the final KE (½ *mv*^{2})
so equate the two and note that the mass cancels, so:

½ *v*^{2} = *gh*

OR:

*v*^{2} = 2*gh*

*v* = (2*gh*)^{½}

*endif
*elseif (guess.lt..8) then
*if (answer(4).eq.68) then
Correct! Conservation of energy assures us that the initial
KE (½ *mv*^{2}) equals the final PE: *mgh*.
*else
Your answer:
*write answer(4) a1
is incorrect; D is correct.
Conservation of energy assures us that the initial
KE (½ *mv*^{2}) equals the final PE: *mgh*,
so equate the two and note that the mass cancels, so:

½ *v*^{2} = *gh*

OR:

*v*^{2}/(2*g*) = *h*
*endif
*else
*if (answer(5).eq.68) then
Correct! Conservation of momentum assures us that the initial
momentum (*mv*_{before}) equals the final momentum: 6*mv*_{after}.
*else
Your answer:
*write answer(5) a1
is incorrect; D is correct.
Conservation of momentum assures us that the initial
momentum (*mv*_{before}) equals the final momentum: 6*mv*_{after}.
Equate the two, and note that *m* cancels:

*v*_{before} = 6*v*_{after}

SO:

(1/6) *v*_{before} = *v*_{after}
*endif
*endif
## Question 2

*let guess=ran
*if (guess.lt..2) then
*if (answer(6).eq.67) then
Correct! In the elastic collision there must be more energy in the final state than
in the final state on the inelastic collision. It turns out that with this unequal a
mass ratio, the red ball rebounds with more speed than in the inelastic collision.
*else
Your answer:
*write answer(6) a1
is incorrect; C is correct.
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:

*mv*_{before} = *mv*_{1} + 5*mv*_{2} (conservation of momentum)

*v*_{before} = -*v*_{1} + *v*_{2} (reversed relative velocity)

*v*_{before} = *v*_{1} + 5*v*_{2} (1)

*v*_{before} = -*v*_{1} + *v*_{2} (2)

2*v*_{before} = 6*v*_{2} [(1)+(2)]

4*v*_{before} = -6*v*_{1} [5×(2)-(1)]

(1/3)*v*_{before} = *v*_{2} [(1)+(2)]

-(2/3)*v*_{before} = *v*_{1} [5×(2)-(1)]

Both of these are faster than the inelastic case.
*endif
*elseif (guess.lt..4) then
*if (answer(7).eq.67) then
Correct! Galileo commented on the nearly isochronous pendulum almost 400 years ago.
*else
Your answer:
*write answer(7) a1
is incorrect; C is correct.
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.
*endif
*elseif (guess.lt..6) then
*if (answer(8).eq.65) then
Correct! In the horizontal direction we have an approximation to
isochronous simple harmonic motion.
*else
Your answer:
*write answer(8) a1
is incorrect; A is correct.
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.
*endif
*elseif (guess.lt..8) then
*if (answer(9).eq.68) then
Correct!
*else
Your answer:
*write answer(9) a1
is incorrect; D is correct.
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.
*endif
*else
*if (answer(10).eq.65) then
Correct! The largest acceleration occurs at the largest amplitude.
*else
Your answer:
*write answer(10) a1
is incorrect; A is correct.
When are the forces large?
Near equilibrium (i.e., near collision) the forces are zero.
The largest acceleration occurs at the largest amplitude.
*endif
*endif

The next random number would have been:
*write ran f6.5
*show