*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 mv2) *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 mv2) so equate the two and note that the mass cancels, so:
½ v2 = gh
OR:
v2 = 2gh
v = (2gh)½
*endif *elseif (guess.lt..8) then *if (answer(4).eq.68) then Correct! Conservation of energy assures us that the initial KE (½ mv2) 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 mv2) equals the final PE: mgh, so equate the two and note that the mass cancels, so:
½ v2 = gh
OR:
v2/(2g) = h *endif *else *if (answer(5).eq.68) then Correct! Conservation of momentum assures us that the initial momentum (mvbefore) equals the final momentum: 6mvafter. *else Your answer: *write answer(5) a1 is incorrect; D is correct. Conservation of momentum assures us that the initial momentum (mvbefore) equals the final momentum: 6mvafter. Equate the two, and note that m cancels:
vbefore = 6vafter
SO:
(1/6) vbefore = vafter *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:

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)]