IPhOO2013-14WOOTSolutions

8/13/2019 IPhOO2013-14WOOTSolutions

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WOOT 2013-14: IPhOO Contest

Official Solutions

Thanks to the following individuals.

Problem Contributors Ahaan S. Rungta

Solution Contributors Ahaan S. Rungta, Cody Johnson, Will Song a

Diagrams and Documenting Evan Chena

Cody and Will were participants at the contest.They posted solutions on AoPS, some of which you see here.

Top 3 :

1. lawrenceli: 9/10

2. codyj: 8/10, winner of Silver Tiebreaker

3. superbob123: 8/10

Answers :

1. 42

2. 1337 nm

3. 6561

4. 155. 0.678

6. 0.949

7. 25 cm

8. 51

9. 100 N

10. 18, 70

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Problem 1 (Ahaan Rungta). A block of mass m on a frictionless inclined plane of angle is connected by a cord over a small frictionless, massless pulley to a second block of massM hanging vertically, as shown. If M = 1 .5m , and the acceleration of the system is g3 ,where g is the acceleration of gravity, what is , in degrees, rounded to the nearest integer?

Solution 1 (Cody Johnson). A force of magnitude mg sin acts on m along the inclinedplane, where Mg = 1 .5mg acts on M downwards. From Newtons second law, we musthave

F = ma = ( m + 1 .5m )g3

= 2.5

3 mg = 1 .5mg mg sin = (1 .5 sin )mg

implying sin = 1 .5 2 .5

3 = 2

3 = 42 .

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Problem 2 (Ahaan Rungta). Light of a blue laser (wavelength = 475 nm) goes througha narrow slit which has width d. After the light emerges from the slit, it is visible on ascreen that is 2.013 m away from the slit. The distance between the center of the screenand the rst minimum band is 765 mm. Find the width of the slit d, in nanometers.

Solution 2 (Ahaan Rungta). Let be the wavelength, d be the slit width, x be thedistance between the screen, and the slit and y be the band spacing. We know that, if is

the angle of diffraction,sin =

md

.

The band order is m = 1, so sin = d

. We also know that

tan = yx

,

so = arctanyx

. Substituting this back into our other equation, we have

sin arctany

x=

d = d =

sin arctanyx

.

We are given = 475 nm, y = 0 .765m and x = 2 .013 m. Substituting these, we have

d = 1337nm .

Problem 3 (Ahaan Rungta). Let the rest energy of a particle be E . Let the work doneto increase the speed of this particle from rest to v be W . If W = 1340 E , then v = kc, wherek is a constant. Find 10000 k and round to the nearest whole number.

Solution 3 (Will Song: soy un chemisto). The relativistic energy of the particle at

speed v is mc 2

1 v2

c2

. Its rest energy is simply E = mc 2 . The problem states that

mc 2

1 v2

c2mc

2 = 1340

mc 2 .

Some rearranging gives 1 v2

c2 = 1600

2809 , which is the same as v = 1209

53 c 0.6561. Hencethe answer is 6561 .

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Problem 4 (Ahaan Rungta). The Iphoon particle, of charge q , is accelerated from restby a potential difference of V . This strange particle then enters a region with a uniformmagnetic eld, B , which is perpendicular to the particles velocity. The Iphoon follows acircular path with radius R . If q = 1 C, V = 1kV, B = 1mT, and R = 2 ft, let the weightof an Iphoon, in Newtons, be w. If w 10 p, where p is an integer, nd p. That is, what isthe order of magnitude of the weight?

Solution 4 (Ahaan Rungta). Let m be the mass of the particle. Let q be its charge. LetV be the potential difference. Let B be the magnitude of the magnetic eld. Let R be thepath radius. Let v be the constant speed at which the particle travels in its circular orbit.

We know that the kinetic energy of the particle is qV , but we also know, from ClassicalMechanics, that it is

12

mv 2 . We set them equal, and we obtain

qV = 12

mv 2 = v = 2qV m .Also, the centripetal force of the particle, which has value

mv 2

R in orbit is provided by the

Lorentz Force, which is qvB . So we set them equal as well:

qvB = mv 2

r = qB 2qV m =

m 2 qV m

R .

Squaring both sides of the equation and re-arranging for m , we have

m = qB 2 R 2

V .

Thus, the weight isw = mg,

where g is the acceleration of gravity on Earth. So, our answer is

w = qgB 2 R 2

V .

We were given q = 1 106 C, V = 1 103 V, B = 1 103 T, and R = 0 .61 m. Also, weknow that g = 9 .81m/s 2 . Substituting these, we havew = 3 .65 10

15 N,

so our answer is p = 15 .

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Problem 5 (Ahaan Rungta). A uniform ladder of mass m and length L is resting on awall. A man of mass m climbs up the ladder and is in perfect equilibrium with the ladderwhen he is 23 L the way up the ladder. The ladder makes an angle of = 30 with thehorizontal oor. If the coefficient of static friction between the ladder and the wall is thesame as that between the ladder and the oor, which is , what is , expressed to thenearest thousandth?

Solution 5 (Ahaan Rungta). Draw a free-body diagram on the ladder, like this oneposted on the AoPS forum. There are six forces acting on the ladder:

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At the ladders point of contact with the wall, a force W acts outward from the wallas the normal force at that point.

At the ladders point of contact with the wall, a frictional force F w acts up the wall.

At the ladders point of contact with the oor, a force N acts up from the oor as thenormal force at that point.

At the ladders point of contact with the oor, a frictional force F f acts in towardsthe wall, so as to restrain the ladder from falling at.

At the middle of the ladder (center of mass), a force mg acts downward.

At the two-third-up point on the ladder, a force mg , from the person, acts downward.

We use Newtons Laws in the x and the y directions, respectively, to get

W = F f ,F w + N = 2mg.

Now, we set the net torque at a certain sensible pivot to be 0, since the ladder is also in

rotational equilibrium. WLOG, let this point be the point at which the ladder is in contactwith the oor. We set the clockwise torque equal to the counterclockwise torque. Doingthis, we obtain

W Lsin + F wLcos = mg L2 cos + mg 2L

3 cos .Note that our unknowns are W , F w , N , F f , and . We have 3 equations and 5 unknowns.We need two more equations to be able to solve. These two equations come from the formulafor friction, which states that F w = W and F f = N . So, the ve equations we have atconsideration are the following:

F w = W ,

F f = N,W = F f ,

F w + N = 2mg,

W sin + F w cos = mg cos 12

+ 23

.

Note that our last equation was a simplication of our previous equation from torque. Wecancelled L on each side and factored an mg cos on the RHS.Now, we solve. Note that

F w = W

= F f = N

= N 2 ,

so

F w + N = N 2 + N

= N 1 + 2 = 2 mg.

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= N = 2mg1 + 2

.

Now, going back to the LHS of our torque equation, we have

W sin + F w cos = F f sin + F f cos = N (sin + cos )

= 2mg1 + 2 (sin + cos ) .

We must now set this equal to the RHS. We get:

2 mg1 + 2 (sin + cos ) = mg cos

76

2 (sin + cos ) = 7

6 cos 1 +

2

12 sin + 12 2 cos = 7 cos + 7 cos 2

(5cos ) 2 + 12sin 7cos = 0

We now substitute = 30. When = 30, we have

sin = 12

, cos = 32

,

so our equation becomes5 3

2 2 + 6

7 32

= 0 .

Now, using the quadratic formula, it is easy to nd that we get solutions of

{2.064, 0.678}.The former solution is clearly meaningless, so the answer is = 0 .678 .

Problem 6 (Ahaan Rungta). A fancy bathroom scale is calibrated in Newtons. Thisscale is put on a ramp, which is at a 40 angle to the horizontal. A box is then put onthe scale and the box-scale system is then pushed up the ramp by a horizontal force F .The system slides up the ramp at a constant speed. If the bathroom scale reads R and thecoefficient of static friction between the system and the ramp is 0 .40, what is F R ? Round tothe nearest thousandth.

Solution 6 (Cody Johnson). Note that since R is equal to the perpendicular force on thescale, its magnitude also equals that of the normal force. Setting up our static equilibrium,we get

F x = F R cos R sin = 0 = F R

= sin + cos = 0 .949

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Problem 7 (Ahaan Rungta). A conical pendulum is formed from a rope of length 0 .50 mand negligible mass, which is suspended from a xed pivot attached to the ceiling. A ping-pong ball of mass 3.0 g is attached to the lower end of the rope. The ball moves in a circlewith constant speed in the horizontal plane and the ball goes through one revolution in 1 .0 s.How high is the ceiling in comparison to the horizontal plane in which the ball revolves?Express your answer to two signicant digits, in cm.

Clarication : During the WOOT Contest, contestants wondered what exactly a conicalpendulum looked like. Since contestants were not permitted to look up information duringthe contest, we posted this diagram. The question is to nd h.

Solution 7 (Ahaan Rungta). Let v be the constant speed at which the ball revolves inits circular motion. Let r be the radius of the circular orbit. Let T be the period of uniformcircular motion. Let be the length of the string and let be the angle the string makeswith the vertical axis. We make a free-body diagram on the ball. The two forces acting aretension up the rope, denoted by F T , and gravity downwards, mg . (We assume the ball hasmass m .) Using Newtons Second Law in the y and x directions, respectively, we have

F T cos = mg,

F T sin = mv 2

r .

Note that the RHS of the second equation is the net centripetal force on the ball. From therst equation, we have

F T = mgcos

.

Plugging this into the second equation, we have

mg tan = mv 2

r =

v2

r = g tan .

But, also, recall that

v = 2r

T = v2 =

4 2 r 2

T 2 .

Thus, we set our two expressions for v2

r equal.

4 2 rT 2

= g tan .

Now, let h be the desired height. We know that tan = rh

. We substitute:

4 2 rT 2

= gr

h = h =

gT 2

4 2.

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Using T = 1 .0s and g = 9 .8m/s 2 , we get

h = 0 .25m = 25 cm .

Problem 8 (Ahaan Rungta). A right-triangulated prism made of benzene sits on a table.The hypotenuse makes an angle of 30 with the horizontal table. An incoming ray of lighthits the hypotenuse horizontally, and leaves the prism from the vertical leg at an acute angleof with respect to the vertical leg. Find , in degrees, to the nearest integer. The indexof refraction of benzene is 1 .50.

Solution 8 (Ahaan Rungta). We use Snells Law. We care about angles with respectto axes normal to the surfaces of impact. The incoming ray hits the prime at an angle of 60 wrt the axis normal to the hypotenuse. Let be the angle at which the ray enters theprism, wrt to the same axis (normal to the hypotenuse). We can evaluate , because weknow, from Snells Law, that

1.00sin60 = 1 .50sin = sin = 3

2

1.50 = 33 .

(Recall that the index of refraction of air is 1.00.) By geometric angle-chasing, we obtainthe following result:

Let be the angle at which the ray of light leaves the prism, wrt the axis normal tothe leg of the triangle which it leaves. We nd that = 60 .

Using this result, we turn back to Snells Law:

1.50sin = 1 .00sin(90

)

cos = 32

sin

cos = 3

2 sin (60 ) .

From our rst equation of Snells Law, we had

sin = 33

,

so we substitute this into our previous equation and solve for . We get = 51 .

Problem 9 (Ahaan Rungta). A massless string is wrapped around a frictionless pulleyof mass M . The string is pulled down with a force of 50 N, so that the pulley rotates dueto the pull. Consider a point P on the rim of the pulley, which is a solid cylinder. Thepoint has a constant linear (tangential) acceleration component equal to the acceleration of gravity on Earth, which is where this experiment is being held. What is the weight of thecylindrical pulley, in Newtons?

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Note : This problem was not fully correct. Within friction, the pulley cannot rotate. Sowe responded: This is very true. To submit, Id say just submit as if it were rotatingand ignore friction. In some effects such as these, Im pretty sure it turns out that frictiondoesnt change the answer much anyway, but, yes, just submit as if it were rotating andyou are just ignoring friction. So do this problem imagining that the pulley does rotatesomehow.

Solution 9 (Ahaan Rungta). Let F be the pulling force. Let be the torque on thepulley cylinder. Let R be the radius of the cylinder. Let I be the moment of inertia for thecylinder about the axis of rotation. Let be the angular acceleration of rotation. Let abe the linear acceleration at any arbitrary point on the rim, P . Let M be the mass of thecylinder, as given in the problem. We seek the value of Mg , where g is the acceleration of gravity.

We know that the force on the rope, F , causes the torque , so

= F R.

But we also know that = I =

1

2MR 2 .

Additionally, we know that =

aR

= a = R.

We are given that a = g, so g = R , giving

2F = M g = 100 N .

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Problem 10 (Ahaan Rungta). A young baseball player thinks he has hit a home runand gets excited, but, instead, he has just hit it to an outelder who is just able to catch theball, and does so at ground level. The ball was hit at a height of 1.5 meters from the groundat an angle above the horizontal axis. The catch was taken at a horizontal distance 30meters from home plate, which was where the batter hit the ball. The ball left the bat ata speed of 21 m/s. Find all possible values of < 90, in degrees, rounded to the nearestinteger. You may use WolframAlpha, Mathematica, or a graphing aid to compute after

you derive an expression to solve for it.Solution 10 (Ahaan Rungta). Let us dene some variables. Let h0 be the height atwhich the bat hit the ball. Let v0 be the speed at which the ball left the bat. Let d be thehorizontal range to the outelder. Let t 1 be the time it takes for the ball to reach the peakof its trajectory. Let t2 be the time it takes for the ball to travel down from the peak tooutelders glove at ground level. Let g be the acceleration of gravity. First of all, observethat

d = ( v0 cos) (t 1 + t 2 ).Next, use the kinematics equation for acceleration:

gt 1 = v0 sin .

We now realize that we want an equation for t 2 . We stick with the upward path of the balland write another kinematics equation:

H = h 0 + (v0 sin )2

2g .

But H is simply 12

gt 22 . Thus, we set the two expressions for H equal:

h 0 + (v sin )2

2g =

12

gt 22 ,

which gives us t2 :

t 2 = 2h 0 + (v0 sin )

2

gg

.

Also, t 1 = v0 sin

g , so we have:

t 1 + t 2 = v0 sin

g +

2h 0 + (v0 sin )2g

g .

But, since D = ( v0 cos) (t 1 + t 2 ), we have:

t 1 + t 2 = D

v0 cos =

v0 sin g

+ 2h 0 + (v0 sin )2

gg

.

The only unknown here is , so we are ready to solve. Substituting D = 30 m, v0 = 21 m/s,g = 9 .81m/s 2 , and h0 = 1 .5 m, we then let a computational aid solve for and we get thetwo approximate solutions:

{ 18, 70 }.

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IPhOO2013-14WOOTSolutions