2 challenge problems for tommorrow(dec 7)

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SUMMARY

The discussion presents two challenge problems: one involving the alignment of clock hands and the other concerning the trajectory of a proton in an electric field. For the clock problem, the minute hand aligns with the hour hand 11 times in a 12-hour period, with specific times calculated as 1:05:27.2727, 2:10:54.5454, and so forth. The second problem requires calculating the projection angles and total time of flight for a proton with an initial speed of V0=8.2*10^5 m/s in an electric field of E=-390j N/C, leading to the determination of forces and accelerations acting on the proton.

PREREQUISITES
  • Understanding of projectile motion and kinematics
  • Knowledge of electric fields and forces (F=qE)
  • Familiarity with trigonometric functions for angle calculations
  • Basic principles of analog clock mechanics
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  • Study the equations of motion for projectiles in electric fields
  • Learn about the forces acting on charged particles in electric fields
  • Explore trigonometric identities and their applications in physics
  • Investigate the mechanics of analog clocks and their mathematical modeling
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Students in physics, educators teaching mechanics, and anyone interested in solving mathematical problems related to motion and forces.

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Second one:
Protons are projected at an angle of 300 with an initial speed, given by V0=8.2*105 m/s, in a region of an electric field E=-390j N/C, is present. The proton is to hit a point 1.27 mm in the pos-x direction from the launch point. Find (a) the two porjections angles (theta) that will result in a hit, and (b) the total time of flight for each of these two trajectories.

Here is what I can get for this one.
(theta)=cos-1(V0x/V0)
(theta)=sin-1(V0y/V0)
F=q*E=-1.15*10-16 N
a=F/m=-6.89*1010 m/s2


First one: Solved
on an analogue clock the minute hand aligns perfectly with the hour hand once and hour. 12:00 is the first time, when are the exact other times this happens. (Hint: in only happens 11 times, not 12 and the hour and minute hand constantly move)

time=3927.272727s*n
where n is the number of times the min hand has ligned up with the hour hand
ex:
1st time...1:05:27.2727
 
Last edited:
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re: the first one

The hands will coincide every [itex]\frac {12}{11}[/itex] hours. Just convert that to hours, minutes and seconds.
 

2nd time...2:10:54.5454
3rd time...3:16:21.8181
4th time...4:21:49.0909
5th time...5:27:16.3636
6th time...6:32:43.6363
7th time...7:38:10.909
8th time...8:43:38.1818
9th time...9:49:05.4545
10th time...10:54:32.7272
11th time...12:00:00

Solving this challenge problem requires knowledge of how the hour and minute hands move on an analog clock. The hour hand completes a full rotation every 12 hours while the minute hand completes a full rotation every 60 minutes. This means that for the minute hand to align perfectly with the hour hand, the minute hand must make 11 full rotations (11 x 60 = 660 minutes) while the hour hand makes 1 full rotation (12 hours). This results in a time difference of 12 hours between each alignment. The first alignment occurs at 12:00, and the last alignment occurs at 12:00 the next day. Using this information, we can determine the other 10 times that the minute hand aligns perfectly with the hour hand. Each time, the minute hand will be at a different minute position, but the hour hand will always be at the 12. Therefore, the exact other times this happens are:

1st time: 12:00
2nd time: 1:05:27.2727
3rd time: 2:10:54.5454
4th time: 3:16:21.8181
5th time: 4:21:49.0909
6th time: 5:27:16.3636
7th time: 6:32:43.6363
8th time: 7:38:10.909
9th time: 8:43:38.1818
10th time: 9:49:05.4545
11th time: 10:54:32.7272
12th time: 12:00:00 (next day)

For the second challenge problem, we are given the initial speed and angle of a proton being projected in an electric
 

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