Physics Contest

Hello everyone!

The purpose of this thread is to both announce and gouge interest for a physics contest on these forums. The contest will consist of 5 approximately High School or Introductory College Level physics questions and will be open to high school students (including those who have graduated this year and are about to enter the workforce, military, college, etc.) only (though anyone may participate! They will simply not be eligible to win the contest.)

Scoring for the competition will work as follows: There will be 2 points for each of these questions. 1 point for having a correct answer and 1 point for showing work and explaining the answer. The student with the most points at the end will receive a prize.

Sure.

This sounds fun. I’m in!

Count me in!

Oh yeah, let’s go!

Problem 1 has been posted!
Answers are due at Midnight(Eastern time) THIS Sunday.

Problem 2 has been posted!

After the deadline for each problem has passed, the correct answer should be posted, in order to help those that may struggle with the problems.

I am planning on doing that!

I am waiting to hear back from 1 participant who was confused by the wording of problem 1 (there was an error in the original wording of the problem). Because of that, the solution for problem 1 will be posted alongside that of problem 2 on Wednesday.

Because I’ve caught a terrible cold, I will post the answers for 1 and 2, as well as problem 3, sometime between Thursday and Saturday.

Problem 3 is up: https://www.chiefdelphi.com/forums/showthread.php?p=1689750#post1689750

Problem 1 Explanation:

It is important to realize that gravity is accelerating down on the sheet, and that a mass times an acceleration results in a force. A torque is the result of applying such a force at a distance, r, from a pivot point.

The bottom right corner of the sheet is serving as its pivot point. Because of this, any mass to the right of the pivot point will result in a torque which attempts to cause a clockwise rotation while any mass to the left will result in a torque which attempts to cause a counterclockwise rotation. When the center of mass (which is the center of the sheet due to the even mass distribution) is above the pivot point, the sheet is in equilibrium and it will not rotate in either direction. The moment that the center of mass is to the right of the pivot, the sheet will rotate clockwise and fall on its face. If a diagram is drawn which compares the original position of the center of mass to the position in which it is over the pivot point, it becomes clear that a triangle with 2 bases l/2 and w/2 is formed between the 2 positions. An inverse tan function may then be used to solve for the angle that demonstrates the displacement between the 2 positions of the center of mass. This angle is complimentary to the outer angle of sheet, which is what we are attempting to solve for. Therefore, the value of the angle is equal to 90 - atan(w/l).

There is a post on stack exchange which demonstrates this using a diagram: https://physics.stackexchange.com/questions/258/what-determines-the-minimum-angle-at-which-a-domino-falls-over

Please PM me if additional clarification is needed, I am more than willing to explain this problem individually :slight_smile:

Problem 2 Explanation:
The important thing to realize with this problem is that positive and negative angular momentums have be defined by the wording of the problem. Because of this, the collision with the clay ball is noted to cause a positive momentum (due to rotating counterclockwise) while the rubber one is noted to cause a negative momentum(due to rotating clockwise). Therefore, the correct answer is that the collision with the clay ball results in a greater positive momentum despite the fact that it is true that the magnitude of the momentum following the collision with the rubber ball is greater.

Please PM me if additional clarification is needed, I am more than willing to explain this problem individually

Problem 4 is up!

**Problem 3 explanation(People seem to have struggled with this one!): **

The key to solving this problem correctly is setting up your free body diagrams correctly, and ensuring that the total forces = MA.

Part A: I think it is fairly intuitive that the objects will be in balance only if they have the same mass. The answer is 5kg.

Part B: Recognize that for M1, T - MG = MA.
Therefore T - 49.05 = 33.5.
T = 82.55 newtons.

Part C: There is more than 1 way to solve this problem, but I feel that this method illustrates the concepts behind the equations best
Again, recognize that F=MA on both sides.
Given that we know acceleration and G, let’s work with M1 since the mass is
known.
T -49.05(M1G) = 40(M1 * Acceleration).
T= 89.05

       Since the string is massless, the tension is uniform throughout.
       Therefore: T - M2G = M2(-8)
       M2 = 49.2kg

       Recall that the system is said to be at rest initially, meaning that both masses were originally equal (5kg)

       Added mass = M2 - 5.
       Added Mass = 44.2Kg

Part D: There are multiple answers(such as transporting to a planet with a gravitational acceleration greater than 9.81) to this portion, the key here was imagination.