Max Angle of Colliding Masses on Strings | Elastic Collision Calculation

[joemama69]

Homework Statement

8. A mass m1 suspended on a string of length L is released from rest with, the string horizontal, as shown in the diagram. At the lowest point in its swing, m1 collides elastically with a stationary mass m2 = 2m1 suspended on a string of length L. Calculate the maximum angle that the string suspending m2 makes with the vertical following the initial collision. noite the picture

Homework Equations

The Attempt at a Solution

m₁gL = .5m₁v_1o²..v_1o = (2gL)^1/2

m₁v_1o = m₁v₁ + m₂v₂ = m₁v₁ + 2m₁v₂..v₂ = ((2gL)^1/2-v₁)/2

.5m₂v₂² = m₂gh... h = L(L-cosQ)

.5(((2gL)^1/2-v₁)/2)² = gL(L-cosQ)

Q = arccos[L - ((((2gL)^1/2-v₁)/2)²)/2gL]

is this correct so far... i will then take the derivative of Q interms of v₁ I believe... if so, is taking the derivative of an inverse function the same as the other trig functions

 

[anathema]

... also, how do I solve for the maximum angle?
Thank you for your question. Your attempt at a solution looks correct so far. To solve for the maximum angle, you will need to take the derivative of Q with respect to v1 and set it equal to 0 to find the critical point. Then, you can plug that value back into your equation for Q to find the maximum angle.

To answer your question about taking the derivative of an inverse function, yes, it is the same as taking the derivative of other trig functions. You can use the chain rule to find the derivative of arccos(x) with respect to x.

I hope this helps. Good luck with your problem solving!

 

[kerimek]

?

I would like to provide a more thorough and detailed response to this content.

Firstly, the problem presented here is a classic example of an elastic collision between two masses suspended on strings. In an elastic collision, kinetic energy is conserved, meaning that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. This can be seen in the equations used in the attempt at a solution, where the initial kinetic energy of the first mass is equal to the sum of the kinetic energy of both masses after the collision.

To solve this problem, the conservation of energy principle can be applied. This principle states that the total energy in a closed system remains constant. In this case, the system is the two masses and the strings, and the only form of energy involved is gravitational potential energy. Therefore, the sum of the initial gravitational potential energy of the masses is equal to the sum of the final gravitational potential energy after the collision.

Using this principle, the maximum angle that the string suspending m2 makes with the vertical can be calculated by equating the initial potential energy of both masses, m1gh, to the final potential energy of both masses, m1gL(1-cosQ) + m2gL(1-cosQ). This can be simplified to m1gh = 2m1gL(1-cosQ), which can then be solved for Q.

The attempt at a solution presented in the content is on the right track, but it is important to note that the derivative of the inverse cosine function is not the same as the derivative of other trigonometric functions. The correct approach would be to solve for Q algebraically, rather than taking the derivative. Additionally, it may be helpful to use the conservation of momentum principle, which states that the total momentum in a closed system remains constant, to solve for the velocities of the masses after the collision.

In conclusion, the problem presented in the content is a good example of an elastic collision and can be solved using the principles of conservation of energy and momentum. It is important to approach the problem with a clear understanding of these principles and to use them correctly in the solution.