Hard Mechanics Discussion Thread

Inspired by the hard integral thread !!!
Topics of discussion : Hard questions from the A level Physics/ Mathematics (Mechanics) and the first-year undergraduate classical mechanics topics including Special Relativity.


The second Question:

The two components of a binary star are observed to move in circles of radii r₁ and r₂. What is the ratio of their masses?
(Note : Binary stars orbits around the combined centre of mass.)

If you have some interesting questions please post them.

A balloon contains a bag of sand of initial mass $m_0$, and the combined mass of balloon+sand is $M_0$. The balloon experiences a constant upthrust R and is initially at rest and in equilibrium, with the upthrust compensating exactly for the gravitational force. At time t = 0 the sand is released at a constant rate dm/dt. It is fully disposed of after a time T.

(i) Determine the velocity of the balloon v(t) at any time between t = 0 and T. (Hint: Derive your expression in terms of $k = m_0/M_0T$)
(ii) Find the height gained by the balloon as a function of time.
(iii) For small values of  $\epsilon= m_0/M_0$ express your results for the velocity and height gain at time T to first order in $\epsilon$ .

Really tough question. My previous solution for (i) was: but now I know that I missed out something. I used F = M(x) dv/dt which is wrong. Need to do it again writing F as F = M(x) dv/dt + dM/dt times v and solve the differential equation again. I am not sure I will be able to solve the differential equation.

Really tough question. My previous solution for (i) was: but now I know that I missed out something. I used F = M(x) dv/dt which is wrong. Need to do it again writing F as F = M(x) dv/dt + dM/dt times v and solve the differential equation again. I am not sure I will be able to solve the differential equation.

Yes you're right there, it's an easy mistake to make as normally the mass isn't changing in mechanics questions. The integration is not too bad actually, I have faith in you

One end of a thin inexstendable, but perfectly flexible, length of string 'L' with uniform mass per unit length is held at a point on a smooth table a distance 'd' (< L) away from a small vertical hole in the surface of the table. The string passes through the hole so that a length L - d of the string hangs vertically. The string is released from rest. Assuming the height of the table is greater than L, find the time taken for the end of the string to reach the top of the hole.

tangotangopapa2 this is a bit easy for you

As I said...a bit easy
Let me find some harder ones for you hmm

Wait a moment, If m_0 is very small compared to M_0. and dm/dt is fairly small, M(x)dv/dt could be reasonable approximation to F. In that case, rewriting above equation using R = M_0 g, we get: and then after integrating this (had to use integration of parts to solve integration of ln(1-kt) by writing it as 1 times ln (1-kt)) we can arrive at the following equation:

I am sorry but I just looked at the solution in the following sheet, last question: http://www2.ph.ed.ac.uk/~egardi/MfP3-Dynamics/Solutions_Workshop3.pdf

Well done, that question is damned brutal (even when you cheat ). I have a similar and possibly harder one which has no answer online if you'd like to try it?

Sure!!!

Actually I just did it and it's not super hard but here

One end of a thin inexstendable, but perfectly flexible, length of string 'L' with uniform mass per unit length is held at a point on a smooth table a distance 'd' (< L) away from a small vertical hole in the surface of the table. The string passes through the hole so that a length L - d of the string hangs vertically. The string is released from rest. Assuming the height of the table is greater than L, find the time taken for the end of the string to reach the top of the hole.

tangotangopapa2 this is a bit easy for you

Hint pls

Are my answers correct?



a) $v = v_o e^{- \frac{bt}{M_o + m_o}}$ Where (v)s are vectors, don't know how to write vectors in LaTex.

b) Momentum is only conserved if there is no external force, as there is external drag momentum is not conserved. The change in energy accounts for the work done against the resistive force.

c) (By conservation of momentum) Mv + delta mu = 0 Differentiating with respect to time and writing, dM/dt = 0 and du/dt = 0, we get the given equation. (I have taken reference frame to be the frame stationary with respect to rocket just before the propulsion.) (This is the famous rocket equation where F = 0.)

d) Integrating the equation given in c) with limits M= M_o + m_o to M = M_o + m_o - m_1 we get the equation in v.

e) As, v_o and u are parallel, we could just write their magnitudes in the equation. Now we simply have to plug in values to obtain:
$m_1 = (M_o + m_o)(1-e^{-\frac{v_o}{10u}})$