# Tutor profile: Annalies K.

## Questions

### Subject: Physics (Newtonian Mechanics)

Two objects are suspended from a pulley, one with mass $$m_1$$ and the other with $$m_2$$ and held up with tension forces $$T_1$$ and $$T_2$$ respectively. Determine the acceleration of object 2.

We know that $$\sum F = Ma$$ for the entire system. The outward forces acting on this system are the forces of gravity on each of the masses, $$F_1 = m_1 g$$ and $$F_2 = m_2 g$$. Let's define the positive direction to be the direction that object 2 is pulled. Because $$F_2$$ is pulling object 2 down and $$F_1$$ is resisting this pull, we define our sum to be $$\sum F = F_2 - F_1$$. Defining the total mass of our system $$M = m_1 + m_2$$ and substituting in our expressions for $$F_1$$ and $$F_2$$ gives us the summation equation $$m_2 g - m_1 g = (m_2+m_1) a$$ which we can solve for $$a$$ to find that $$a = g(\frac{m_2 - m_1}{m_2+m_1})$$

### Subject: Linear Algebra

Let $$A$$ be a 4x4 matrix with eigenvalues $$\lambda = 1, 3, -4$$ and you know that the eigenspace for $$\lambda = 1$$ is two dimensional. Is $$A$$ diagonalizable?

We know that there are two linearly independent vectors $$v_1$$ and $$v_2$$ that span the eigenspace corresponding to $$\lambda = 1$$. We also know that there is at least one eigenvector each for $$\lambda = 3$$ and $$\lambda = 4$$ -- we can call these $$v_3$$ and $$v_4$$. Because the eigenvectors for distinct eigenvalues of a given matrix are always linearly independent, we know that {$$v_1, v_2, v_3, v_4$$} is an independent set. Because $$A$$ is 4x4 each of its eigenvectors must exist in $$R^4$$. There can be no more than four linearly independent vectors in $$R^4$$, and so we know that $$v_3$$ and $$v_4$$ are in fact the only eigenvectors for $$\lambda = 3$$ and $$\lambda = 4$$. Therefore we know that the eigenspaces for $$\lambda = 3$$ and $$\lambda = 4$$ are one-dimensional. An n x n matrix $$A$$ is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to n. $$2+1+1 = 4$$, so we can conclude that yes, this matrix is diagonalizable!

### Subject: Physics

Relativistic protons travel with a speed $$v$$ in the laboratory frame are selected by measuring the time it takes a proton to travel between two detectors a distance $$L$$ apart. Each detector releases a short pulse when a proton passes through it. A coincident circuit delays the pulse from the first detector by $$L/v$$. If the pulses from the first and second detectors arrive at the same time in the laboratory frame, calculate the time difference between the arrival of the pulses in the rest frame of the proton.

In the rest frame of the proton, distances are Lorentz contracted by a factor of $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$. If the detectors are a distance $$L$$ apart in the rest frame, they are a distance $$L' = L / \gamma$$ apart according to the proton. The time spent travelling between the two detectors according to the proton is then $$L'/v = L / v \gamma = \frac{L \sqrt{1-\frac{v^2}{c^2}}}{v}$$ (Remember, velocity $$v$$ is conserved between frames!) This is the time when the second pulse will be released. However, the delay time of the first pulse is also shortened in the proton's rest frame, such that the first pulse is delayed by $$L'/v = L/ v\gamma = \frac{L \sqrt{1-\frac{v^2}{c^2}}}{v}$$. Because these two times are equal, the pulses should arrive at the same time in either rest frame!

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