# Tutor profile: Ashmeet S.

## Questions

### Subject: Physics (Newtonian Mechanics)

A firefighter, standing at a distance $$d$$ from a burning building wants to extinguish a fire which is burning at a height $$h$$ above the ground. She holds the hose at an angle $$\theta$$ from the horizontal and the net speed of water as it leaves the hose is $$u$$. Find a quadratic equation in $$(\tan \theta)$$ whose solutions(assuming they are real) would be the two possible angles for which the firefighter would be able to extinguish the fire?

This is a problem of projectile motion in two dimensions under the influence of gravity. Set up your coordinate axis $$x-y$$ such that the origin coincides with the firefighter's position and $$+x$$ - axis is horizontal (parallel to the ground and pointing towards the building) and $$y$$ is vertical. The initial speed is $$u$$ and it makes an angle, say $$\theta$$ with the horizontal. Now, the position of the rock can be written as a function of time, $$x(t) = (u \cos \theta) t $$ and $$y(t) = (u \sin \theta) t - \frac{1}{2} g t^2 $$ where $$g$$ is the magnitude of the acceleration due to gravity. Now we want to eliminate time from the first equation above, which can be done as follows, $$t = \frac{x(t)}{u \cos \theta} $$ For our case of interest, we want to water stream from the hose to reach the fire in the building which is at a height $$y = h$$ and a distance $$x = d$$ away from the firefighter. Thus, it takes a time $$t_{0}$$ for the water to reach the fire, $$ t_{0} = \frac{d}{u \cos \theta} $$ and we can now substitute this along with $$y = h$$ to get an equation in $$\theta$$, $$ h = tan(\theta) d - \frac{1}{2} g \left( \frac{d^2}{u^2 \cos^2 \theta} \right) $$ Now getting the quadratic equation in $$\tan \theta$$ is a matter of trigonometric manipulation. Identify $$\sec^2 \theta = (1 + \tan^{2} \theta)$$ and we obtain, $$ \frac{g d^2}{2 u^2} \tan^2 \theta - (d) \tan \theta + \left( h + \frac{g d^2}{2 u^2} \right) \: = \: 0 $$

### Subject: Physics (Electricity and Magnetism)

An alpha particle with a kinetic energy of $$1.7 \times 10^{-14}$$ J is shot directly towards a platinum nucleus from a very large distance. What will be the distance of closest approach? The electric charge of the alpha particle is $$2e$$ and that of the platinum nucleus is $$78e$$. Treat the alpha particle and the nucleus as point charges for this problem and disregard the motion of the nucleus.

This is a simple problem of energy conservation. Energy is conserved because the involved force is purely Coulomb's force (we ignore relativistic effects, magnetic fields etc. and we also treat the nucleus to be very massive so it effectively does not move) which is a conservative force. The energy at any point in the trajectory is the sum of kinetic and potential energy terms and the potential energy of the system when the alpha particle (charge $$2e$$) at a separation of $$r$$ from the nucleus(charge $$78e$$) is simply, $$V(r) = \frac{1}{4 \pi \epsilon_{0}} \frac{(78e) (e2)}{r} $$ and hence the total energy at this generic point in the trajectory is, $$E = KE + V(r) $$ where $$KE$$ is the kinetic energy at that point. Now the alpha particle begins very far away, thus the system has negligible initial potential energy to give the total final energy as $$E_{i} = KE_{i} = 1.7 \times 10^{-14} J$$. Finally, at the closest approach with distance $$d$$, the alpha particle stops momentarily, thus $$KE_{f} = 0$$ and the total energy is purely potential, $$E_{f} = V(d)$$, and thus equating the two energies, we get, $$d = \frac{1}{4 \pi \epsilon_{0}} \frac{(78e)(2e)}{KE_{i}} $$ and numerical values can be substituted to get the answer.

### Subject: Physics

Let us apply ideas in circular motion and non-inertial forces to estimate some very cool quantities in astrophysics! (a.)The Sun is about $$25,000$$ light years from the center of the galaxy and travels approximately in a circle with a period of $$1.7 \times 10^8$$ years. The Earth is $$8$$ light min. from the Sun and also goes in an approximate circular orbit. From this data alone, find the approximate gravitational mass of the galaxy in units of the Sun's mass. You may assume that all the relevant galactic mass for the gravitational force on the Sun is located at the center of the galaxy. (b.) A spherical object rotates with angular velocity $$\omega$$. If the only force preventing centrifugal disintegration is gravity, what minimum density should the object have? Use this to estimate the minimum density of the Crab Pulsar which rotates 30 times per sec.

(a.) For the motion of Earth around the Sun, $$\frac{mv^2}{r} = \frac{GmM_{\odot}}{r^2} $$ where $$m$$ is the Earth's mass and $$M_{\odot}$$ is the Sun's mass and we have taken $$r$$ to be the Earth-Sun distance and $$v$$ is the tangential Earth's speed around the Sun. For the motion of Sun around the galactic center, $$\frac{M_{\odot}V^2}{R} = \frac{GMM_{\odot}}{R^2}$$ where $$M$$ is the galactic mass and we the Sun is at a distance $$R$$ from the galaxy's center and $$V$$ is the Sun's speed. Combining the above two equations, we get, $$M = \frac{RV^2}{G} \: = \: \frac{R}{r} \left(\frac{V}{v}\right)^{2} M_{\odot} $$ Using $$V = 2 \pi R/T$$ and $$v = 2 \pi r/t$$, where $$T$$ and $$t$$ are the revolution periods of the Sun and Earth respectively, we have, $$M = \left(\frac{R}{r}\right)^{3} \left(\frac{t}{T}\right)^{2} M_{\odot} $$ And, from the given data, we obtain, $$M \: \approx \: 1.53 \times 10^{11} \: M_{\odot} $$ (b.) Consider the limiting case that the Crab Pulsar (of mass $$M$$, radius $$R$$) is just about to disintegrate. Then, the centrifugal force on a test body (mass $$m$$) at the equator of the Crab is just smaller than the gravitational force, $$\frac{mv^2}{R} \: = \: m R \omega^{2} \leq \frac{GmM}{R^2} $$ which gives us, $$\frac{M}{R^3} \geq \frac{\omega^2}{G} $$ In the above equation, $$v$$ is the tangetial speed of the test body and $$G$$ is the Universal Gravitational Constant. Hence, the minimum density of the pulsar is, $$ \rho \: = \: \frac{M}{\frac{4}{3} \pi R^3} \geq \frac{3\omega^2}{4 \pi G} \: = \: \frac{3(2 \pi \times 30)^2}{4 \pi \times 6.67 \times 10^{-11}} \approx 1.4 \times 10^{14} \: \rm{kg/m^3} $$

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