Physics Guru

At t = 0, four particles A, B, C and D are situated at the vertices of a square ABCD of side d. Each particle moves with constant speed v such that A always has its velocity along AB, B along BC, C along CD and D along DA. At what time will the particles meet? Solution Let us consider particles A and B. The relative velocity of B w.r.t. A can be obtained from the right triangle formed by the two velocity vectors: $v_{rel}=\sqrt 2 .v$ The component of $v_{rel}$ along BA = $v_{rel} cos45^\circ $ = v This component will decrease the distance d. So, $t=\frac {d}{v}$

A small drone launched at an angle of $45^\circ$ with horizontal moves with constant velocity of $\sqrt {200}$ m/s. Its power shuts down just 4 second after the launch. Find the horizontal range of the drone. [Ignore air, $g=10 m/s^2$] Solution Motion with constant velocity $d = ucos\theta.t = 40m$ $h = usin\theta.t = 40m$ Projectile motion $y = -40 = usin\theta .t - \frac {1}{2} g t^2 = 10.t-\frac {1}{2}.10.t^2=10t-5t^2$ $\therefore 5t^2-10t-40=0$ $\Rightarrow t^2-2t-8=0$ $\Rightarrow (t-4)(t+2) = 0$, t = 4 sec $R=ucos\theta .t=10.4 = 40 m$ Range = d+R = 40+40 m = 80 m

A charge q is placed at a distance of $\frac {a}{\sqrt {24}}$ above the centre of a horizontal, equilateral triangular surface of edge a. Find the flux of the electric field through the equilateral triangular surface. Solution Let us evaluate if we can have regular tetrahedron as the closed Gaussian surface. The altitude of tetrahedron having side a is given by $H = a \sqrt {\frac {2}{3}}$ Distance of centroid from any face = $\frac {H}{4} = \frac {a}{\sqrt {24}}$ So, we can imagine a regular tetrahedron as the closed Gaussian surface with q placed at the centroid. Total flux $= \frac {q}{\epsilon_0} = 4 \times $ flux though one surface So, flux through one triangular surface $= \frac {q}{4\epsilon_0}$

Find a if ${S_n} = {1^4} + {2^4} + .......... + {n^4} = a{n^5} + b{n^4} + c{n^3} + d{n^2} + en + f$, where n is a natural number and a, b, c, d, e, f are constants. Solution We have, ${S_{n - 1}} = {1^4} + {2^4} + .......... + {(n - 1)^4}$ $\therefore {S_n} - {S_{n - 1}} = {n^4}$ $\therefore a\{ {n^5} - {(n - 1)^5}\}  + b\{ {n^4} - {(n - 1)^4}\}  + c\{ {n^3} - {(n - 1)^3}\}  + ..... = {n^4}$ In LHS, $n^4$ can only come from $a\{ {n^5} - {(n - 1)^5}\} $ as other terms have lower power on n. Using binomial expansion of ${(n - 1)^5}$ and equating the coefficient of $n^4$, $a.5 = 1$ $ \Rightarrow a = \frac{1}{5}$

There are two identical right-triangular ramps. Their vertical faces are parallel and facing each other, separated by a horizontal distance of 5 m, and their upper ends are at the same height. The inclined surface of each ramp makes an angle of 15° with the horizontal. A small toy car approaches the first ramp along a horizontal road and moves onto the inclined surface with constant speed. What is the minimum speed with which the car should enter the ramp so that it clears the gap and lands safely on the inclined surface of the second ramp? Neglect air resistance. Take $𝑔=10 m/s^2$. Solution The car leaves the first ramp with speed u at $15^\circ$. For the minimum speed, its projectile range must be 5 m. $R=\frac {u^2sin 2\theta}{g}$ $\therefore 5=\frac {u^2sin30}{10}$ $\Rightarrow u = 10m/s$

A negative point charge -q is brough near an isolated metal cube. (A) The cube becomes positively charged (B) The cube becomes negatively charged (C) The interior of the cube remains charge free and the surface gets nonuniform charge distribution (D) Protons inside the metal cube move towards -q making the interior of the cube negatively charged and the surface close to -q positively charged Answer The charge does not stay inside the metal cube. The charge has to reside on the surface. Moreover, the free electrons move farther away from -q so there is nonuniform surface charge distribution. Hence, (C)

At t = 0, three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side 1.5d. Each particle moves with constant speed v such that A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet? Solution Velocity of A along AB = $v$ Velocity of B along BA = $v cos 60^\circ = 0.5v$ Relative velocity of B towards A = $v+0.5v=1.5v$ Displacement covered by B to reach A with respect to A = 1.5d time = $\frac {d_{rel}}{v_{rel}}=\frac {1.5d}{1.5v}=\frac {d}{v}$