Entrance Exams Q&A

July 31, 2008

Q. 1.19 - "Problems in General Physics" by I.E. Irodov

July 30, 2008

Q. 1.17 - "Problems in General Physics" by I.E. Irodov

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July 29, 2008

Q. 1.16 - "Problems in General Physics" by I.E. Irodov

$\begin{justify} \textbf{Question}: Two particles, 1 and 2, move with constant v... ...ce between the particles become the smallest? What is it equal to? \end{justify}$

$\begin{justify} \noindent\textbf{Answer}: At any given time $t$, the $1^{st}$\... ...rom $O$ and the $2^{nd}$ particle is at distance $l_2 - v_2 t$. \end{justify}$

$\begin{justify} At this time the distance between the two particles is \begin{... ...c{(v_1 l_2 - v_2 l_1)}{\sqrt{v_1^2 + v_2^2}} \vert \end{eqnarray*}\end{justify}$

Q. 1.15 - "Problems in General Physics" by I.E. Irodov

$\begin{justify} \textbf{Question}: An elevator car whose floor-to-ceiling dista... ...the free fall in the reference frame fixed to the elevator shaft. \end{justify}$

$\begin{justify} \noindent\textbf{Answer}: When the bolt gets detached from th... ...as $t_0$. Also, let us denote the elevator's acceleration as $a$. \end{justify}$

$\begin{justify} After detaching, the bolt experiences an acceleration of magnit... ...his is in direction opposite to the acceleration of the elevator. \end{justify}$

$\begin{justify} Thus the bolt's acceleration relative to the elevator floor is ... ...ion. And the bolt's initial speed relative to elevator floor is 0. \end{justify}$

$\begin{justify} The bolt travels $l$, the floor-to-ceiling distance in the elev... ...gin{eqnarray*} d & = & a t_0 t - \frac{1}{2}gt^2 \end{eqnarray*}\end{justify}$

$\begin{justify} Now, to calculate the total distance covered by the bolt, we sh... ...ay*} The negative sign indicates that the bolt traveled downwards. \end{justify}$

$\begin{justify} Now, to get the total distance covered by bolt, we need to get ... ...} 2h + d & = & 2 \times 0.3 + 0.7 \\ & = & 1.3m \end{eqnarray*}\end{justify}$

Q. 1.14 - "Problems in General Physics" by I.E. Irodov

Question: A train of length= 350 m starts moving rectilinearly with constant
acceleration= $3.0 \times 10^{-2}$ ;= 3 s after the start the locomotive headlight
is switched on (event), and $\tau$ = 60 s after that event the tail signal light is switched
on (event). Find the distance between these event in the reference frames fixed to the train
and to the earth. How and at what constant velocityrelative to the Earth must a certain
reference framemove for the two events to occur in it at the same point.

Answer:
In the reference frame fixed to the train, obviously the distance between the two points is=
350 m.

For the reference frame fixed to earth, lets fix the initial position of headlight as the origin and
the direction of train movement as given by unit-vector $\vec{i}.$

In this reference frame, position of headlight at timeis $(\frac{1}{2}wt^2)\vec{i}.$ This is
the position of event.

After $\tau + t$ , the position of headlight is $(\frac{1}{2}w(\tau+t)^2)\vec{i}.$

Thus, the position of taillight at time $\tau + t$ is $(\frac{1}{2}w(\tau+t)^2 - l)\vec{i}.$ This is
the position of event.

Thus, distance between these two events:

$\begin{eqnarray*} & = & \frac{1}{2}(w(\tau+t)^2 - t^2) - l \\ & = & \frac{1}{2}... ...& \frac{1}{2}w(\tau+2t)\tau - l \\ & = & w\tau (t + \tau/2) - l \end{eqnarray*}$

June 20, 2008

Q. 1.13 - "Problems in General Physics" by I.E. Irodov

Question: Point A moves uniformly with velocity v so that the vector

is continually ”aimed”at point B which in its turn moves rectilinearlyand uniformly with velocity u . At the initial moment of time

⊥

and the points are separated by a distance l. How soon will the pointsconverge?

Answer: Lets consider an arbitrary time when the point A is at a distance of x from point B.