Entrance Exams Q&A: June 2008

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.

Taking point B’s velocity component along the direction of A velocity, the two
points are coming closer at a speed given by

dx- - dt = v - ucos ϕ

Integrating this equation,

∫ 0 ∫ τ ∫ τ - dx = vdt - u cos ϕdt l 0 0

where τis the required time. This gives:

∫ τ ⇒ l = vτ - u 0 cosϕdt

Also, when the points meet, the two points must have traveled an equal
distance along the direction in which point B travels. Thus,

∫ ∫ τ vcos ϕdt = τudt 0 0

∫ τ ⇒ v cosϕdt = u τ 0

v-τ --l ⇒ v u = u τ

⇒ τ(v2 - u2) = lu

⇒ τ = ---lu---- v2 - u2

I.e. the two points will meet after a time given by v2lu-u2 .

June 14, 2008

Q. 1.12 - "Problems in General Physics" by I.E. Irodov

$\textbf{Question}$ :

Three points are located at the vertices of an equilateral triangle whose side equals . They all start moving simultaneously with velocity constant in modulus, with the first point heading continually for the second, the second for the third and the third for the first. How soon will the points converge?

$\textbf{Answer}$ :

By symmetry, the particles all converge when they reach the center of the triangle. Also, by symmetry, the particles maintain their relative configuration i.e. they always remain at the vertices of an equilateral triangle whose side keeps decreasing with time.

Consider an observer fixed to the first point. For this observer, the relative speed with which the second particle is coming towards the first particle is $v + v\cos(60)^o$ . This speed remains constant through-out the duration of traversal since the particles maintain their relative configuration as mentioned above.

Now, the initial distance between the first and second particle is . So the time taken for the first and second particle to collide is $= \frac{a}{(v+v\cos(60)^o)} = \frac{a}{\frac{3v}{2}} = \frac{2a}{3v}$ .

June 13, 2008

Q. 1.11 - "Problems in General Physics" by I.E. Irodov

$\textbf{Question}$ :

Two particles move in a uniform gravitational field with an acceleration . At the initial moment the particles were located at one point and moved with velocities = 3.0 m/s and = 4.0 m/s horizontally in opposite directions. find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

$\textbf{Answer}$ :

Since there is no horizontal force acting on both the particles, the horizontal component of their velocity remains same.

Representing the particle velocities as vectors,

Velocity vector of the first particle at time = $v_1\vec{i} + gt\vec{j}$ (assuming $\vec{j}$ as the direction in which the gravitational force is acting).

Velocity vector of the second particle at time = $-v_2\vec{i} + gt \vec{j}$ .

Angle between these two velocity vectors at time =

$\begin{displaymath}\tan^{-1}\frac{gt}{v_1} - \tan^{-1}\frac{gt}{-v_2}\end{displaymath}$

$\begin{displaymath}= \tan^{-1}\frac{gt(v_1 + v_2)}{g^2 t^2 - v_1 v_2} \end{displaymath}$

When this angle is $\pi/2$ , . Therefore, $t = \frac{1}{g}\sqrt{v_1 v_2}$ .

And at this time, the distance between the two particles is = $\frac{(v_1 + v_2)\sqrt{v_1 v_2}}{g}$

Using values, we get the distance = $\frac{7 \times \sqrt{12}}{9.8}$ = 2.47m

Q. 1.10 - "Problems in General Physics" by I.E. Irodov

$\textbf{Question}$ :

Two bodies were thrown simultaneously from the same point; one, straight up, and the other, at an angle of $\theta = 60^o$ to the horizontal. The initial velocity of each body is equal to . Neglecting the air drag, find the distance between the bodies at later.

$\textbf{Answer}$ :
We will use vector notation for this problem. Let $\vec{j}$ denote the upward direction and $\vec{i}$ denote the horizontal direction towards which the second object was thrown.

Since the first object was thrown vertically, its position vector after is = $(v_0 t - 1/2 gt^2)\vec{j}$ .

The second object was thrown with its initial velocity as $v_0\cos\theta\vec{i} + v_0\sin\theta\vec{j}$ . Since there is no horizontal force acting on the second body, its horizontal position after time is $v_0 t\cos\theta\vec{i}$ . And its vertical position at this time is $(v_0 t\sin\theta - 1/2gt^2)\vec{j}$