Let's discuss a beautiful problem useful for Physics Olympiad based on Particle Motion.

The Problem: Particle Motion

A particle Q is moving +Y axis. Another particle P is moving in XY plane along a straight line x=-d (d>0) with a uniform speed v parallel to that of Q. At time t=0, particles P and Q happen to be along X-axis whereas a third particle R situated at x=+d starts moving opposite to P with a constant acceleration a. At all further instants, the three particles happen to be collinear. Then Q

(A) has an initial speed v/2

(B)will come to rest after a time interval v/a

(C)has an acceleration –a/2

(D) will return to its initial position after a time interval 2v/a

Discussion:

$$y_2-y_1=\frac{y_2-y_1}{x_2-x_1}(x_2-x_1)$$
$$ y-vt=\frac{\frac{-1at^2}{2}-vt}{2d}(x+d)$$
On differentiating  taking x=0
$$v'-v=\frac{-at-v}{2d}(d)$$

At t=0,
$$ v'-v=\frac{0-v}{2}....(i)$$.
$$\Rightarrow v'=\frac{v}{2}$$
For $$v'=0,$$

we have,
$$ 0-v=\frac{-at-v}{2}$$
$$ \Rightarrow t=\frac{v}{a}$$
Differentiating (i)
$$ a'-0=\frac{-a}{2}$$
$$\Rightarrow a'=\frac{-a}{2}$$

Hence, the correct options will be a,b,c,d.

Let's discuss a beautiful problem useful for Physics Olympiad based on Motion in an Electric Field.

The Problem: Motion in an Electric Field

A particle moves rectilinearly in an electric field E=E₀-ax where a is a positive constant and x is the distance from the point where the particle is initially at rest. Let the particle have a specific charge q/m.

Find:

(I) the distance covered by the particle till the moment at which it once again comes to rest, and

(II) acceleration of the particle at this moment.

Solution:

A particle moves rectilinearly in an electric field $$E=E_0-ax$$ where a is a positive constant and x is the distance from the point where the particle is intially at rest.
The particle has a specific charge q/m.
Now,

$$ F=q(E_0-ax)$$
$$or, a = \frac{q(E_o-ax)}{m}$$
At x=0,

$$a=\frac{qE_0}{m}$$
Particle will move in the x direction
$$\frac{vdv}{dx}=a$$

$$v\frac{dv}{dx}=\frac{q(E_0-ax))}{m}$$
$$vdv=\frac{q(E_0-ax)}{m}dx$$
$$\int_{0}^{0} vdv=\int_{0}^{x_0} \frac{q(E_0-ax)}{m}dx$$
$$ 0=\frac{q(E_0x-\frac{ax^2}{2})}{m}$$
Now, $$ v=0, x=x_0$$
Hence,

$$E_0x_0=a\frac{x_0^2}{2}$$
$$x_0=\frac{2E_0}{a}$$
Distance covered by the particle before coming to rest =
$$\frac{2E_0}{a}$$
Acceleration before coming to rest will be
$$ a=\frac{-qE_0}{m}$$
The direction of the particle will be towards the negative x-axis.

Some Useful Links:

• Angular Momentum – Problem
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Let's discuss a beautiful problem useful for Physics Olympiad based on Power and Acceleration.

The Problem: Power and Acceleration

Assume that a constant power P is supplied to an electric train and it is fully used in accelerating the train. Obtain relation giving the velocity of the train and distance traveled by it as functions of time.

Solution:

We know, power (P)= force(F)*velocity (v).

Therefore,

m dv/dt=P/v

or, vdv=P/m dt

Integrating both sides, we have

∫vdv= P/m ∫dt

v²/2=P/m t+c

where c is a constant of integration.

Now, at t=0, v=0 so c=0.

Therefore,

Or, v = √(2Pt/m)……… (i)

We know, v= dx/dt

So, from equation (i)

dx/dt= √(2Pt/m)

Integrating both sides,

∫dx= √(2Pt/m) dt

or, x= (2/3)√(2P/m)t^3/2