"It is
unwise to be too sure of one's own wisdom. It is healthy to be reminded that
the strongest might weaken and the
wisest might err”

– Mahatma Gandhi

Today we shall discuss three questions which appeared in JEE (Advanced)
2013 question paper. JEE
(Advanced) is the examination conducted for admitting students to under
graduate engineering programmes in IITs, IT-BHU and ISM Dhanbad (in place of the
earlier IIT-JEE).

(1) A particle of mass

*m*is projected from the ground with an initial speed*u*_{0}at an angle α with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upwards from the ground with the same initial speed*u*_{0}. The angle that the composite system makes with the horizontal immediately after the collision is
(a) π/4

(b) π/4 + α

(c) π/4 – α

(d) π/2

At the highest point (P in fig.) of the trajectory, the velocity of the
first particle is

*u*_{0 }cos α.
[The vertical component of the velocity of projection of
the first particle is zero at P. The horizontal component (

*u*_{0 }cos α) of the velocity of projection is retained through out the trajectory]
The momentum of the first particle at the highest point P is

*m**u*_{0 }cos α and is directed*horizontally rightwards*.
The velocity (

*v*) of the second particle at the point P is given by the equation of linear motion,*v*^{2}= u_{0}^{2}– 2*gH*. This gives*v =*√(u

_{0}

^{2}– 2

*gH*)

But

*H =*(u_{0}^{2}_{ }sin^{2}α)/2*g*
Therefore we have

*v =*√[u

_{0}

^{2}– 2

*g*(

*u*

_{0}

^{2}

_{ }sin

^{2}α)/2

*g*] = √[u

_{0}

^{2}(1 – sin

^{2}α) = u

_{0}cos α

Therefore, the momentum of the second particle at the point P is

*m**u*_{0 }cos α and is directed*vertically upwards*.
The momentum of the composite particle
immediately after the inelastic collision is the resultant of the momenta of
the two particles. The resultant of the

*horizontally rightward*momentum*mu*_{0 }cos α and the*vertically upward*momentum*mu*_{0 }cos α is inclined at 45º with respect to the horizontal.
Therefore, the correct option is (d).

Two more questions are discussed below. These
questions too are single correct answer type multiple choice questions and are
based on a given paragraph.

Here is the paragraph and the two questions
related to it:

A small block of mass 1 kg is released from rest
at the top of a rough track. The track is a circular arc of radius 40 m. The
block slides along the track without toppling and a frictional force acts on it
in the direction opposite to the instantaneous velocity. The work done in
overcoming friction up to the point Q, as shown in the figure below, is 150 J.
(Take the acceleration due to gravity, g = 10 ms

^{–2})**(i)**The speed of the block when it reaches the point Q is

(a) 5 ms

^{–1}
(b) 10 ms

^{–1}
(c) 10√3 ms

^{–1}
(d) 20 ms

^{–1}
The block reaches the point Q after falling
through a height

*h*equal to*R*sin30º =*R*/2. Therefore it loses gravitational potential energy*mgh*. Out of this energy 150 joule is used up against friction and the remaining portion is transferred to the block as kinetic energy ½*mv*^{2}where*v*is the speed of the block at the point Q. Therefore, we have*mgh*– 150 = ½

*mv*

^{2}

Therefore, 10×20

*– 150 =**v*^{2}/2
This gives

*v =*10 ms^{–1}**(ii)**The magnitude of the normal reaction that acts on the block at the point Q is

(a) 7.5 N

(b) 8.6 N

(c) 11.5 N

(d) 22.5 N

With reference to the adjoining figure, if the normal reaction acting on
the block at the point Q is

*N*, we have*N*–

*mg*cos 60 =

*mv*

^{2}/

*R*

[

Therefore

*mv*^{2}/*R*is the centripetal force required for the circular motion of the block]Therefore

*N =**mv*^{2}/*R + mg*cos 60 = (1×10^{2})/40 + 1×10× ½ = 7.5 newton.