Thursday, August 22, 2013

JEE (Advanced) 2013 Questions Involving Two Dimensional Motion



"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 u0 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 u0. 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 u0 cos α.
[The vertical component of the velocity of projection of the first particle is zero at P. The horizontal component (u0 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 mu0 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, v2 = u02 – 2gH. This gives
            v = √(u02 – 2gH)
But H = (u02 sin2α)/2g
Therefore we have
            v = √[u02 – 2g(u02 sin2α)/2g] = √[u02(1 – sin2α) = u0 cos α
Therefore, the momentum of the second particle at the point P is mu0 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 mu0 cos α and the vertically upward momentum mu0 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 mv2 where v is the speed of the block at the point Q. Therefore, we have
            mgh – 150 = ½ mv2
Here m = 1 kg, g = 10 ms–2 and h = R/2 = 40/2 = 20 m
Therefore, 10×20 – 150 = v2/2
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 = mv2/R

[mv2/R is the centripetal force required for the circular motion of the block]
Therefore N = mv2/R + mg cos 60 = (1×102)/40 + 1×10× ½  = 7.5 newton.



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