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).

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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.

NEET 2013 Questions (MCQ) from Work and Energy

“The world is a dangerous place, not because of those who do evil, but because of those 

who look on and do nothing.” 

– Albert Einstein 

The following questions on work and energy appeared in the National Eligibility Cum Entrance Test (NEET) 2013 which replaced AIPMT for admitting students to MBBS and BDS courses:.

(1) A uniform force of (3 i + j) newton acts on a particle of mass 2 kg. Hence the particle is displaced from position (2 i + k) metre to position (4 i + 3 j – k) metre. The work done by the force on the particle is

(1) 6 J

(2) 13 J

(3) 15 J

(4) 9 J

Work W is the scalar product (dot product) of force F and displacement s.

Or, W = F.s

Since the particle is displaced from position (2 i + k) metre to position (4 i + 3 j – k) metre, the displacement is given by

            s = (4 i + 3 j – k) – (2 i + k) = (2 i + 3 j – 2 k)

Therefore work W = F.s = (3 i + j) . (2 i + 3 j – 2 k) = 6 + 3 = 9 joule [Option (4)].

(2)The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by

(1) μ = 2/tanθ

(2) μ = 2 tanθ

(3) μ = tanθ

(4) μ = 1/tanθ

This question is popular among question setters (See AIEEE 2005 question paper).

The component of gravitational force along the incline is mg sin θ where m is the mass of the object and g is the acceleration due to gravitaty.

The work done by the gravitational force when the object moves along the incline is mgL sinθ since the force mg sinθ acting along the incline moves the object through the length L of the incline.

The work done by the gravitational force imparts kinetic energy to the object. But the entire kinetic energy is used up in doing work against the frictional force acting along the lower half of the incline. Therefore, we have

            mgL sinθ = μmg cosθ (L/2)

[Note that the frictional force is μmg cosθ since the normal force exerted (by the incline) on the object is mg cosθ and the coefficient of friction is μ]

From the above equation we obtain μ = 2 sinθ/cosθ = 2 tanθ [Option (2)].

[You may argue in an equivalent manner like this too:

The object falls through a height L sinθ, thereby losing gravitational potential energy mgL sinθ. The entire gravitational potential is used up in doing work against the frictional force acting along the lower half of the incline. Therefore, we have

            mgL sinθ = μmg cosθ (L/2) from which μ = 2 tanθ

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You can work out the above problem equally well by using the equation of linear motion,

            v² = u² + 2as

Considering the uniformly accelerated motion of the object down the upper smooth half of the inclined plane, the final velocity ‘v’ of the object is given by

            v² = 0 + 2 g sinθ (L/2) ………..(i)

Considering the uniformly retarded motion of the object down the lower rough half of the inclined plane, we have

            0 = v² + 2(g sinθ – μg cosθ) (L/2) ………(ii)

Substituting for v² from Eq. (i), we have

            2 gL sinθ = μgL cosθ, from which μ = 2 tanθ]

Questions on Atomic Physics Including JEE (Advanced) 2013 Questions

“Happiness comes when your work and words are of benefit to others.”
– Buddha

Today we shall discuss a few multiple choice questions on atomic physics. Questions in this section are simple and interesting. You can work out most of the questions in this section without consuming much time and hence you will be justified in giving some preference to them. Here are the questions with their solution:

(1) An electron and a positron moving along a straight line in opposite directions with equal speeds suffer a head-on collision and get annihilated, producing two photons. Along which directions will the photons travel?

(a) Along straight lines at right angles

(b) Along straight lines inclined at 45 º

(c) Along straight lines inclined at 120 º

(d) Along a straight line in opposite directions

(e) Along straight lines arbitrarily oriented

The net momentum of the system consisting of an electron and a positron moving with the same speed in opposite directions is equal to zero since the electron and the positron have the same mass. Therefore, the net momentum of the products produced in the collision process must be zero in accordance with the law of conservation of momentum. Cancellation of the momenta of the photons (for achieving the condition of zero net momentum) is possible only if the photons travel along a straight line in opposite directions [Option (d)].

(2) Suppose the wave length of one of the photons produced in the pair annihilation process considered in the above question is λ. The wave length of the other photon is

(a) λ

(b) 2 λ

(c) 3 λ

(d) an integral multiple of λ

(e) any thing, which can not be theoretically predicted.

Since the net momentum of the system is zero, the photons must possess equal and opposite momenta. Momentum p of a photon is related to its energy E by

            p = E/c where c is the speed of light in free space.

The photons generated must therefore be of the same energy. In other words their wave lengths must be the same [Option (a)].

The following single correct answer type multiple choice question was included in the JEE (Advanced) 2013 question paper:

(3) A pulse of light of duration 100 ns is completely absorbed by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3×10⁸ ms^–1. The final momentum of the object is

(a) 0.3×10^–17 kg ms^–1

(b) 1×10^–17 kg ms^–1

(c) 3×10^–17 kg ms^–1

(d) 9×10^–17 kg ms^–1

Since the light pulse is completely absorbed by the object, the entire momentum of the pulse is transferred to the object. If E is the energy of the pulse, the momentum p is given by

            p =  E/c where c is the speed of light in free space.

But E = Pt where P is the power and t is the duration of the pulse.

Therefore, we have

             p = Pt/c = (30×10^–3)×(100×10^–9) /( 3×10⁸) = 1×10^–17 kg ms^–1

The following single digit integer answer type question (in which the answer is an integer ranging from 0 to 9) also was included in the JEE (Advanced) 2013 question paper:

(4) The work functions of silver and sodium are 4.6 and 3.2 eV respectively. The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is:

Ans : ?

The maximum kinetic energy KE_max of the photo electron is given by

            KE_max = hν – φ where h is Planck’s constant, ν is the frequency of the incident radiation and φ is the work function of the photo emitting surface.

If the stopping potential is V, we have

            KE_max = eV where e is the electronic charge.

Therefore, eV = hν – φ from which

            V = (h/e)ν – φ/e

The above equation shows that if the stopping potebtial V is plotted against the frequency ν, a straight line graph of slope h/e is obtained.

Since h/e is a constant, the slope is the same for all photo emitters. Therefore, the  ratio of  slopes = 1.

Thus Ans. = 1 

You will find some useful multiple choice questions from atomic physics and nuclear physics here as well as here.