IIT JEE , AIEEE Forum - Physics , Mechanics

himanshu2006

Q2:- A SHELL FLYING WITH A VELOCITY 'V=500 M/S' BURSTS INTO THREE IDENTICAL PIECES SO THAT THE K.E. OF THE SYSTEM INCREASES
BY 1.5 TIMES. WHAT MAX. VELOCITY CAN ONE OF THE FRAGMENTS
OBTAIN ?

vineet

is it 661.5m/s for 2pieces,&other-500m/s?

gorakavipraveen

vineet, how did you solve it.
Give the solution if you can here friend

krishna.gopal

Hi friends. First of all i am sorry for such a delayed reply. I and the reason for this delay was the fact that i am yet not very sure of my solution as i am unable to generalize the problem and hence get an absolute maxima of velocity. Still i am trying to answer it using some general understanding of physics.

First thing that as the system already has a momemtum along orignal direction of motion the particle which has maximum velocity afer the burst must continue to move in that direction only (Largest increase in velocity may com if the impulse on this particle is along initial direction of motion)

In general after the bursts the other two pieces may have momentums along the orignal line of motion as well as perpendiculr to it. But to maximize velocity of third partice the momentum perpendicular to motion will be zero.(Total KE is constant. If there is some momentum in perpendicular to motion, it will take some KE and hence velocity of third particle is reduced.

Third point that i want to make that the remaining two particles should move with equall velocity (that will give them minimnum KE for same momentum)

On the basis of these three logics for maximum velocity one particle should have velocity V after collision and other two should have v

Let Orignal mass of particle be 3m
Momentum balance
3m*500 = m(V+2v)
V+2V=1500
1.5*(1/2)*3m*500^2=(1/2)*m*V^2+m*v^2

Solving these two we get V=1000m/s
and v= 250 m/s

So according to me maximum velocity can be 1000m.s

cvramana

We have 3m V = mv₁ + mv₂+ mv₃ , where the velocities are in vector form.

Let one particle continue moving in the same direction as the 3m mass and let the other two travel at angles q₂ and q₃ respectively with the original direction.

Therefore 3v = v₁ + v₂ cos q₂ + v₃ cos q₃ ---(1)

In the perpendicular direction we have

v₂ sin q₂ = v₃ sin q₃ ----(4)

Also 3 m v² (3/2) = m (v²₁ + v²₂ + v²₃) ?(5)

From 1 for V₁ to be maximum q₂ and q₃ should be p each and therefore we get

3v = v₁ - v₂ - v₃ from 1.

Eliminating v₃ from the above equations we get

9v² / 2 = v₁² + v₂² + (3v - v₁ - v₂)²

make this a quadratic equation in v₂ and apply the condition for real roots we get the value of v₁ (maximum).

fundoonikhil

i've not understood y v2=v3

cvramana

It is not mentioned that v2 and v3 are equal. Check up!

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