1i agree 2 :)
17 Answers
62initial momentum is mv0 in the vertical direction..
When colission occurs.. the velocity in the y direction is equal for all three balls.
SO mv0=3mvy
Thus vy=v0/3
Now can we conserve energy to find the velocity of the balls along x axis?
by:
1/2 m v02 = 1/2 mvy2 x 3 + 1/2 m vx2 x 2
We have vy in terms of v0
So 1/2 m v02 = 1/2 mv02/9 x 3 + 1/2 m vx2 x 2
So v02 = v02/3 + vx2 x 2
So 2/3 v02 = vx2 x 2
So vx = v0/√3
So net speed will be given by √( v0/√3)2+( v0/3)2= 2/3 v0
1yes nishant sir is right .....
after finding the vertical comp velocity as v0 /3 we can apply conservation of energy to get the horizontal comp of velocity anf finally find the resultant of the two...
getting 2v0/3
1three identical balls are connected by light inextensible with each other as shown.. and rest over a smooth horizontal table .at the moment t=0,ball B is imparted a velocity v0 .calculate the velocity of A when it collides with C..
i have written the exact language dude..
1the qn is very confusin believe me i have taken 2 hours no answer i got ..plz write the full qn
1
this velocity 'u'.. is variable..it depends upon the angle θ..for finding the acceleration ..multiply d(u)/d(θ) with d(θ)/d(t)..
so acc. will be v0cos2θ * d(θ)/d(t)..the body at the middle has travelled 'p' distance in time 't'.. keeping that in mind ..we can find d(θ)/d(t)..bcoz..θ≈p/L..so.. and we know d(p)/d(t)=v0
/..i am still not sure try this way.. lez c .aaj hi khatam karenge isko :)
1the above approach of momentum..is incorrect because sum of final kinetic energies will rely upon the x component of velo. of 1st and 3rd mass..and this is going to be dependend upon the angle
θ a variable...so the velocities will keep adjustin so that the final k.e.= initial k.e...and v0/3 is not possible it will readjust as the balls 1 and 3 approach towards e/odr..this is obvious if u think what will happen after collision ..!!..although if there is a proof that θ will not vary then answer is v0/√3.. by energy conservation taking y velocties as v0/3 for all ....................and v0/√3 for body 1 and 3 toward e/odr ..HOWEVER IN THIS CASE COLLISION CANNOT TAKE PLACE
1At the time of collision all the balls align themselves in the Y-direction.
So they have a common velocity vy at the time of collison.
But at this instant velocities in the X-direction is nil !!!because at the instant the balls colllide and the leading ball also moves in Y-direction.
So i don't think there should be any velocity in the X-direction DURING collision. It's true that there will be a velocity of approach in X-direction of ball A towards B....but during collision it must be 0.
so ans becomes v0 / 3.
@ Nishant Sir ... i don't get this :
"Now can we conserve energy to find the velocity of the balls along x axis?"
62initial momentum is mv0 in the vertical direction..
When colission occurs.. the velocity in the y direction is equal for all three balls.
SO mv0=3mvy
Thus vy=v0/3
Now can we conserve energy to find the velocity of the balls along x axis?
1it is kept on the table.....and the middle one is imparted a velocity v0 and then the question goes as mentioned
1terror please post the question with original statements as given in book...........tumhara yeh question mujhe samajh mein hi a nahi raha hai.
1is it vertical or horizontal ? question is not clear.
i mean is it moving on a table ? or moving vertically up in the sky ?