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` `two projectiles are fired simultaneously one is at 60 with + X-axes and other is 30 with -X-axis,(u=20*(*3)**^(1\2) * ).the minimum distance between them during their time of flight is?

-two beads of equal masses are attached by string of length 1.414a and are free to move in smooth circular ring lying in vertical plane**(positions of beads A,B are at 90 and 180)**here a is radius of ring.find the tension and accleration of **B** just after the released to move

7 years ago

dear akash

projectile question is not clear

initial distance between position of two projectile is not given

please post it again clearly

Regards

Badiuddin

7 years ago

Dear akash

Question is still not clear ,2 cases are possible.I will discuss both

for first case minimum distance is initial seperation i.e 20 m

for second case

general point on curve first is (u/2 t ,u√3/2 t -1/2gt

^{2 )}general point on second curve (20-u√3/2 t ,u/2 t-1/2gt

^{2 })now find the distance between two points ,and for minimum distance differentiate and put it equal to zero

minimum distance will come =5√2(√3-1) m

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Badiuddin

7 years ago

its the second case and answer you doesnt match (ans 10m)

see i did this prob by using relative velocity concept velocity of 1 w.r.t 2 will be horizontal (since magnitude is same for both and angle between velocity vectors is 90)then i simply calculated perpendicular distance from 2 but i got wrong answer

i dont know how draw

u will not believe me but i got this question while drawing in paint

also do the second question please??????

7 years ago

answer doesnt match

(ans 10m)well i did using relative velocity concept velocity of 1 w.r.t 2 is horizontal then i simply found perpendicular distance but i got wrong answer

idont know how to draw

u wont believe me but i got this problem now while drawing in paint i made amistake by assuming resultant velocity is horizontal but it was 45 with v2...

please solve the second one (i dont understand if resolve T(tension) along mg i.e Tcos45=mg=>T=1.414mg and when i resolve mg along T iget T=mg/1.414

(correct answer) where am i wrong?and explain what happens when beads are released

7 years ago

Force equs for 2 beads

For bead A:

T/(1.414) = mdv/dt (1)

T/(1.414) + N1 + mg = mv*v/a (2)

For bead B:

mg - T/(1.414) = mdv/dt (3)

T/(1.414) + N2 = mv*v/a (4)

Solving equs (1) & (3) we get,

T = mg/(1.414).

Now at the time of release v = o,so no centripetal acc. only tangential acc.which for bead B can be calculated from equ (3)

dv/dt = g - g/2 = g/2

7 years ago

Sir can you please tell me how do you differentiate it step by step ?

7 years ago

I'm getting this quation for some other values but isd the same type !

This is the distance which i found between the 2 points :

Root of the whole euqation √170000+v

^{2}t^{2}+ 200vt

So do i differentiate it ?

7 years ago

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