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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

Grade:Upto college level

## 8 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 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

akash kashyp
8 Points
11 years ago
intial distance is 20m
Badiuddin askIITians.ismu Expert
147 Points
11 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 -1/2gt2   )

general point on second curve (20-u√3/2 t ,u/2 t-1/2gt2   )

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

akash kashyp
8 Points
11 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??????
akash kashyp
8 Points
11 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

RAJEEV CHOUDHARY
9 Points
11 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

541 6339
18 Points
10 years ago

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

541 6339
18 Points
10 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+v2t2 + 200vt

So do i differentiate it ?

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