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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
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
6 years ago
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dear akash projectile question is not clear initial distance between position of two projectile is not given please post it again clearly Regards Badiuddin
dear akash
projectile question is not clear
initial distance between position of two projectile is not given
please post it again clearly
Regards
Badiuddin
intial distance is 20m
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 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards, Askiitians Experts Badiuddin
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
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation.
All the best. Regards, Askiitians Experts Badiuddin
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??????
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
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
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
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
Sir can you please tell me how do you differentiate it step by step ?
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 ?
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 ?
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