sign up I login
 advanced
refer a friend - earn nickels!!

Ask & Discuss Questions with Community & Experts

 Moderation Team
Ask iit jee aieee pet cbse icse state board experts Discussion Response Post to: QUE IN MECH... PLZ HELP ME OUT..RATES ASSURED..
Forum Index -> Mechanics -> View Full Question like the article? email it to a friend.  
Author Message
[Post New]26 Apr 2008 08:19:48 IST
    Subject: Re:QUE IN MECH... PLZ HELP ME OUT..RATES ASSURED..
[Up]
anchitsaini (4352)

Blazing goIITian

Olaaa!! Perrrfect answer. 796  [982 rates]
anchitsaini's Avatar
total posts: 1242    
offline Offline

\mbox{1) Let the force being applied upwards be F } \\ \\<br/>Mg - F=Ma \\ \\<br/>F=M(g - a)......1\\ \\<br/>\mbox{Taking remaining mass to be m}\\ \\<br/>F-mg = ma\\ \\<br/>M(g-a)-mg=ma....from \ 1 \\ \\<br/>or \ m = \frac{M(g-a)}{g+a} \\ \\<br/>\mbox{Hence mass removed }\\ \\<br/>=M-m\\ \\<br/>=\frac{2Ma}{g+a}


\mbox{2) }\\ \\<br/>p_1=p \ j \\ \\<br/>p_2=\sqrt{3}p[\frac{\sqrt{2}}{\sqrt{3}}\ i- \frac{1}{\sqrt{3}}j]\\ \\<br/>=\sqrt{2}p \ i - p \ j \\ \\<br/>Final \ momentum = Initial \ momentum = \\ \\<br/>p_1+p_2=\sqrt{2}p \ i


\mbox{3)}\\ \\<br/>h=ut \ - \ \frac{1}{2}gt^2 \\ \\<br/>or \ gt^2 - 2ut + 2h=0\\ \\<br/>Hence \ \\ \\<br/>t_1+t_2=Sum \ of \ roots=\frac{2u}{g}\\ \\<br/>t_1*t_2=Product \ of \ roots=\frac{2h}{g}


4)\\ \\<br/>y=x\tan \theta-\frac{gx^2}{2u^2\cos^2 \theta}\\ \\<br/>=a\tan 45 - \frac{ga^2}{2u^2\cos^2 45}\\ \\<br/>=a - \frac{ga^2}{u^2}...1 \\ \\<br/>Also \\ \\<br/>Range=a+b\\ \\<br/>=\frac{u^2\sin (2*45)}{g}=\frac{u^2}{g}\\ \\<br/>Plugging \ this \ in \ 1 \\ \\<br/>y=a - \frac{a^2}{a+b}\\ \\<br/>=\frac{ab}{a+b}


5) \\ \\<br/>At \ highest \ point \ minimum \ Tension \\ \\<br/>T_1=\frac{mu^2}{l}-mg \\ \\<br/>At \ bottommost \ point \ maximum \ Tension \\ \\<br/>T_2=mg + \frac{mv^2}{l} \\ \\<br/>Also \ v^2=u^2=2g(2l) \\ \\<br/>Hence \\ \\<br/>T_2=5mg + \frac{mu^2}{l} \\ \\<br/>Given \ \frac{T_2}{T_1}=4 \\ \\<br/>5mg + \frac{mu^2}{l}=4[\frac{mu^2}{l}-mg]\\ \\<br/>u^2=3gl = 3g\frac{10}{3} \\ \\<br/>u=10m/s
Impossible To be Impossible is Impossible
 this reply: 35 points  (with 7 Olaaa!! Perrrfect answer.   in 7 votes )   [?]
 
You have to be logged on to rate
  
 

Top Offers for goIITians
Correspondence Courses
Brilliant Tutorials
Narayana Institute
Aakash Institute
Classroom/Crash Courses
Narayana - Kota , Delhi , Others
Brilliant Tutorials - Class , Crash
Aakash Institute - Medical , Engg
Online Test Series
Brilliant Tutorials
Narayana Institute
Aakash Institute
Mahesh Tutorials
AMITY      Sri Chaitanya