11he meant that the direction of change should be in the normal direction i dont know to derive it mathematically
(and in ur figure just rotate that blue line a bit to make it normal :p)
a particle is approaching with a velocity of 5i^+3j^
and it returns after colliding with a velocity of -3i^+4j^
find the coefficient of restitution
-
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37 Answers
11
62hmm... now i got it ...
ok ok got it :)
good work :)
I din realise the first step was that simple :)
gr8 man.. :)
i was trying to calculate the normal reaction in all the dirty way of the world...
33I was wondering that...
angle of v1 and v2 with N should be equal naaa..
and how v1-v2 gave Normal..
:(
:(
chemistry ate my mind.. :(
33 V1 - V2
|v1| |v2|
should give the normal naa..?
:(
:(
chemistry..
1so yes see my post !! they cancel out na......
(in fact he's a genius) evryone know so no need for this
33yeah.. rohan is a superman..
but i am unable to understand thats it :(
1oh sorry i think i misunderstood u
but well first of all its so old even i forgot ...
u talking the diff thing i see sorry for misunderstanding...
lets see again
1damn that blasted power cut
tum bhi soch rahe hoge kya paagal admi hai adhe par hi chhod gaya [1]
ok now see abhishek this is one of the few days you are doing a mistake...
1just check ur self again im sure you'll find the mistake
well wat rohan pointed out has to be correct not bcoz he is Rohan(this itself can be taken as a valid reason in many cases[6])
but because from the N L M ...
F = m( v2 - v1 )/t
isn't it so abhi[1] don't u get it even now....
1ok
direction of normal reaction=-3i^+4j^ - 5i^- 3j^=-8i^ + j^
v1=v initial
v2=vfinal
e=-v2cosβ/v1cosθ
v1.N=-40+3 = > v1cosθ* √65 = -37
v2.N=24+4 = >(-v2vosβ)*√65= 28
dividing we get e=28/37
33shit...
damn...
this is not my day...
studying chemsitry.. these days.. and it ate my mind up....
[2][2][2][2]
[3]
1lol same happened to me abhishek had to put in a few hrs of maths to get myself done with that headache it was causiing [4] lol
Subhash well think a bit [1][1]
11Newtons Laws Of Motion
is the only thing coming to my sleepy mind right now
1you see that force exerted by the wall is only along the normal because the component of the velocity of object along the wall doesnt change(as it is frictionless)
further force = dP/dt
=m(v2-v1)/dt
dt is scalar
m is scalar
so F is along v2-v1
but we know F is just the normal
9if we dont assume point of incidence as origin then
there are infinite ans acc to eq that i derived
at any plane if vector makes θ
then vcosθ remains same and vsinθ becoms evsinθ
62celestine.. we cant assume a wall at the x axis!!
i think we are talking about a smooth wall (this was a very good quesiton that you asked!)
62Its rohan's quesiton.. i think he will know the answer better than I do!
62I mean the method or what he really intends to ask.. (but as i see it. it is a frictionless wall with the direction of the wall not known!)
62there is no origin!! i dont see any!
are u talking about origin as the point of reflection?