1yeah u r right Qwerty. thank you.
1.. Two particles are thrown from a certain height along horizontal direction mutually opposite to each other with velocities v1 and v2.find the distance b/w the particles when their velocity vector becomes mutually perpendicular.
[ans=(v1+v2)/√v1v2/g ]
2. A body of mass m is thrown up with velocity u0. find u' the velocity with which the body strike the surface if air drag @ any instant is given by cu2 , where u is the instantaneous velocity of that instant and c is a positive constant.
[u' = u0/√1+(cu02/mg)
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19 Answers
341Choosing the conventional coordinate system, we have
v_{1x}(t) = v_1, v_{1y} = -gt which gives \tan \theta_1(t) = -\frac{gt}{v_1}
Similarly, \tan \theta_2(t) = \frac{gt}{v_2}
Since the vectors are perpendicular, we get \tan \theta_1(t)\tan \theta_2(t) = -1 \Rightarrow t = \frac{\sqrt{v_1v_2}}{g}
The vertical displacement is same for the particles and hence the distance between them is the same as the horizontal distance between them which is obviously (v_1+v_2)t = (v_1+v_2)\frac{\sqrt{v_1v_2}}{g}
211st one is very easy...
so (v1 i + gt j) (v2 i + gt j) = 0 ( as 2 vectors r perpendicular)
so t =√v1 v2 /g
distance = √v1 v2 /g (v1 + v2)
23are ha , dont confuse x with horizontal coordinate, i have taken it as vertical coordinate only
23i didnt get wich is the surface?
any way
while going up , net force = - (mg+cu2) ( j )
and while going down net force = (cu2 - mg ) ( j )
now while going up ,
accn a = mg+ cu2m = - dudt= - ududx
so
mudumg+cu2 = -dx
i.e mclog(mg+cu2)= -x
where u changes from uo to u , and x changes from 0 to x
(x measured from lowest point )
so, - cx = mlog(mg+cu2mg+cuo2)
i m assuming that the surface is the one frm which the particle is thrown
while going down
accn a = mg-cu2m = +dudt= + ududx
mudumg-cu2 = dx
i.e
- mclog(mg-cu2) = x
where u changes from 0 to some u
and x changes from 0 to some x (measured from highest point )
so - mclog(mg-cu2mg) = x
so we got 2 eqns
- x = mclog(mg+cu2mg+cuo2) ........(1)
- mclog(mg-cu2mg) = x ...........(2)
in eqn 1 , put u = 0 , and x =xmax
in eqn 2 , u is the u required , and put x = xmax
and equate xmax in the 2 eqns
so mgmg+cuo2 = mg-cu2mg
m2g2 = m2g2 +cmg(uo2-u2) - c2uo2u2
so mg(uo2-u2)=cuo2u2, i.e mguo2=u2(cuo2+mg)
u2 = mguo2cuo2+mg
wats the ans given ?
6for 2nd draw the FBD
write the corresponding eqns
only 2 forces r acting
W and air drag
find the acc
and hence the vel..
1Yeah, i see. it's dimensionally incorrect. But this is what the ans is given
1for 1st the ans that u hav posted is dimensionally not correct.... plz check out