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Find the degree of the D.E (y’)^2+5y=0. ...

Find the degree of the D.E `(y’)^2+5y=0`.
a) 1
b) 2
c) 4
d) 3

Text Solution

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The correct Answer is:
A, B, C, D
Two ball A and B each of mass m are joined rigidly to the ends of a light rod of length L. The system moves with a velocity `V_0` in a direction perpendicular to the rod. A particle P of mass m kept at rest rod. A particle OP of masss m kept at rest on teh surface sticks to the ball A as the ball collides with it.
a. The light rod wil exert a force on the bal B only along its length. So collision will not affect its velocity.
B has a velocity `=V_0`
if we consider the three bodies to be a system
Applying L.C.L.M.
Therefore `mv_0=2mxxv`
`rarr v=v_0/2`
Therefore A has velocity `=v_0/2`
b. If we consider the three bodies tobe a system
net exterN/Asl force =0`
Therefore
`V_(VCM)=(mxxv_0+2mxx(v_0/2))/(m+2m)` ltbr.gt `=(mv_0+mv_0)/(3m)`
`=(2v_0)/3`
(along the iniltial velocity as befoe collision)
c. The velocity of (A+P) w.r.t
`the center of pass ={((2v_0)/3)}-(v_0-2)`
`=v_0/6`
and the velocity of B w.r.t the centre of mass
`=v_0-(2v-0)/3=v_0/3`
(only magnitude has been taken)
Distance of the (A+P) from centre of mass
`=l/3 and for B it is ((2l)/3)`
therfore `P_(cm)=L_(cm)xxomega`
`rarr 2mxxV_0/6xxl/3+mxxV_0/3xx(2l)/3
`{2m(l/3)^2+((2l)/3)^2m}xxomega`
rarr 6m(V_0l)/18=18((6ml)/9)omega`
`rarr omega=(v_0/(2l))`
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