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A particle is released at point P(2 m,2 m) on a smooth parabolic track given by y=x22. It slides down the track and leaves the contact with the track at point Q(1 m,0.5 m), and then slides till point R along QR. Distance QR in m equals (g=10 m/s2)

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Solution

y=x22
(T.M.E)P=(T.M.E)Q
mg×32=12mv2
v2=3g

Also a=μg
v(dvdx)=μg03gvdv=μgx0dx
[v22]03g=μg[x]x0μx=32
x=32μ=32×12=3 m

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