1.

A block of mass m=2kg with a semicircular track of radius r=1.1m rests on horizontal frictionless surface

Answer»

Answer :-

MOMENTUM is conserved for the system

Pm = PM

( in opposite directions )

mV1 = (m+M) V2

k.E = 1/2 mv^2 ( M' ,= m + M )

so,

√2KmM = √2Km'M'

Km / KM' = M'/m

= KM' = m/ (m+M) ×Km ---------------(I)

Now kinetic energy will be Changed in potential energy

CHANGE in potential energy = MG (r-R) ------(II)

using equation (I) and (ii)

1/2 MV^2 = m/ ( m+m )×(mg (r-R))

V^2 = 2m^2g (r-R) / M ( m+M)

V^2 = 2×10×1.1-.1 / 2×3

=> V = √10/3 Ans.

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@GauravSaxena01


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