Answer fast anyone
Q.23. A bead starts sliding from a point P on a frictionless wire with initial velocity of 5 m$s^{-1}$. Find the velocity of bead at point R (take g = 10 m$s^{-2}$)

(1) 7 m/s
(2) 5$\sqrt{2}$ m/s
(3) 6$\sqrt{2}$ m
(4) 6 m/s

hi, i think it help u,

In the first instance , let’s take it that the wire is a straight line sloping downwards . The problem here is that the equation of the line is not known . If it were known then considering equiangular triangles then the heights of P and R can be calculated . Anyway

The total energy apt P equals that at R

At P , PE + KE = PE +KE at R

At P ; mgh[1] + [1/2]mv[1]^2 = mgh[2] + [1/2]mv[2]^2 at R

2gh[1] + v[1]^2 = 2mgh[2] + v[2]^2

v[2]^2 = 2g ( h[1] -h[2]) +25

v[2] = sq rt [ 20 ( h1 - h2 ) +25]

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In a second instance the wire could be circular with a radius r

Measure the angles from the vertical diameter.

Let P subtend an angle x with the centre and diameter

Let R subtend an angle y with the centre and diamerer. This should be easy to draw.

The height of P above the origin is h[1] = r cos x

The height 0f R above the origin is h[2] =r cos y

Again energy is conserved at P and R

At P ; [1/2]mv[1]^2 + mgh[1] = [1/2]mv[2]^2 + mgh[2] , cancel m and multiply by 2

v[1]^2 + 2gh[1] = v[2]^2 + 2gh[2] . Given v[1] , h[1] and h[2] above , then

v[2]^2 = 2g( r cos x - r cos y + 25

v = sq rt [ 20r( cos x - cos y) +25]

Again the radius [ equation of the circle) and the actual positions of P and R are not known to give a particular answer.

The wire could take on many other shapes ; isochrone , etc.

Hope this helps you.