See the attachment for the diagram. 'A' is a fixed point at a height 'H' above a smooth horizontal plane. An ideal string of length 'L' (>H) has one end attached to 'A' and the other to a particle. The particle is held horizontally (the string is taut) and released. The collision between the particle and wall is inelastic. We have to find the height when the particle is at momentary rest again.
I assumed the following unknowns :
'[itex]\theta[/itex]' = The angle shown in the diagram.
'h' = Answer
'v' = Velocity of particle wrt groung just before the collision.
g= Acceleration due to gravity at the earth's surface.
Frame of reference = ground (inertial)
v = (2gH)1/2
The velocity along the common normal just after the collision is 0 and the velocity along the common tangent is vsin[itex]\theta[/itex], just after the collision. Hence the magnitude of velocity at this point is vsin[itex]\theta[/itex].
Applying conservation of energy to particle + earth system just after collision to max. height point,
h = (v sin[itex]\theta[/itex])2/2g
[External forces do no work]
The Attempt at a Solution
The answer comes out at H3/L2, but the answer given is H5/L4.