Question

In the figure shown when the massless spring is in relaxed state, its free end is at point B. A very small block is pressed against the spring by a distance `delta` and then released from rest. Except the portion BC where coefficient of kinetic is `mu`, track is smooth everywhere. Determine the spring compression `delta` so that the block enters a small hole at E. Consider all value shown in the figure.If `delta` is given by `delta=x[(mgd(5+mu))/(2K)]^((1)/(2))`, find x. Block never leaves contact with surface

Answer 1

Correct Answer - 9
Mg is parallel to centre line and normal reaction passes through it so Torque about centre line is zero. So angular momentum is conserved about centre line.
`mv_(0)r_(0)=m(vcostheta)r`
`vcostheta=(v_(0)r_(0))/(r )=(v_(0)r_(0))/(sqrt(r_(0)^(2)-h^(2)))=v_(0)sqrt(2)`
Energy conservation
`(1)/(2)mv^(2)=(1)/(2)mv_(0)^(2)+mgh`
`v^(2)=v_(0)^(2)+2gh`
`=(10)^(2)+2(10)(15)=400`
`V=20`
`costheta=(10sqrt(2))/(20)=(1)/(sqrt(2))`