Slider on rod strikes spring

Assumed geometry (figure not shown): the rod is a quarter circle of radius $R=0.6$ m, so the slider starts at A at the level of the circle centre and drops $h=0.6$ m to the lowest point B, where it meets the horizontal spring ($K=20$ kN/m). $m=4$ kg, rod smooth.

a) Velocity at the spring

Only gravity does work from A to B:

$$mgh=\tfrac12mv^{2}\ \Rightarrow\ v=\sqrt{2gh}=\sqrt{2(9.81)(0.6)}=3.431\ \text{m/s}$$

b) Maximum deflection of the spring

All kinetic energy at B is stored in the spring when the slider stops (horizontal compression, no further drop):

$$\tfrac12mv^{2}=\tfrac12Kx^{2}$$ $$x=v\sqrt{\frac{m}{K}}=3.431\sqrt{\frac{4}{20000}}=0.0485\ \text{m}$$

Answer: $v\approx3.43$ m/s; maximum spring deflection $x\approx48.5$ mm. (If the slider moves down an additional distance $x$ while compressing the spring, use $mg(h+x)=\tfrac12Kx^{2}$.)