A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface, collides with a…

The surface is smooth, so mechanical energy is conserved. At maximum compression the mass momentarily stops, and all its kinetic energy is stored as elastic potential energy of the spring.
According to the question: $\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2$
$\Rightarrow x = v\sqrt{\dfrac{m}{k}}$
$x = 1.5\sqrt{\dfrac{0.5}{50}} = 1.5\sqrt{0.01} = 1.5 \times 0.1$
$x = 0.15$ m
So the maximum compression of the spring is 0.15 m.