Can you integrate?

Clearly, what we need to find is the final momentum of the two block system as only normal force (by vertical wall) is the cause for change in momentum for the two block system. The momentum of the system will not change after A looses contact with the wall. Hence to find the final momentum of the system let us consider the moment just after A loses contact with the ball At that instant there will be almost negligible elongation on the spring (think why?). speed of A would be zero. Let the speed of B be v. and hence from energy conservation, $\frac{1}{2} mv^2 = \frac{1}{2} kx^2$ Hence final momentum = mv

Hence the value of the integral is $\sqrt{km}x$ .

This is how I did,

The velocity of the COM when the Normal reduces to 0 is $\dfrac{v}{2}$

So total change in momentum = $\int_{0}^{\infty} N.dt = 2m \times (v-\dfrac{v}{2}) = mv = 3.5$ .

Here v can be easily found out using $\dfrac{1}{2} kx^2 = \dfrac{1}{2} mv^2$

$\implies v=x\sqrt{k}{m}$

$\implies v=5x$

So we put value and get $\int_{0}^{\infty} N.dt = 3.5 kg ms^{-1}$

All we have to do is find the change in momentum