Can Work-Energy Formula Be Used to Solve Ball-Spring Collision?

[walking]

Homework Statement

A ball collides with a spring and they stick together at maximum compression of the spring. Find the final speed of the spring.

Relevant Equations

Conservation of momentum

I completely missed the collisions approach when I first tried to solve this and tried using the work-energy formula. I am wondering if this approach can be made to work? Here is my attempt:

So I let the work done on the ball be $W_b$ and work done on spring be $W_s$. Then $$W_b=\Delta K = \frac{1}{2}m_b v^2-\frac{1}{2}m_bv_i^2$$ and $$W_s=\Delta K=\frac{1}{2}m_sv^2-0=\frac{1}{2}m_sv^2$$, where the final velocities are equal due to sticking together. Now I said that since the only force acting on spring is from the ball and the only force acting on the ball is the spring, then intuitively $W_b=-Ws$. So $$\frac{1}{2}m_b(v_i^2-v^2)=\frac{1}{2}m_sv^2$$ or $$(m_s+m_b)v^2=m_bv_i^2$$ or $$v=v_i\sqrt{\frac{m_b}{m_s+m_b}}$$. But this is wrong unfortunately (answer is $\frac{v_im_b}{m_s+m_b}$ which I know how to obtain using the collisions approach).

 

[Doc Al]

While you think you found the work done on the spring, what you actually expressed was the change in KE of the gun. And you set the decrease in KE of the bullet equal to the increase in KE of the gun. Ah, but some of that energy gets stored in the spring -- you left that out.

(You do need to use conservation of momentum as well.)