# Conservation of Work and Energy

1. Nov 13, 2015

### CivilSigma

1. The problem statement, all variables and given/known data
Hello, I am having trouble understanding the logic behind the solution to the posted problem. How did they deduce the equation for "Just after impact" ? I don't see how
$$v_y = \sqrt{2g(0.6)}$$

What assumptions did they make or how did they get this simplified equation?

I understand how they got the formula for "Just before impact"

In the y direction:
$$mg(1.6)+0.5m(0)^2 = mg(0)+0.5mv_f^2$$
$$v = \sqrt{2g(1.6)}$$

But I don't see how they applied this formula for "just after impact"

Any guidance is really appreciated.

2. Nov 13, 2015

### honlin

They are applying the formula for motion under constant acceleration because there is a resultant force acting on the plate. It is actually v2=u2+2as

3. Nov 13, 2015

### CivilSigma

So what happened to u2 ? Why is it assumed to be 0 here?

4. Nov 13, 2015

### Student100

Apply the same logic for the energy of the system after collision.

5. Nov 13, 2015

### honlin

u is not assumed to be 0. Look back at the previous diagram and u will know why =)

6. Nov 13, 2015

### ehild

Use the same equation for conservation of energy mgh+0.5mvy2=constant, but with initial height 0 and velocity unknown and final velocity zero at height 0.6 m.