Homework Help: 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.