Conservation of Work and Energy

[CivilSigma]

Homework Statement

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.

Thank you for your time.

 

[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

 

[CivilSigma]

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

 

[Student100]

Homework Statement

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
$$v_y = \sqrt{2g(0.6)}$$I understand how they got the formula for "Just before impact"

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

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

 

[honlin]

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

 

[ehild]

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"

Use the same equation for conservation of energy mgh+0.5mv_y²=constant, but with initial height 0 and velocity unknown and final velocity zero at height 0.6 m.