How Does Kinetic Energy Change in a Partially Inelastic Collision?

[rolodexx]

[SOLVED] Partial Inelastic Collision

Is it ok if I keep this here for my own reference (and others if it helps them)? I don't know why it helps to type everything out before I realize what I did wrong! But looking back on my work seems to reinforce what I learned.

Homework Statement

As shown in the figure (attached), a superball with mass m equal to 50 grams is dropped from a height of $$h_i$$ = 1.5 m. It collides with a table, then bounces up to a height of $$h_f$$ = 1.0 m. The duration of the collision (the time during which the superball is in contact with the table) is $$t_c$$ = 15 ms. In this problem, take the positive y direction to be upward, and use g = 9.8$$m/s^2$$ for the magnitude of the acceleration due to gravity. Neglect air resistance.

The problem has 5 parts; with some logic and work I solved the fist 4, but the last one is really getting to me. It's weird, because it should be the easiest:
Find $$K_a - K_b$$, the change in the kinetic energy of the ball during the collision, in joules.

Homework Equations

1 kg/1000 g

.5m$$v^2_a$$ - .5m$$v^2_b$$

.5m$$v^2$$ = mgh (This was used for both v before and after)

The Attempt at a Solution

Within the part of the problem I correctly solved were questions asking for the momentum immediately before and after the collision. Since p is solved for by multiplying mass and velocity, I used conservation of energy to relate gravitational potential energy (at $$h_i$$) to kinetic energy after the collision. I got a velocity $$v_b$$ and multiplied it by the mass (.05 kg) to get a correct momentum for that part of the problem. The same logic was followed for the "after" momentum. The velocities I got from these two correct answers are what I used in this last question. ($$v_b$$=5.4 m/s; $$v_a$$=4.43 m/s)

I solved for the kinetic energies by multiplying .5 * .05kg * 19.6$$m^2/s^2$$ and .5 *.05kg * 29.4$$m^2/s^2$$. I then subtracted the "before" K (.729 J) by the "after" K (.491 J) to get .238 J.

But... it said "The ball bounces up to a lower height than that from which it was dropped. This implies that the kinetic energy after the collision was less than the kinetic energy before the collision." So I thought I had mixed up the two terms, and threw a negative sign in front of my answer, but that warranted a "Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures."

*edit* - It really was the wrong number of significant figure-rounding when calculating... that one is going to eat me forever if I don't stop it:shy:

 

[kuruman]

You can answer the last part quite easily by considering that
1. Just before the collision, the kinetic energy is $K_i=mgh_i$ because all the potential energy is converted to kinetic.
2. After the collision when the ball reaches maximum height all its kinetic energy immediately after the collision is converted to potential. Therefore, $mgh_f=K_f$.
Therefore, the change in kinetic energy is $$\Delta K=K_f-K_i=mg(h_f-h_i).$$