Homework Help: Partial Inelastic Collision

1. Feb 23, 2008

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.

1. The problem statement, all variables and given/known data
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.

2. Relevant 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)

3. 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:

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• collision.jpg
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Last edited: Feb 23, 2008