How Does Kinetic Energy Change in a Partially Inelastic Collision?

• rolodexx
In summary, the problem involves a superball being dropped from a height of 1.5 m and bouncing up to a height of 1.0 m after colliding with a table for 15 ms. The problem asks to find the change in kinetic energy of the ball during the collision. By using the conservation of energy, it can be determined that the change in kinetic energy is equal to the difference in the potential energies before and after the collision, which is equal to the weight of the ball times the difference in the heights.
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:

Attachments

• collision.jpg
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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 $$\Delta K=K_f-K_i=mg(h_f-h_i).$$

1. What is a partial inelastic collision?

A partial inelastic collision is a type of collision where the objects involved do not stick together after impact, but some energy is lost in the form of heat or deformation. This means that the final kinetic energy of the system is less than the initial kinetic energy.

2. How is momentum conserved in a partial inelastic collision?

In a partial inelastic collision, momentum is still conserved. This means that the total momentum before the collision is equal to the total momentum after the collision. However, some of the kinetic energy is converted into other forms of energy, such as heat or sound.

3. What factors affect the amount of energy lost in a partial inelastic collision?

The amount of energy lost in a partial inelastic collision depends on several factors, including the masses and velocities of the objects involved, the type of material they are made of, and the angle of impact.

4. Can a partial inelastic collision be perfectly inelastic?

Yes, a partial inelastic collision can be perfectly inelastic. This means that the objects involved stick together after impact and move together as one mass. In this case, all of the kinetic energy is lost and the final velocity of the system is zero.

5. How is a partial inelastic collision different from an elastic collision?

In an elastic collision, both momentum and kinetic energy are conserved. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In a partial inelastic collision, only momentum is conserved and some kinetic energy is lost. Additionally, in an elastic collision, the objects involved do not deform or lose any energy due to heat or friction, while in a partial inelastic collision, some energy is lost in these forms.

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