Physics acceleration problem: falling to ground

[mileena]

Homework Statement

Free-Falling Objects. Motion in One Dimension:

Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally an acceleration less than 800 m/s² lasting for any length of time will not cause injury; whereas an acceleration greater than 1,000 m/s² lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.40 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 2.0 mm. If the floor is carpeted this stopping distance is increased to about 1.0 cm. Calculate the magnitude and duration of the deceleration in both cases to determine the risk of injury. Assume the child remains horizontal during the fall to the floor.

Homework Equations

The Attempt at a Solution

What does the 2.0 mm and 1.0 cm above mean? Does it mean the head bounces after hitting the floor?? I don't think we have even studied that yet. This is only the second week of the course. Thanks!

 

[voko]

These are the displacements of the head after the initial contact with the surface, during which the head decelerates and comes to a full stop.

 

[rodriguez1gv]

The two distances refer to two separate cases as well, it looks like you need to solve for each case.

 

[mileena]

Wow, I forgot that the floor actually deforms when it is hit by a head. I was so nervous doing the problem this morning that I totally forgot about that. Thank you!

I will use this formula to find the final velocity on hardwood:

v² = v₀² + 2a\Deltax

v² = 0² m/s + 2(9.820 m/s²)(0.040 m + 2.0 mm)

v² = 0² m/s + 2(9.820 m/s²)(0.4020 m)

v² = 7.89528 m/s

v = +2.809854089 m/s and v = -2.809854089 m/s

Since the child was falling downward, the velocity must be negative. Therefore, final v = -2.810 m/s, with four significant digits.

I hope this is right. This was the hardest problem on the problem set. If it is right, then I will also do the deceleration for carpet.

By the way, I hope deceleration is the same as negative acceleration.

 

[voko]

I will use this formula to find the final velocity on hardwood:

The final velocity is zero: the head is stopped by the floor. You are to find the acceleration while the head is being stopped by the floor. Re-read what the problem is about.

 

[mileena]

Thank you. But I thought when the child hits the floor, it is at a velocity greater than zero, otherwise there would be no impact to the head? I think I already know the acceleration, which is g??

I am sorry; I know this probably isn't right.

Maybe g decreases once the floor is hit and begins its deformation by 2.0 mm. I can figure out the time since I know the distance is 2.0 mm and I can probably find the two velocities at impact and at rest since I know g, the distance, and the time from above (at rest, the velocity would be 0 m/s)??

 

[voko]

Thank you. But I thought when the child hits the floor, it is at a velocity greater than zero, otherwise there would be no impact to the head?

Just when the head touches the floor, the velocity is indeed greater than zero. This is the velocity that is acquired during the free fall from the bed.

I think I already know the acceleration, which is g??

This is the acceleration during the free fall. You are asked to find the acceleration while the head is being stopped by the floor; this is obviously not a free fall.

Maybe g decreases once the floor is hit and begins its deformation by 2.0 mm.

g does not change, it simply does not apply to the situation which is not a free fall. It is correct to say that when the floor is hit, its reaction force greatly exceeds the force of gravity ( = mg), and the resultant force stops the head, decelerating it a rate that you need to find.

I can figure out the time since I know the distance is 2.0 mm and I can probably find the two velocities at impact and at rest since I know g, the distance, and the time from above (at rest, the velocity would be 0 m/s)??

Just make sure you do not confuse the free fall distance with the impact distance.

 

[mileena]

Ok, thanks for your explanation!

I know then that:

Vf = 0 m/s
Vi = -2.809854089 m/s (moment of impact)
\Deltax = 2.0 mm
t₀ = 0 s
a = \Deltav
/
\Deltat

Instantaneous acceleration at a point p also equals the slope of the tangent at the point p for constant acceleration.

Unfortunately, since I don't know t_f or v_f, I can't find the negative acceleration from the impact.

 

[voko]

What about the formula used in #4?

 

[haruspex]

a = Δv/Δt

That certainly gives average acceleration, but as you note you do not know time. What other kinematic equations do you know for constant acceleration? You need one relating two speeds, acceleration and distance.
Btw, although the problem setter probably expects you to assume constant acceleration, it won't be true in practice. Whether carpet or wood, deceleration increases steadily, being zero at initial contact. It will be a partially elastic collision.

 

[mileena]

Ok, thanks to both of you!

I will use:

v² = v₀² + 2aΔx

a = v² - v₀²
/
2Δx

a = (0 m/s)² - (2.809854089 m/s)²
/
2(2.0 mm)

a = (0 m/s)² - 7.895280001 m/s²
/
2(2.0 mm)

a = -1.97382 mm/s²

a = -.0019738 m/s²

a = -2.0 x 10^-3 m/s² (2 significant digits in answer)

This is less than the 1,000 m/s² required for injury, so I don't have to calculate if it lasted for at least 1 ms??

The answer above looks wring to me though. I expected an acceleration closer to 1,000 m/s².

 

[voko]

Ok, thanks to both of you!

I will use:

v² = v₀² + 2aΔx

a = v² - v₀²
/
2Δx

a = (0 m/s)² - (2.809854089 m/s)²
/
2(2.0 mm)

Until here, it is all correct. The you start making mistakes with units:

a = (0 m/s)² - 7.895280001 m/s²
/
2(2.0 mm)

The units for velocity squared are not m/s² but m²/s². Secondly, if the units of velocity are meters and seconds, the units for the distance must be meters, not millimeters. Correct this and you should get the correct answer.

 

[mileena]

Ok, I am really an idiot to do that mistake. The truth is this homework set is due tomorrow, and I completely freak out under pressure, and because I know there are at least a few other problems I got wrong, and I don't have the time with all of my other coursework to post them here. Ugh. I hope the professor goes over the problems in class before we turn them in, like last time. I need to learn better time management techniques. Three of my four courses have labs (calculus, physics, and human biology), which is also hard, and they are held in cities 25 minutes and 40 minutes away.

So here it is:

a = 0 m²/s² - 7.895280001 m²/s²
/
2(2.0 mm)

a = 0 m²/s² - 7.895280001 m²/s²
/
2(0.0020 m)

a = -1,973.82 m/s²

a = -2,000 m/s² (two significant digits because of 2.0 mm above)

This is greater than the 1,000 m/s² threshold for injury.

Now I wonder if I also have to figure out the time for this period of impact, since it has to be at least 1 ms??

And I have to do the problem again for carpet.

Well, anyway, so far thank you again voko!

 

[voko]

Yes, the problem wants you to find the duration of deceleration. With the magnitude of the acceleration now determined, that should be easy.

 

[mileena]

Thanks so much voko or all of your help! I disappeared yesterday because I had a three-hour class. I worked from 6 AM to 10:30 AM this morning finishing my physics homework in my car before class. I didn't dare go in, because I fear that some people might have asked me for help or the answers in the lab. I guess I am considered somewhat intelligent there, but from the looks here, I have a long way to go. If I was at a good college, I would be near the middle or bottom of my class. Maybe I will improve though. I just have to learn not to panic.

Also, the physics homework was actually not due today. I learned before class it was actually due next Wednesday. Ugh. But now I have more time to catch up on my reading in bio and nutrition.

Oh, and I caught an error above. The distance I calculated in #4 used to calculate the velocity at impact due to g in #11 was wrong! I included the 2.0 mm distortion of the hardwood floor in the distance, in addition to the 0.40 m free fall. I shouldn't have included the 2.0 mm at that point, since I was just calculating the velocity from the free fall at impact. I had forgotten to take it out in my calculations in #11. It was so small though, that it didn't change my final answer by much!