Impulse problem about a medical biopsy needle.

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem related to impulse and the penetration depth of a biopsy needle used in medical procedures. Participants are analyzing the forces and kinematics involved in the needle's motion as it penetrates tissue.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are calculating impulse and using kinematic equations to determine the penetration depth of the needle. There are discussions about the correct interpretation of the given values, including the units of force and mass.

Discussion Status

Several participants have shared their calculations and results, but there is no consensus on the correct answer. Questions about the units of force and the assumptions made in the calculations are being raised, indicating a productive exploration of the problem.

Contextual Notes

There is confusion regarding the units of the stopping force, with participants clarifying whether it is in milliNewtons or microNewtons. The requirement for significant figures in the answer is also under discussion, alongside the implications of rounding errors in their calculations.

tbarker5
Messages
10
Reaction score
0

Homework Statement



Physicians perform needle biopsies to sample tissue from internal organs. A spring-loaded gun shoots a hollow needle into the tissue; extracting the needle brings out the tissue core. A particular device uses 8.4 needles that take 81 to stop in the tissue, which exerts a stopping force of 38 .
How far into the tissue does the needle penetrate?

Homework Equations



Just F=ma and J=Ft, the same thing really.

The Attempt at a Solution



So I calculated J by multiplying the force through the time. I got 3.1x10^-3 m/s
Then I did the following:
Δp=J
m(v2-v1)=J
v1=-J/m ; assuming v2 equals zero since the needle stops
v1=-2.604x10^-5 m/s

Then I used a kinematic to solve for the acceleration:
v2=v1+aΔt
0=v1+aΔt ; once again assuming v2 equals zero since the needle stops
a=v1/Δt
a=3.2148x10^-4 m/s^s

Using another kinematic now that i know the acceleration:
Δd=v1Δt+(1/2)a(Δt)^2
Δd=1.054x10^-6 m.

Anyways I rounded it to 1.1*10^-6 m since it wanted the answer to two significant figures. This was wrong and I really don't know where I went wrong. Any help would be greatly appreciated
 
Physics news on Phys.org
tbarker5 said:

Homework Statement



Physicians perform needle biopsies to sample tissue from internal organs. A spring-loaded gun shoots a hollow needle into the tissue; extracting the needle brings out the tissue core. A particular device uses 8.4 needles that take 81 to stop in the tissue, which exerts a stopping force of 38 .
Please specify all units and say what the numbers refer to. 8.4 needles - one was broken? 8.4km long needles? 8.4 sq mm cross-section needles?
I got 3.1x10^-3 m/s
Wrong units for an impulse. Also, please show working.
 
haruspex said:
Please specify all units and say what the numbers refer to. 8.4 needles - one was broken? 8.4km long needles? 8.4 sq mm cross-section needles?
Wrong units for an impulse. Also, please show working.

sorry. 8.4 mg needles. It took 81 ms to stop and my impulse was 3.1x10^-3 kgm/s

Here's my work through with numbers:

ΔP=J
m(v2-v1)=J
v2-v1=J/m
v1=-J/m + v2
v1=-(3.1x10^-3)/(8.4x10^-3)+0
v1=-0.369m/s

v2=v1+aΔt
a=(v2-v1)/Δt
a=(0+0.369)/(81x10^-3)
a=4.56 m/s^2

Δd=v1Δt+(0.5)aΔt^2
Δd=(-0.369)(81x10^-3)+(0.5)(4.56)(81x10^-3)^2
Δd=-1.4943x10^-2 m
Δd=-1.5x10^-2 m ; rounded to 2 sig figs.

Anyways, that still isn't the right answer
 
I used a slightly simpler approach. I would normally show all units, but you still haven't told me the units for the '38' number, so I can't do that.
accn = -38/84
time = 81
final velocity = 0
s = vt - at2/2 = 1484
So to two sig figures I confirm your result. The question then is the order of magnitude. If the 38 is in microNewtons, I get 1.5 10-3 m.
 
haruspex said:
I used a slightly simpler approach. I would normally show all units, but you still haven't told me the units for the '38' number, so I can't do that.
accn = -38/84
time = 81
final velocity = 0
s = vt - at2/2 = 1484
So to two sig figures I confirm your result. The question then is the order of magnitude. If the 38 is in microNewtons, I get 1.5 10-3 m.

So sorry, yeah the units on the 38 is mN. I'll try that. I got a 1.5x10^-2 so I know that I've got the right digits, just a problem with my powers. Thanks a lot for the confirmation/help!
woah, just tried 1.5x10^-3 and it was wrong. And it was my last guess. I have no idea what was wrong
 
tbarker5 said:
woah, just tried 1.5x10^-3 and it was wrong. And it was my last guess. I have no idea what was wrong
No, sorry - the power of 10 error was mine.
Are you sure the answer is required to 2 sig digits only? And in what units?
You said mN (milliNewtons). Did you mean microNewtons (μN)? mN gives me 15m!
If the answer is required in mm, try 15 or 14.8.
 
haruspex said:
No, sorry - the power of 10 error was mine.
Are you sure the answer is required to 2 sig digits only? And in what units?
You said mN (milliNewtons). Did you mean microNewtons (μN)? mN gives me 15m!
If the answer is required in mm, try 15 or 14.8.

It's definitely milliNewtons (mN). Anyways I had three guesses, my first was an arithmetic error. And guess two and three were 1.5x10^-2 and 1.5x10^-3 respectively. It was also definitely asking for the answer to two sig figs, and when you get close to an answer but not exactly it tells you you're close and to check rounding etc. That message never came up. Also the program I'm doing these questions on is called "masteringphysics" and it doesn't care what units you use as long as they're equivalent. Ie I could use 10cm or 0.1m and it wouldn't care

Anyways, I'm just really curious as to what the answer would be, incase a question like this came up on a test or something. Thanks for the help by the way.
 

Similar threads

Replies
7
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
3K
Replies
1
Views
3K
  • · Replies 34 ·
2
Replies
34
Views
3K
  • · Replies 8 ·
Replies
8
Views
6K
Replies
9
Views
3K