Newton's laws of motion problem -- Knife penetrating cardboard

AI Thread Summary
A knife edge of mass 200 g is dropped from a height of 5 m onto cardboard, penetrating 2 m into it. Initial calculations suggested an impact speed of 10 m/s, but the discussion highlighted the need to consider the continued descent of the knife due to gravity after impact. The average resistance force from the cardboard was recalculated to be 7 N after addressing the oversight regarding gravitational effects. The method of calculating average force by dividing energy by distance was debated, with emphasis on the importance of understanding momentum and energy conservation principles. The conversation concluded with a request for further clarification on calculations, indicating a collaborative learning environment.
Shivansh Mathur
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An open knife edge of mass 200 g is dropped from height 5 m on a cardboard. If the knife edge penetrates distance 2 m into the cardboard, the average resistance offered by the cardboard to the knife edge is:
 
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Where is your attempt of the question?
 
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For finding the initial speed
V(2)-u(2) = 2xgxh (g=10m/s(2)
V(2)= 2x10x5
V= 10m/s
So,speed of knife before it hits the cardboard is 10m/s. Now?
 
I know we just have to find the negative acceleration by the cardboard. If there was a mention about how much time it takes for the knife to stop, I would have done it.
 
here is the complete attempt, you may verify it
20150905_142358.jpg
 
oops!
fvf.PNG
 
Please do not post images of handwritten algebra. Images are fine for diagrams and textbook extracts, but take the trouble to type in your working.

Your calculation overlooks the fact that the blade continues to descend after hitting the top of the cardboard. Gravity doesn't suddenly switch off at that time. Also, it is unnecessary to find the impact velocity. There's no point in translating energy into velocity then back to energy again.

However, my main complaint is with the question itself. The method you use, dividing energy by distance, is clearly the one intended (since it is the only information you have) but it is not generally valid. It is OK if the force is constant during the deceleration, but it won't be. Average force is momentum divided by time, ##\frac{\Delta mv}{\Delta t}##, which need not be equal to ##\frac{\Delta E}{\Delta s}##.
See section 3 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/.
 
i'm sorry, wasn't aware about the handwritten part.

However, you are suggesting that it is unnecessary to find the impact velocity. But if i don't do so, how am i going to find the negative acceleration produced on the knife because of the cardboard? And if i don't get the acceleration, i won't get the force also. Am i right?
Thanks for the correction regarding average force. But please answer my above question.
 
Shivansh Mathur said:
However, you are suggesting that it is unnecessary to find the impact velocity. But if i don't do so, how am i going to find the negative acceleration produced on the knife because of the cardboard?
.
after finding the velocity, you used the SUVAT equation ##v^2-u^2=2as##. If we multiply that equation through by m/2 we get ##\frac 12 mv^2-\frac 12 mu^2=mas=Fs##. Clearly this is an energy equation, ##\Delta KE =## work done in accelerating the mass. Before that you used ##\Delta KE=-\Delta PE##. By eliding the KE bit, we can go straight from PE to work done against the force.
However, there is one point I made to which you did not respond. The blade continues to lose PE after it strikes the cardboard.
 
  • #10
Yes now I've understood what you were trying to say. So the only problem was that I didn't take gravity into account after hitting the cardboard. 'So gravity does not switch off :)'

Well, modifying the calculation gives me a different answer i.e 7N . Now I'm confident about it.
Thanks.
 
  • #11
Shivansh Mathur said:
Yes now I've understood what you were trying to say. So the only problem was that I didn't take gravity into account after hitting the cardboard. 'So gravity does not switch off :)'

Well, modifying the calculation gives me a different answer i.e 7N . Now I'm confident about it.
Thanks.
Yes.
 
  • #12
am new here so can you elaborate on your calculations
 
  • #13
bell Cranel said:
am new here so can you elaborate on your calculations
Please post your own attempt in a new thread.
 
  • #14
Ok
 
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