Calculating Potential Energy and Velocity in Newton's Cradle Lab

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SUMMARY

The discussion centers on calculating potential energy and velocity in a Newton's Cradle lab experiment. The initial potential energy (PE) was calculated to be 1.5 x 10^-2 J using the formula PE = mgh, where the mass is 0.05 kg and the height is 0.03 m. The conservation of energy equation Eki + epi = Ekf + Epf was applied to find the velocity before impact, leading to a misunderstanding regarding the rearrangement of the equation. The correct approach resulted in a velocity of 0.77 m/s, confirming that energy input closely matches energy output, with minor losses attributed to sound and heat.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of potential energy and kinetic energy
  • Knowledge of the conservation of energy principle
  • Basic algebra for rearranging equations
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  • Study the derivation of the conservation of energy formula in physics
  • Learn about energy loss mechanisms in mechanical systems
  • Explore the dynamics of elastic and inelastic collisions
  • Investigate the effects of mass and height on potential energy calculations
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Students and educators in physics, particularly those focusing on mechanics and energy conservation principles, as well as anyone conducting experiments with Newton's Cradle.

gdhillon
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1. This lab is based on 'Newtons cradle' () I can't link to the actual lab video because that requires my password and username...not that I don't trust you guys/gals but its a public site. Anyway, he let's the first ball go at 3.0 cm above it's equilibrium position and asks for me to calculate the potential energy (the ball on the side reaches a height of 2.6cm from the initial bounce). I found that to be 1.5*10^-2J. the equation I used was ep=mgh...=.05(9.8).03. Then I am asked to Use the law of conservation of energy to calculate its velocity before impact. This where I am confused as to what numbers I should use.



I tried Eki+epi=Ekf+Epf, I plugged in the numbers;

1/2(.05)v^2+(.05*9.8*03)=0
.025v^2=-.0147
v=-2.424...I set this equation to equal zero b/c they ask me for the velocity before impact, and before there was an impact there was no energy transferred. Am I correct in this thinking? If I am not please correct me. The negative velocity I got doesn't quite make sense to me. And I am a little stuck, pleas help!


Thanks!

Gavin
 
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Your equation...

Eki+epi=Ekf+Epf

is correct but you then made a mistake rearranging it.

The initial KE at release Eki=0
The final PE at the bottom Epf=0

so you are left with

Epi=Ekf
or
mgh = 0.5mV2

The mass then cancels and you can solve for V.
 
okay thank you, I got 7.7x10^-1 J

Now for question 5) they ask for the velocity of the ball on the other side after the collision. So I would use the formula :Eki+epi=Ekf+Epf. And for the height I would have .026m. So the initial PE would this time be zero and the final KE would be zero to.
And I would have Eki=Epf. Plugging numbers in I would have 1/2v^2=9.8(.026) and I would end up with 7.1x10^-1J . Does that sound right? It sounds right to me because the energy in is very close to the energy out (I assume a little bit of energy was lost to sound and heat)

Thanks,

Gavin
 
Can anyone chime in?
 
The reasoning seems correct to me. But did they ask for v or E?

You have to know the mass of ball bearings to find E?

You gave an answer in Joules. Where are the m/s?
 
Last edited:
Should the 0.77 be in m/s ?
 
pgardn said:
Should the 0.77 be in m/s ?

You mean .71, right?
 
haruspex said:
You mean .71, right?
okay thank you, I got 7.7x10^-1 J

I was referring to his first answer.
 

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