What is the speed of the 100g ball in Newton's Cradle after being lifted 3.0cm?

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SUMMARY

The speed of a 100 g ball in Newton's Cradle, lifted to a height of 3.0 cm, can be calculated using the principles of energy conservation. The gravitational potential energy (Eg) at the height is 0.0294 J, derived from the equation Eg = mgh, where m = 0.1 kg and h = 0.03 m. This potential energy converts to kinetic energy (Ek) at the lowest point, represented by the equation Ek = 1/2 mv². By equating Eg and Ek, the final speed of the ball can be determined accurately.

PREREQUISITES
  • Understanding of gravitational potential energy (Eg = mgh)
  • Knowledge of kinetic energy (Ek = 1/2 mv²)
  • Familiarity with the conservation of energy principle
  • Basic algebra for solving equations
NEXT STEPS
  • Calculate the final speed of the ball using the equation derived from energy conservation.
  • Explore the implications of energy conservation in different physical systems.
  • Investigate the effects of varying mass and height on the speed of the ball.
  • Learn about the dynamics of Newton's Cradle and its applications in physics.
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Homework Statement


In David and Mervin's Newton's Cradle the 100 g ball was lifted a height of 3.0 cm. With what speed would it collide with the other balls at the bottom of its motion?


Homework Equations


Eg=mgh
Ek=1/2mv^2
100g=0.1 kg
3.0cm=0.03m


3. The Attempt at a Solution [/b]
Eg=mgh
Eg=(0.1kg)(9.8m/s^2)(0.03m)
Eg=0.0294J
Ek=1/2mv^2
0.0294J=1/2(0.1kg)(v)^2

Am i on the right track? I think i am doing something wrong
 
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What do you think you're doing wrong? Or why do you think something is wrong?
 
ok so what if i use the conservation of energy equation
Et=Ek=Eg

would the final equation be
mgh = 1/2mv2^2
 

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