Calculating Kinetic Energy of a 15g Nail Struck by 550g Hammer

AI Thread Summary
The discussion revolves around calculating the kinetic energy of a 15 g nail struck by a 550 g hammer moving at 3.7 m/s, with the collision being approximately elastic. Participants clarify the use of the correct formula for elastic collisions, emphasizing that conservation of energy principles apply differently than inelastic collisions. One user derives the kinetic energy using the formula Kf = 0.5(m2) * (2m1 / (m1 + m2))^2 * vo^2, resulting in a value of K = 0.3933 J. The importance of distinguishing between elastic and inelastic collisions is highlighted, as it affects the application of conservation laws. The discussion concludes with a focus on ensuring the correct principles are applied for accurate calculations.
Hughey85
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Hi! Here's the question: I really need a starting block for this because I really don't know where to start.

The collision between a hammer and a nail can be considered to be approximately elastic. Estimate the kinetic energy acquired by a 15 g nail when it is struck by a 550 g hammer moving with a speed of 3.7 m/s.

Do I use this equation: (M1 * Vo) = (M1 * V1Final) + (M2 * V2Final)

My book doesn't help anymore than this! Please help!
 
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Hughey85 said:
Hi! Here's the question: I really need a starting block for this because I really don't know where to start.

The collision between a hammer and a nail can be considered to be approximately elastic. Estimate the kinetic energy acquired by a 15 g nail when it is struck by a 550 g hammer moving with a speed of 3.7 m/s.

Do I use this equation: (M1 * Vo) = (M1 * V1Final) + (M2 * V2Final)

My book doesn't help anymore than this! Please help!


What does it mean fopr a collision to be elastic? There is one more concept that you need to use.
 
Ok. I got it. I came up with this formula that substitutes the inelastic equation because like you said, there is more than once concept. Here is what I used:

Kf = .5(m2) * (2m1 / m1 +m2)^2 * vo^2

m2 is the weight of the nail in kg
m1 is the the weight of the hammer in kg

I got K= .3933 J

Thanks for the help!
 
Hughey85 said:
Ok. I got it. I came up with this formula that substitutes the inelastic equation because like you said, there is more than once concept. Here is what I used:

Kf = .5(m2) * (2m1 / m1 +m2)^2 * vo^2

m2 is the weight of the nail in kg
m1 is the the weight of the hammer in kg

I got K= .3933 J

Thanks for the help!

The problem said the collision was elastic not inelastic and you definitely can't use conservation of energy with inelastic collisions.
 
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