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

In summary, the conversation is about estimating the kinetic energy of a 15 g nail when it is struck by a 550 g hammer moving at 3.7 m/s in an elastic collision. The formula used to solve this problem is Kf = .5(m2) * (2m1 / m1 +m2)^2 * vo^2, where m2 is the weight of the nail in kg and m1 is the weight of the hammer in kg. The answer obtained is K = .3933 J. The concept of elastic collisions is also discussed in the conversation.
  • #1
Hughey85
14
0
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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  • #2
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.
 
  • #3
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!
 
  • #4
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.
 

1. How do you calculate the kinetic energy of a 15g nail struck by a 550g hammer?

The formula for calculating kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is the velocity. In this case, the mass of the nail is 15g and the velocity is the speed at which the hammer strikes the nail. Once you have the velocity, you can plug in the values into the formula to find the kinetic energy.

2. What is the unit of measurement for kinetic energy?

Kinetic energy is typically measured in joules (J), which is a unit of energy. In the formula, the mass is measured in kilograms (kg) and the velocity is measured in meters per second (m/s), so the unit for kinetic energy is kg * m^2/s^2.

3. How does the mass of the hammer affect the kinetic energy of the nail?

The mass of the hammer does not directly affect the kinetic energy of the nail. However, the mass of the hammer will determine the force with which it strikes the nail, which in turn will affect the velocity of the nail and thus the kinetic energy. A heavier hammer will likely result in a higher velocity and therefore a higher kinetic energy.

4. Can you calculate the kinetic energy of the nail if you only know the velocity?

Yes, you can calculate the kinetic energy of the nail if you only know the velocity. As mentioned before, the formula for kinetic energy is KE = 1/2 * m * v^2. If you know the velocity and the mass of the nail, you can plug in those values and solve for the kinetic energy.

5. Does the angle at which the hammer strikes the nail affect the kinetic energy?

Yes, the angle at which the hammer strikes the nail can affect the kinetic energy. The angle can determine the direction and force of the impact, which will affect the velocity of the nail. For example, if the hammer strikes the nail at a more direct angle, it will likely result in a higher velocity and therefore a higher kinetic energy.

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