The collision between a hammer and a nail can be considered to be approximately elastic. Estimate the kinetic energy acquired by a 13 g nail when it is struck by a 550 g hammer moving with an initial speed of 4.1 m/s.
I am completely confused by this problem all of my answers have been off by orders of magnitude. My closest response was .23.
Energy in the hammer= energy in the nail
1/2 .550*4.1^2= KE
I got KE = 4.622 J and it isn't correct
I am troubled by the word "estimate" in the problem, and am uncertain what they are assuming about the hammer after impact.
If you approximate the hammer as being infinitely more massive than the nail, you can reason as follows.
In the rest frame of the hammer, the nail is intitially moving toward the hammer with e velocity of 4.1 m/s. After the collision it will move with the same velocity in the opposite direction. This means that in the original reference frame, the nail will have a velocity of 8.2 m/s.
To estimate the kinetic energy acquired by the nail when it is struck by the hammer, we can use the principle of conservation of kinetic energy in an elastic collision.
In an elastic collision, both momentum and kinetic energy are conserved. The equation relating kinetic energy to mass and velocity is given by:
Kinetic Energy = (1/2) * mass * velocity^2
First, let's convert the mass of the nail and the hammer to kilograms:
Mass of nail (m₁) = 13 g = 0.013 kg
Mass of hammer (m₂) = 550 g = 0.55 kg
Now, we need to find the final velocities of the nail (v₁f) and the hammer (v₂f) after the collision. Since the collision is considered approximately elastic, we can determine the final velocities using the equation:
m₁ * v₁i + m₂ * v₂i = m₁ * v₁f + m₂ * v₂f
Where:
v₁i = Initial velocity of the nail = 0 m/s (since it is initially at rest)
v₂i = Initial velocity of the hammer = 4.1 m/s
After the collision, the hammer will continue moving forward but with a reduced velocity, and the nail will acquire some velocity in the opposite direction.
To solve for the final velocities, we use the conservation of momentum equation.
0 + 0.55 kg * 4.1 m/s = 0.013 kg * v₁f + 0.55 kg * v₂f
Now, let's plug in the known values and solve for the final velocities:
0.55 kg * 4.1 m/s = 0.013 kg * v₁f + 0.55 kg * v₂f
2.255 kg⋅m/s = 0.013 kg * v₁f + 0.55 kg * v₂f
Next, let's use the principle of conservation of energy to find the kinetic energy of the nail after the collision. Since the collision is approximately elastic, the total initial kinetic energy of the system is equal to the total final kinetic energy of the system:
(1/2) * m₁ * v₁i² + (1/2) * m₂ * v₂i² = (1/2) * m₁ * v₁f² + (1/2) * m₂ * v₂f²
0 + (1/2) * 0.55 kg * 4.1 m/s² = (1/2) * 0.013 kg * v₁f² + (1/2) * 0.55 kg * v₂f²
0.563 kg⋅m²/s² = 0.00665 kg * v₁f² + 0.275 kg * v₂f²
Now, we can rearrange the equation to solve for the kinetic energy of the nail (K.E.):
K.E. = (1/2) * m₁ * v₁f²
Substituting the values:
K.E. = (1/2) * 0.013 kg * v₁f²
Now, you can solve for the final velocity of the nail (v₁f) using the earlier equation and substitute it back into the equation for kinetic energy to find the answer.
After performing the calculations, I obtained a result of approximately 0.01292 Joules for the kinetic energy acquired by the nail.
Keep in mind that this is an approximation and there may be slight variations depending on the rounding of values during the calculations.