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 an initial speed of 5.1 m/s.
got it
v=[(2*m1)/(m1+m2)]Vo
then take your answer and insert into 1/2mv^2 and divide by 1000
Why did the nail go to the party with the hammer? Because it wanted to have a "striking" good time!
Now, let's get to the math. To calculate the kinetic energy acquired by the nail, we need to consider the principle of conservation of kinetic energy. In an elastic collision, the total initial kinetic energy is equal to the total final kinetic energy.
The initial kinetic energy of the hammer is given by:
KE1 = (1/2)mv^2
where m is the mass of the hammer (550 g = 0.55 kg) and v is its initial velocity (5.1 m/s).
KE1 = (1/2)(0.55 kg)(5.1 m/s)^2
Now let's calculate the final kinetic energy when the hammer strikes the nail:
KE2 = (1/2)mv^2
where m is the mass of the nail (15 g = 0.015 kg) and v is the final velocity acquired by the nail.
KE2 = (1/2)(0.015 kg)(???^2)
Since the collision is considered to be approximately elastic, the final velocity of the nail can be approximated to the initial velocity of the hammer:
v = 5.1 m/s
Plugging in the values, we have:
KE2 = (1/2)(0.015 kg)(5.1 m/s)^2
Now, let's solve for KE2. Give me a moment while I crunch the numbers...
*Clown Bot pulls out a calculator and starts crunching numbers with fanfare music in the background*
After calculating, we get:
KE2 ≈ 0.388 J
Therefore, the kinetic energy acquired by the 15 g nail when struck by the 550 g hammer is approximately 0.388 Joules. The nail surely had a "nailing" good time at that party!
To estimate the kinetic energy acquired by the nail when struck by the hammer, we can use the principle of conservation of kinetic energy in an elastic collision. The principle states that the total kinetic energy before the collision should be equal to the total kinetic energy after the collision.
First, let's consider the initial kinetic energy of the hammer. We can calculate this using the formula:
Kinetic Energy (K.E.) = (1/2) * mass * velocity^2
Given that the initial mass of the hammer is 550 g (which is equivalent to 0.55 kg) and the initial velocity is 5.1 m/s, we can calculate the initial kinetic energy of the hammer:
K.E. of the hammer = (1/2) * 0.55 kg * (5.1 m/s)^2
Next, we need to consider the final kinetic energy after the collision. Since the collision is approximately elastic, we assume that the total kinetic energy before and after the collision is the same.
Now, let's consider the kinetic energy acquired by the nail. Given that its mass is 15 g (which is equivalent to 0.015 kg) and we assume it starts with negligible initial velocity, we can consider its initial kinetic energy to be zero.
Thus, the total kinetic energy after the collision is equal to the kinetic energy of the nail:
K.E. of the nail = K.E. of the hammer
So, to estimate the kinetic energy acquired by the 15 g nail, we can simply calculate the initial kinetic energy of the hammer using the formula provided earlier.
You might consider the collision as nearly elastic for the brief instant before the nail moves into the board. You need to consider conservation of momentum also, to figure out what fraction of the hammer's kinetic energy is transferred to the nail.
By assuming both momentum and total kinetic energy conservation, you should obtain
V = Vo (M-m)/(M+m) = (535/565)*5.1 m/s = 4.83 m/s for the hammer velocity after impact.
For the nail velocity v,
v = (M/m)(Vo -V) = 36.67*.27 m/s
= 9.9 m/s
Try the calculations yourself and see if you agree.