The collision between a nail and hammer can be considered to be approximately elastic. Calculate the kinetic energy acquired by a 12gm nail when it is struck by a 550 gm hammer moving with an initial speed of 4.5 m/sec.
To calculate the kinetic energy acquired by the nail, we can make use of the principle of conservation of kinetic energy in an elastic collision. In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Let's denote the mass of the nail as 'm1' (12 gm = 0.012 kg) and the mass of the hammer as 'm2' (550 gm = 0.55 kg), while the initial speed of the hammer is 'v1' (4.5 m/s).
The initial kinetic energy of the system, before the collision, is given by:
KE_initial = (1/2) * (m1 * v1^2 + m2 * v1^2)
Since the nail is initially at rest, the initial kinetic energy of the nail is zero:
KE_initial_nail = 0 * v1^2 = 0
Now, let's find the initial kinetic energy of the hammer:
KE_initial_hammer = (1/2) * (m2 * v1^2)
Substituting the values:
KE_initial_hammer = (1/2) * (0.55 kg) * (4.5 m/s)^2
KE_initial_hammer = 5.83 J (rounded to two decimal places)
After the collision, the nail and the hammer move together with a common final velocity 'v_final'.
The final kinetic energy of the system, after the collision, is given by:
KE_final = (1/2) * (m1 + m2) * v_final^2
Since the nail and the hammer are moving together, their final kinetic energies are equal:
KE_final_nail = KE_final_hammer​
Substituting the values:
KE_final_nail = (1/2) * (0.012 kg + 0.55 kg) * v_final^2
We can equate the initial and final kinetic energies to find the final velocity 'v_final'. However, we need one additional equation to solve for 'v_final'.
The principle of conservation of momentum states that in an elastic collision, the initial momentum is equal to the final momentum.
The initial momentum of the system is given by:
P_initial = m1 * v1 + m2 * v1
The final momentum of the system is given by:
P_final = (m1 + m2) * v_final
Since the initial and final momenta are equal, we can set them equal to each other:
m1 * v1 + m2 * v1 = (m1 + m2) * v_final
Substituting the values:
(0.012 kg) * (4.5 m/s) + (0.55 kg) * (4.5 m/s) = (0.012 kg + 0.55 kg) * v_final
Simplifying:
0.0675 kg m/s + 2.475 kg m/s = 0.562 kg * v_final
2.5425 kg m/s = 0.562 kg * v_final
Now, we can solve for 'v_final':
v_final = 2.5425 kg m/s / 0.562 kg
v_final = 4.525 m/s (rounded to three decimal places)
Finally, we can calculate the final kinetic energy of the nail (which is equal to the final kinetic energy of the hammer):
KE_final_nail = (1/2) * (0.012 kg + 0.55 kg) * (4.525 m/s)^2
KE_final_nail = 5.92 J (rounded to two decimal places)
Therefore, the kinetic energy acquired by the 12 gm nail when struck by a 550 gm hammer moving with an initial speed of 4.5 m/sec is approximately 5.92 Joules.