A 32.0 kg block at rest on a horizontal frictionless table is connected to the wall via a spring with a spring constant k= 32.0 N/m. A 2.00E-2 kg bullet travelling with a speed of 5.500E+2 m/s embeds itself in the block. What is the amplitude of the resulting simple harmonic motion? Recall that the amplitude is the maximum displacement from equilibrium.
m1v1 = (m1+m2)vf
1/2 (m1 + m2) vf^2 = 1/2 kx^2
solve for x
To determine the amplitude of the resulting simple harmonic motion, we need to consider the conservation of momentum.
1. First, let's find the initial momentum of the bullet-block system.
Initial momentum = mass of the bullet * velocity of the bullet
= (2.00E-2 kg) * (5.500E+2 m/s)
2. Since the bullet embeds itself in the block, the final system momentum will be zero.
Final momentum = 0
3. The total momentum of the system is conserved. Hence, the initial momentum is equal to the final momentum.
(2.00E-2 kg) * (5.500E+2 m/s) = 0
4. Solve the equation above for the velocity of the block:
Velocity of the block = Initial momentum / Mass of the block
The mass of the block can be calculated as:
Mass of the block = Mass of the bullet + Mass of the block
= (2.00E-2 kg) + (32.0 kg)
5. Next, let's calculate the kinetic energy of the block after the bullet embeds itself. The kinetic energy is given by:
Kinetic energy = (1/2) * Mass of the block * Velocity of the block^2
6. The maximum potential energy of the block is equal to the kinetic energy of the block, as there is no energy loss in this system.
Potential energy = Kinetic energy
7. The equation for potential energy of a spring is:
Potential energy = (1/2) * spring constant * Amplitude^2
8. Set the potential energy obtained in step 6 equal to the potential energy in step 7, and solve for the amplitude:
(1/2) * spring constant * Amplitude^2 = Potential energy
9. Calculate the amplitude:
Amplitude = √(2 * Potential energy / spring constant)
By following these steps, you can calculate the amplitude of the resulting simple harmonic motion.