91Ó°ÊÓ

The total energy of simple harmonic motion can be given by

$E=\frac{1}{2}k{A}^2$ $\left(1\right)$

Kinetic energy

$K=\frac{1}{2}m{v}^2$ $\left(2\right)$

Conservation of momentum

${\stackrel{→}{\mathbf{p}}}_i={\stackrel{→}{\mathbf{p}}}_f$

$m_1{\stackrel{→}{\mathbf{v}}}_{1i}+m_2{\stackrel{→}{\mathbf{v}}}_{2i}=m_1{\stackrel{→}{\mathbf{v}}}_{1f}+m_2{\stackrel{→}{\mathbf{v}}}_{2f}$ $\left(3\right)$

The mass of the block can be $m_1=1.00kg.$

The elastic constant of the spring is can be $k=2500N/m.$

The initial speed of the block is: .

The mass of the sphere can be: .

The ball is shot in the face opposite the spring and the club. The amplitudes of the following oscillations are:

The objective is to findout the initial speed of the bullet .

Define the system as block, sphere, and spring. Assume the initial state of the system is at the instant just before the ball hits the block and the final state is at the moment when the ball sinks into the block.

Since there are no external forces acting on the system, the mechanical energy of the system is conserved. Therefore, the total energy of the system, found from equation (1), is equal to the energy of the system in the final state:

$E_{\text{total}}=E_f$

Where the energy of the system in the final state is the sum of the kinetic energies of the mass and the sphere found from equation (2):

$\frac{1}{2}k{A}^2=\frac{1}{2}\left(m_1+m_2\right)v_f^2$

where v_f is the joint velocity of the ball and the block after the collision. Notice that the elastic potential energy of the spring is zero because it has not been compressed by the ball. Jumpled and settle for

$v_f=\sqrt{\frac{k{A}^2}{m_1+m_2}}$

$v_f=\sqrt{\frac{(2500\mathrm{N}/\mathrm{m})(0.1\mathrm{m}{)}^2}{1.00\mathrm{kg}+0.01\mathrm{kg}}}$

$=4.975\mathrm{m}/\mathrm{s}$

By applying principle of momentum conservation from equation $\left(3\right)$

$m_1{v}_{1i}+m_2{v}_{2i}=\left(m_1+m_2\right)v_f$

$m_2{v}_{2i}=\left(m_1+m_2\right)v_f$

$v_{2i}=\frac{\left(m_1+m_2\right)v_f}{m_2}$

$v_{2i}=\frac{(1.00\mathrm{kg}+0.01\mathrm{g})(4.975\mathrm{m}/\mathrm{s})}{0.01\mathrm{g}}$

$=5.0×{10}^2\mathrm{m}/\mathrm{s}$