Question

Chapter 15, Problem 075 A 8.2 kg block is suspended from a spring with k 251 N/m. A 35 g bullet is fired into the block from directly below with a speed of 160 m/s and becomes embedded in the block. (a) Find the amplitude of the resulting simple harmonic motion. (b) What percentage of the original kinetic energy of the bullet is transferred to mechanical energy of the harmonic oscillator? (a) Number Units (b) Number Units

Answer 1

Answer 2

(a) Amplitude of the resulting simple harmonic motion:

1. Conservation of momentum (inelastic collision):

   $$m_{\text{bullet}}{v}_{\text{bullet}}=(m_{\text{block}}+m_{\text{bullet}})v_f$$$$0.035\text{kg}\times 160\text{m/s}=(8.2\text{kg}+0.035\text{kg})v_f$$$$v_f=\frac{5.6}{8.235}\approx 0.680\text{m/s}$$

2. Energy conservation (kinetic energy converts to spring potential energy at amplitude $A$):

   $$\frac{1}{2}(m_{\text{block}}+m_{\text{bullet}})v_f^2=\frac{1}{2}k{A}^2$$$$A=v_f\sqrt{\frac{m_{\text{block}}+m_{\text{bullet}}}{k}}=0.680\sqrt{\frac{8.235}{251}}\approx 0.039\text{m}$$

Answer:
The amplitude is $\boxed{{\displaystyle 0.039\text{m}}}$.

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(b) Percentage of bullet's kinetic energy transferred to the harmonic oscillator:

1. Initial kinetic energy of the bullet:

   $$K_{\text{bullet}}=\frac{1}{2}{m}_{\text{bullet}}{v}_{\text{bullet}}^2=\frac{1}{2}\times 0.035\times 160^2=448\text{J}$$

2. Mechanical energy of the oscillator (same as maximum spring potential energy):

   $$E_{\text{oscillator}}=\frac{1}{2}k{A}^2=\frac{1}{2}\times 251\times 0.039^2\approx 0.191\text{J}$$

3. Percentage transferred:

   $$\text{Percentage}=\left(\frac{E_{\text{oscillator}}}{K_{\text{bullet}}}\right)\times 100=\left(\frac{0.191}{448}\right)\times 100\approx 0.0426\mathrm{\%}$$

Answer:
The percentage transferred is $\boxed{{\displaystyle 0.0426\mathrm{\%}}}$