Question

please help

A 6.85 g bullet traveling at 495 m/s embeds itself in a 1.85 kg wooden block at rest on a frictionless surface. The block is attached to a spring with k = 87.0 N/m

Find the amplitude of the subsequent simple harmonic motion.
A=?

Answer 1

Let mass of block be m. Total energy is constant. So,
\[E= (1/2)mv ^{2} + (1/2)kx ^{2}\]
differentiate it, we get
\[0= v(ma+kx)\]
\[a=-(kx/m)\]
\[-(kx/m)=-\omega ^{2}x\]
\[\omega=\sqrt{(k/m)}\]
Now,
\[Total energy= (1/2)m \omega ^{2}A ^{2}\]
\[(1/2) 6.85\times10^{-3}\times495^{2}= (1/2)m(k/m)A ^{?}\]
\[A=4.39 meters\]