Answer
The speed of the bullet is 312.6 m/s
Work Step by Step
The potential energy stored in the spring is equal to the kinetic energy of the block just after the bullet strikes the block. We can find the speed of the block $v_f$ just after the bullet strikes the block;
$KE = U_s$
$\frac{1}{2}(M+m)v_f^2 = \frac{1}{2}kx^2$
$v_f^2 = \frac{k}{M+m}~x^2$
$v_f = \sqrt{\frac{k}{M+m}}~x$
$v_f = \sqrt{\frac{162.7~N/m}{(4.148~kg)+(0.007870~kg)}}~(0.09460~m)$
$v_f = 0.5919~m/s$
We can use conservation of momentum to find the initial speed $v$ of the bullet;
$m~v = (M+m)~v_f$
$v = \frac{(M+m)~v_f}{m}$
$v = \frac{(4.148~kg+0.007870~kg)(0.5919~m/s)}{0.007870~kg}$
$v = 312.6~m/s$
The speed of the bullet is 312.6 m/s
