Answer
(a) $897m/s$
(b) $0.0687s$
Work Step by Step
(a) We can find the initial speed of the bullet as follows:
$v_{\circ}=\sqrt{\frac{KA^2(m+M)}{m^2}}$
We plug in the known values to obtain:
$v_{\circ}=\sqrt{\frac{(785N/m)(5.88\times 10^{-2}m)^2(2.25\times 10^{-3}Kg)+1.50Kg}{(2.25\times 10^{-3}Kg)^2}}$
$v_{\circ}=897m/s$
(b) We can find the required time as follows:
$t=\frac{1}{2}\pi \sqrt{\frac{m+M}{K}}$
We plug in the known values to obtain:
$t=\frac{1}{2}(3.1416)\sqrt{\frac{2.25\times 10^{-3}Kg+1.50Kg}{785N/m}}$
$t=0.0687s$
