Answer
(a) Just after impact, the magnitude of the block's velocity is 2.60 m/s.
(b) The initial speed of the bullet is 325 m/s.
Work Step by Step
(a) We can find the force constant of the spring.
$kx = F$
$k = \frac{F}{x} = \frac{0.750~N}{0.00250~m}$
$k = 300~N/m$
The initial kinetic energy of the block will be equal to the potential energy stored in the spring.
$K = U_s$
$\frac{1}{2}mv_2^2 = \frac{1}{2}kx^2$
$v_2^2 = \frac{kx^2}{m}$
$v_2 = \sqrt{\frac{kx^2}{m}}$
$v_2 = \sqrt{\frac{(300~N/m)(0.150~m)^2}{1.00~kg}}$
$v_2 = 2.60~m/s$
Just after impact, the magnitude of the block's velocity is 2.60 m/s.
(b) We can use conservation of momentum to find the bullet's initial speed $v_1$.
$m_1v_1 = m_2v_2$
$v_1 = \frac{m_2v_2}{m_1}$
$v_1 = \frac{(1.00~kg)(2.60~m/s)}{0.00800~kg}$
$v_1 = 325~m/s$
The initial speed of the bullet is 325 m/s.
