## University Physics with Modern Physics (14th Edition)

(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.