Answer
(a) The pendulum rises through a vertical height of 2.94 cm.
(b) $K_1 = 866~J$
(c) $K_2 = 1.74~J$
Work Step by Step
(a) We can use conservation of momentum to find the speed of the block after the bullet hits it.
$m_2~v_2 = m_1~v_1$
$v_2 = \frac{m_1~v_1}{m_2}$
We can equate the kinetic energy of the block at the bottom of the swing to the potential energy at the top.
$PE = K$
$(m_1+m_2)gh = \frac{1}{2}(m_1+m_2)v_2^2$
$gh = \frac{1}{2}(\frac{m_1~v_1}{m_2})^2$
$h = \frac{1}{2g}(\frac{m_1~v_1}{m_2})^2$
$h = \frac{1}{(2)(9.80~m/s^2)}(\frac{(0.0120~kg)(380~m/s)}{6.012~kg})^2$
$h = 0.0294~m = 2.94~cm$
The pendulum rises through a vertical height of 2.94 cm.
(b) $K_1 = \frac{1}{2}m_1~v_1^2$
$K_1 = \frac{1}{2}(0.0120~kg)(380~m/s)^2$
$K_1 = 866~J$
(c) $K_2 = \frac{1}{2}(m_1+m_2)~v_2^2$
$K_2 = \frac{1}{2}(m_1+m_2)( \frac{m_1~v_1}{m_2})^2$
$K_2 = \frac{1}{2}(6.012~kg)(\frac{(0.0120~kg)(380~m/s)}{6.00~kg})^2$
$K_2 = 1.74~J$
