Answer
(a) The speed of the baseball is 12.1 m/s in the opposite direction as it was originally traveling.
(b) Before the collision, the kinetic energy is 56.4 J.
After the collision, the kinetic energy is 13.7 J.
Work Step by Step
(a) Let $m_A$ be the mass of the baseball and let $m_B$ be the mass of the brick.
We can use conservation of momentum to find the speed of the baseball after the collision.
$m_A~v_A + 0 = m_A~v_A' + m_B~v_B'$
$v_A' = \frac{m_A~v_A - m_B~v_B'}{m_A}$
$v_A' = \frac{(0.144~kg)(28.0~m/s) - (5.25~kg)(1.10~m/s)}{0.144~kg}$
$v_A' = -12.1~m/s$
The speed of the baseball is 12.1 m/s in the opposite direction as it was originally traveling.
(b) Before the collision:
$KE = \frac{1}{2}(0.144~kg)(28.0~m/s)^2 = 56.4~J$
After the collision:
$KE = \frac{1}{2}(0.144~kg)(-12.1.~m/s)^2 + \frac{1}{2}(5.25~kg)(1.10~m/s)^2$
$KE = 13.7~J$
Before the collision, the kinetic energy is 56.4 J.
After the collision, the kinetic energy is 13.7 J.
