Answer
(a) The second ball has a mass of 0.840 kg.
(b) 0.75 of the original kinetic energy gets transferred to the second ball.
Work Step by Step
(a) Let $m_A = 0.280~kg$. Note that $2v_B' = v_A$.
We can use conservation of momentum to set up an equation.
$m_A~v_A + 0 = m_A~v_A' + m_B~v_B'$
$2m_A~v_B' = m_A~v_A' + m_B~v_B'$
Since the collision is elastic, we can use Equation 7-7 to set up another equation.
$v_A - 0 = v_B' - v_A'$
$2v_B' = v_B' - v_A'$
$v_A' = -v_B'$
We can use this expression for $v_A'$ in the first equation.
$2m_A~v_B' = m_A~v_A' + m_B~v_B'$
$2m_A~v_B' = -m_A~v_B' + m_B~v_B'$
$m_B = 3m_A = (3)(0.280~kg) = 0.840~kg$
The second ball has a mass of 0.840 kg.
(b) We can find the original KE of the croquet ball.
$KE_A = \frac{1}{2}(0.280~kg)(v_A)^2$
$KE_A = 0.140~v_A^2 ~J$
We can find the KE of the second ball.
$KE_B = \frac{1}{2}(0.840~kg)(\frac{v_A}{2})^2$
$KE_B = 0.105~v_A^2 ~J$
$\frac{KE_B}{KE_A} = \frac{0.105~v_A^2}{0.140~v_A^2} = 0.75$
Therefore, 0.75 of the original kinetic energy gets transferred to the second ball.
