Answer
The kinetic energy of particle B is $~~40~J$
Work Step by Step
Note that the mass of particle A is $2.00$ times the mass of particle B. We can express this as: $~~m_A = 2.00~m_B$
The initial momentum of the system is 0.
We can use conservation of momentum to find a relationship between the velocity $v_A$ of particle A and the velocity $v_B$ of particle B after they are released:
$p_f = p_i$
$m_A~v_A+m_B~v_B = 0$
$m_B~v_B = -m_A~v_A$
$v_B = -\frac{m_A~v_A}{m_B}$
$v_B = -\frac{2.00~m_B~v_A}{m_B}$
$v_B = -2.00~v_A$
We can find an expression for the kinetic energy $K_A$ of particle A:
$K_A = \frac{1}{2}m_A~v_A^2$
We can find an expression for the kinetic energy $K_B$ of particle B:
$K_B = \frac{1}{2}m_B~v_B^2$
$K_B = \frac{1}{2}(\frac{m_A}{2.00})~(-2.00~v_A)^2$
$K_B = m_A~v_A^2$
$K_B = 2~K_A$
We can find the kinetic energy of particle A:
$K_A+K_B = 60~J$
$K_A+2~K_A = 60~J$
$3~K_A = 60~J$
$K_A = 20~J$
We can find the kinetic energy of particle B:
$K_B = 2~K_A$
$K_B = (2)(20~J)$
$K_B = 40~J$
The kinetic energy of particle B is $~~40~J$
