Answer
The tension in the wire just after the collision is 13.6 N.
Work Step by Step
We can use conservation of momentum to find the speed of the ornament.
$m_2~v_2 = m_1~v_1$
$v_2 = \frac{m_1~v_1}{m_2}$
$v_2 = \frac{(0.200~kg)(12.0~m/s)}{1.00~kg}$
$v_2 = 2.40~m/s$
We can use the speed to find the tension in the wire.
$\sum F = \frac{mv^2}{R}$
$T-mg = \frac{mv^2}{R}$
$T = \frac{mv^2}{R}+mg$
$T = \frac{(1.00~kg)(2.40~m/s)^2}{1.50~m}+(1.00~kg)(9.80~m/s^2)$
$T = 13.6~N$
The tension in the wire just after the collision is 13.6 N.
