Answer
The minimum speed of the dart is 58.6 m/s
Work Step by Step
To make a complete loop, the minimum speed $v_t$ at the top of the loop is such that the required centripetal force is equal to the force of gravity. Let $m_2$ be the mass of the ball and dart.
$\frac{m_2~v_t^2}{r} = m_2~g$
$v_t^2 = g~r$
$v_t = \sqrt{g~r}$
$v_t = \sqrt{(9.80~m/s^2)(2.80~m)}$
$v_t = 5.24~m/s$
The minimum kinetic energy at the bottom of the loop is equal to the sum of the potential energy and the minimum kinetic energy at the top of the loop.
$\frac{1}{2}m_2v_2^2 = m_2~gh+\frac{1}{2}m_2~v_t^2$
$v_2^2 = 2gh+v_t^2$
$v_2 = \sqrt{2gh+v_t^2}$
$v_2 = \sqrt{(2)(9.80~m/s^2)(5.60~m)+(5.24~m/s)^2}$
$v_2 = 11.71~m/s$
We can use conservation of momentum to find the minimum initial speed of the dart.
$m_1v_1 = m_2v_2$
$v_1 = \frac{m_2v_2}{m_1}$
$v_1 = \frac{(25.00~kg)(11.71~m/s)}{5.00~kg}$
$v_1 = 58.6~m/s$
The minimum speed of the dart is 58.6 m/s
