Answer
The kinetic energy of block 3 is less than the initial kinetic energy of block 1
Work Step by Step
We can find an expression for the initial kinetic energy of block 1:
$K_1 = \frac{1}{2}m_1~v_1^2$
$K_1 = \frac{1}{2}m_1~(4.00)^2$
$K_1 = (8.00~m_1)~J$
In part (a), we found that the speed of block 3 is $7.11~m/s$
Note that $m_3 = 0.25~m_1$
We can find an expression for the kinetic energy of block 3:
$K_3 = \frac{1}{2}m_3~v_3^2$
$K_3 = \frac{1}{2}(0.25~m_1)~(7.11)^2$
$K_3 = (6.32~m_1)~J$
The kinetic energy of block 3 is less than the initial kinetic energy of block 1
