Answer
a) $30^o$
b) $v'_T=-v'_N=\frac{v}{\sqrt{3}}$
c) $\frac{2}{3}$ of total kinetic energy
Work Step by Step
a) In the horizontal direction
$m_Nv_N=m_Tv'_{Tx}$
$v_N=2v'_{Tx}$
In the vertical direction
$0=m_Nv'_N+m_Tv'_{Ty}$
$v'_N=-2v'_{Ty}$
$\frac{m_Nv_N^2}{2}+\frac{m_Tv_T^2}{2}=\frac{m_Nv_N^{'2}}{2}+\frac{m_Tv_T^{'2}}{2}$
$v_N^2+2v_T^2=v_N^{'2}+2v_T^{'2}$
$(2v'_{Tx})^2+2v_T^2=(-2v'_{Ty})^{2}+2v_T^{'2}$
$4v_{Tx}^{'2}=4v_{Ty}^{'2}+2v_{Tx}^{'2}+2v_{Ty}^{'2}$
$2v_{Tx}^{'2}=6v_{Ty}^{'2}$
$\sqrt{2}v'_{Tx}=\sqrt{6}v'_{Ty}$
$\frac{v'_{Ty}}{v'_{Tx}}=\frac{1}{\sqrt{3}}$
$\arctan\Big(\frac{1}{\sqrt{3}}\Big)=30.0^o$
b) $v'_{Tx}=\frac{v}{2}$
$v'_T=\frac{\frac{v}{2}}{\cos(30^o)}=\frac{v}{\sqrt{3}}$
$v'_{Ty}=\frac{v}{\sqrt{3}}\sin(30^o)=\frac{v}{2\sqrt{3}}$
$v'_N=-2\frac{v}{2\sqrt{3}}=-\frac{v}{\sqrt{3}}$
c) $v^2=\frac{v^2}{3}+2\frac{v^2}{3}$
$\frac{2}{3}$ of total kinetic energy