Answer:
Velocity of ball B after impact is [tex]0.6364v_0[/tex] and ball A is [tex]0.711v_0[/tex]
Explanation:
[tex]v_0[/tex] = Initial velocity of ball A
[tex]v_A=v_0\cos45^{\circ}[/tex]
[tex]v_B[/tex] = Initial velocity of ball B = 0
[tex](v_A)_n'[/tex] = Final velocity of ball A
[tex]v_B'[/tex] = Final velocity of ball B
[tex]e[/tex] = Coefficient of restitution = 0.8
From the conservation of momentum along the normal we have
[tex]mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0[/tex]
Coefficient of restitution is given by
[tex]e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0[/tex]
[tex](v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0[/tex]
[tex]v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0[/tex]
Adding the above two equations we get
[tex]2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0[/tex]
[tex]\boldsymbol{\therefore v_B'=0.6364v_0}[/tex]
[tex](v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0[/tex]
From the conservation of momentum along the plane of contact we have
[tex](v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}[/tex]
[tex]v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}[/tex]
Velocity of ball B after impact is [tex]0.6364v_0[/tex] and ball A is [tex]0.711v_0[/tex].
