Question

Ball A with mass 0.85 kg is rolling at 3.5 m/s [N] when it collides with stationary Ball B with mass 1.15 kg. After the collision, Ball A is moving 2.4 m/s [N40oW]. Find the velocity of Ball B after the collision.

Answer 1

$m_A\vec{v_A}+m_B\vec{0}=m_A\vec{u_A}+m_B\vec{u_B},$

$\vec{u_B}=\frac{m_A}{m_B}(\vec{v_A}-\vec{u_A}),$

$|\vec{v_A}-\vec{u_A}|=\sqrt{v_A^2+u_A^2-2v_Au_Acos(90°-40°)}=2.69~\frac ms,$ $u_B=\frac{0.85}{1.15}\cdot 2.69=1.99~\frac ms,$

$\frac{u_A}{sin(90°-p)}=\frac{u_B}{sin50°},\implies$

$p=90°-arcsin(sin50°\cdot\frac{u_A}{u_B})=22.5°~NofE.$