Question

Ball A with a mass of 8 kg moves with a velocity of 3 m/s at a 90° angle and collides with a stationary Ball B with a mass of 6 kg. After the collision Ball A continues on at a velocity of 2 m/s at an angle of 60°. What is the velocity and angle of Ball B after the collision?

Answer 1

$m_1v_1=m_1v_1'\cos60°+m_2v_2'\cos\alpha$

$m_1v_1'\sin60°=m_2v_2'\sin\alpha$

$8\cdot 3=8\cdot 2\cdot \cos60°+6\cdot v_2'\cos\alpha$

$8\cdot 2\cdot \sin60°=6\cdot v_2'\sin\alpha$

$2.67=v_2'\cos\alpha$

$2.31=v_2'\sin\alpha$

$\tan\alpha=0.8652\to\alpha=40.87°$

$v_2'=2.31/\sin40.87°=3.53\ (m/s)$