Answer to Question #151139 in Physics for Hendrix Lagasca

Question

A 60 g mass with a horizontal velocity of 50 cm/s collides with an 85 g mass at rest. After collision, the 60 g mass travel 30 O below the horizontal and the 85 g mass travels 45 O above the horizontal. Find the speed of each mass after impact.

Answer 1

So,

X: $m_1v_1=m_1v_1'\cos30°+m_2v_2'\cos45°$

Y: $m_1v_1'\sin30°=m_2v_2'\sin45°$

We get

$v_2'=\frac{m_1v_1}{\frac{m_2\sin45°}{\sin30°}+m_2\sin45°}=\frac{0.06\cdot 0.5}{\frac{0.085\cdot\sin45°}{\sin30°}+0.085\cdot\sin45°}=0.17(m/s)$

$v_1'=\frac{m_2v_2'\sin45°}{m_1\sin30°}=\frac{0.085\cdot 0.17\cdot\sin45°}{0.06\cdot\sin30°}=0.34(m/s)$