Ball collision speed

Two inelastic/plastic balls move towards each other with equal velocities of 2 m/s. The weight of one ball is 3kg, the other 1kg. Determine the speed of the balls after the collision and the loss of kinetic energy.

Final Answer:

v =  1 m/s
E =  6 J

Step-by-step explanation:

$$\begin{gathered} v_1=2\text{m/s} \\ {v}_2=v_1=2\text{m/s} \\ {m}_1=3\text{kg} \\ {m}_2=1\text{kg} \\ \Sigma h=0 \\ |m_1\cdot v_1-m_2\cdot v_2|=(m_1+m_2)\cdot |v| \\ v=\frac{m_1\cdot v_1-m_2\cdot v_2}{m_1+m_2}=\frac{3\cdot 2-1\cdot 2}{3+1}=1\text{m/s} \end{gathered}$$

$$\begin{gathered} E_1=\frac{1}{2}\cdot m_1\cdot v_1^2=\frac{1}{2}\cdot 3\cdot 2^2=6\text{J} \\ {E}_2=\frac{1}{2}\cdot m_2\cdot v_2^2=\frac{1}{2}\cdot 1\cdot 2^2=2\text{J} \\ {E}_3=\frac{1}{2}\cdot (m_1+m_2)\cdot v^2=\frac{1}{2}\cdot (3+1)\cdot 1^2=2\text{J} \\ E=E_1+E_2-E_3=6+2-2=6\text{J} \end{gathered}$$

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