Answer :
[tex]m_1=2 \\ m_2=3 \\ v_1=4 \\ v_2=1 \\ v\text{ =speed after collision (to be determined)}.[/tex]
The momentul of the system preserves:
[tex]m_1v_1-m_2v_2=(m_1+m_2)v \ \ \ \ \ \Rightarrow \ \ \ \ \ v=\dfrac{m_1v_1-m_2v_2}{m_1+m_2}.[/tex]
Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with [tex]E[/tex].
Now, we write the energy conservation law:
[tex]\dfrac{m_1v_1^2}{2}+\dfrac{m_2v^2_2}{2}=\dfrac{(m_1+m_2)v^2}{2}+E[/tex]
From the above equation, you find [tex]E[/tex], and then conclude that the sound energy can certainly not be greater than this.
The momentul of the system preserves:
[tex]m_1v_1-m_2v_2=(m_1+m_2)v \ \ \ \ \ \Rightarrow \ \ \ \ \ v=\dfrac{m_1v_1-m_2v_2}{m_1+m_2}.[/tex]
Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with [tex]E[/tex].
Now, we write the energy conservation law:
[tex]\dfrac{m_1v_1^2}{2}+\dfrac{m_2v^2_2}{2}=\dfrac{(m_1+m_2)v^2}{2}+E[/tex]
From the above equation, you find [tex]E[/tex], and then conclude that the sound energy can certainly not be greater than this.
