Respuesta :
Given:
The mass of your first friend's bumper boat is
[tex]m_{1i}=\text{ 210 kg}[/tex]The speed of your first friend's boat before collision towards the left is
[tex]v_{1i}=3.5\text{ m/s}[/tex]The mass of your second friend's bumper boat is
[tex]m_{2i}=221\text{ kg}[/tex]The speed of your second friend's boat before collision towards the right is
[tex]v_{2i}=\text{ 1.8 m/s}[/tex]The boats collided perfectly elastic.
Required:
(a) Speed of first friend's boat
Explanation:
According to the conservation of momentum,
[tex]\begin{gathered} m_1v_{1i}-m_2v_{2i}=m_1v_{1f}+m_2v_{2f} \\ v_{2f}=\frac{m_1v_{1i}-m_2v_{2i}-m_1v_{1f}}{m_2} \\ =\frac{337.2-210v_{1f}}{221} \end{gathered}[/tex]As the collision is perfectly elastic, the kinetic energy will also be conserved
[tex]\begin{gathered} \frac{1}{2}m_1(v_{1i})^2+\frac{1}{2}m_2(v_{2i})^2=\frac{1}{2}m_1(v_{1f})^2+\frac{1}{2}m_2(v_{2f})^2 \\ m_1(v_{1\imaginaryI})^2+m_2(v_{2\imaginaryI})^2=m_1(v_{1f})^2+m_2(v_{2f})^2 \end{gathered}[/tex]On substituting the values, the speed of the first friend's boat will be
[tex]\begin{gathered} 210\times(3.5)^2+221\times(1.8)^2=210(v_{1f})^2+221\times\frac{(337.2-210v_{1f})^2}{(221)^2} \\ 3288.54=210(v_{1f})^2+514.5+199.5(v_{1f})^2-640.83v_{1f} \\ 409.5(v_{1f})^2-640.83v_{1f}-2774.04=0 \\ v_{1f}=\text{ -1.94 m/s} \end{gathered}[/tex]Final Answer: The speed of the first friend boat is -1.94 m/s after the collision.
