Answer:
[tex]\Delta v= 4.66\frac{m}{s}[/tex]
Explanation:
In this case we have to use the Principle of conservation of Momentum:
This principle says that in a system the total momentum is constant if no external forces act in the system. The formula is:
[tex]m_1v_1+m_2v_2=m_1u_1+m_2u_2[/tex]
Where:
[tex]m_1:[/tex] Mass of the first object.
[tex]m_2:[/tex] Mass of the second object.
[tex]v_1:[/tex] Initial velocity of the first object.
[tex]v_2:[/tex] Initial velocity of the second object.
[tex]u_1:[/tex] Final velocity of the first object.
[tex]u_2:[/tex] Final velocity of the second object.
In this problem we have:
[tex]m_1=81kg\\m_2=40kg\\v_1_2=2.3\frac{m}{s}[/tex]
[tex]u_1=0\frac{m}{s}[/tex]
Observation: [tex]v_1_2:[/tex] Is because the system has the same initial velocity.
First we have to find [tex]u_2[/tex],
[tex]m_1v_1+m_2v_2=m_1u_1+m_2u_2[/tex]
We can rewrite it as:
[tex](m_1+m_2)v_1_2=m_1u_1+m_2u_2[/tex]
Replacing with the data:
[tex](m_1+m_2)v_1_2=m_1u_1+m_2u_2\\\\(81kg+40kg)2.3\frac{m}{s}=81kg(0\frac{m}{s})+40kg(u_2)\\\\(121kg)2.3\frac{m}{s}=40kg(u_2)\\\\\frac{(121kg)2.3\frac{m}{s}}{40kg}=u_2\\\\\frac{278.3}{40}\frac{m}{s}=u_2\\\\6.96\frac{m}{s}=u_2[/tex]
We found the final velocity of the cart, but the problem asks for the resulting change in the cart speed, this means:
[tex]\Delta v=u_2-v_2\\\Delta v=6.96\frac{m}{s}-2.3\frac{m}{s}\\\Delta v= 4.66\frac{m}{s}[/tex]
Then, the resulting change in the cart speed is:
[tex]\Delta v= 4.66\frac{m}{s}[/tex]
