Answer:
The second car was moving at 2.1 m/s to the east
Explanation:
Conservation Of Linear Momentum
The total momentum of a system of bodies is constant if no external force is applied to it. The momentum of a body with mass m and velocity v is p=mv. When two objects collide and join together afterward, the total mass is the sum of the individual masses, and the final speed is common to both.
Let's say
[tex]\displaystyle m_1,\ m_2\ ,\ v_1,\ v_2\ ,\ p_1,\ p_2[/tex]
are the masses of the objects 1 and 2, their speeds, and their linear momentums respectively before the collision, and
[tex]\displaystyle v_1',\ v_2'\ ,\ p_1',\ p_2'[/tex]
are the speeds of each object and their linear momentums after the collision.
The principle of conservation of linear momentum states that
[tex]\displaystyle p_1+p_2=p_1'+p_2'[/tex]
This means that
[tex]\displaystyle m_1\ v_1+m_2\ v_2=m_1\ v_1'+m_2\ v_2'[/tex]
Since both cars remain together after the collision,
[tex]\displaystyle v_1'=v_2'=v'[/tex]
The relation becomes
[tex]\displaystyle m_1\ v_1+m_2\ v_2=(m_1+m_2)\ v'[/tex]
Solving for [tex]v_2[/tex]
[tex]\displaystyle v_2=\frac{(m_1+m_2)v'-m_1\ v_1}{m_2}[/tex]
The given data is
[tex]\displaystyle m_1=2150\ kg\ ,\ m_2=3250\ kg\ ,\ v_y=10\ m/s,\ v'=5.22\ m/s[/tex]
Let's assume all the speeds are positive towards East
Replacing those values
[tex]\displaystyle v_2=\frac{(2150+3250)5.22-2150(10)}{3250}[/tex]
[tex]\displaystyle v_2=\frac{6688}{3250}=2.06\ m/s[/tex]
The second car was moving at 2.1 m/s (to the nearest tenth) to the east
