Answer:
Both cars travel at < 10 , 4 > m/s
Explanation:
Conservation of Linear Momentum
The total momentum of a system of particles of masses m1 and m2 traveling at velocities v1 and v2 (vectors) is given by
[tex]\vec p_1=m_1\vec v_1+m_2\vec v_2[/tex]
When the particles collide, their velocities change to v1' and v2' while their masses remain unaltered. The total momentum in the final condition is
[tex]\vec p_2=m_1\vec v'_1+m_2\vec v'_2[/tex]
We know the collision is perfectly inelastic, which means both cars stick together at a common final velocity v'. Thus
[tex]\vec p_2=m_1\vec v'+m_2\vec v'=(m_1+m_2)\vec v'[/tex]
Both total momentums are equal:
[tex]m_1\vec v_1+m_2\vec v_2=(m_1+m_2)\vec v'[/tex]
Solving for v'
[tex]\displaystyle v'=\frac{m_1\vec v_1+m_2\vec v_2}{m_1+m_2}[/tex]
The data obtained from the question is
[tex]m_1=1200\ kg[/tex]
[tex]m_2=3000\ kg[/tex]
The first car travels north which means its velocity has only y-component
[tex]\vec v_1=<0,5>[/tex]
The second car travels east, only x-component of the velocity is present
[tex]\vec v_2=<5,0>[/tex]
Plugging in the values
[tex]\displaystyle v'=\frac{1200<0,5>+3000<5,0>}{1200+3000}[/tex]
[tex]\displaystyle v'=\frac{<0,6000>+<15000,0>}{1500}[/tex]
[tex]\displaystyle v'=\frac{<15000,6000>}{1500}=<10,4>\ m/s[/tex]
The magnitude of the velocity is
[tex]|v'|=\sqrt{10^2+4^2}=10.77\ m/s[/tex]
And the angle
[tex]\displaystyle tan\alpha=\frac{4}{10}=0.4[/tex]
[tex]\alpha=21.8^o[/tex]
