Respuesta :
Answer:
The final velocity has a magnitude of 21.69 m/s and is at 24.66° north of east.
Explanation:
Taking north direction as positive y axis ([tex]\vec{j}[/tex]) and east direction as positive x axis ([tex]\vec{i}[/tex]).
Given:
Mass of first car is, [tex]m_1=950\ kg[/tex]
Initial velocity of first car is, [tex]u_1=20\vec{j}\ m/s[/tex]
Mass of second car is, [tex]m_2=1150\ kg[/tex]
Initial velocity of second car is, [tex]u_2=36\vec{i}\ m/s[/tex]
Let the combined final velocity after collision be [tex]\vec{v}=(v_x\vec{i}+v_y\vec{j})[/tex] m/s where [tex]v_x\ and\ v_y[/tex] are components of final velocity along east and north directions respectively.
Now, as the net external force is zero, momentum is conserved for the two car system along the east and north directions.
Conserving momentum along the east direction, we have:
Initial momentum = Final momentum
[tex]m_1u_{1x}+m_2u_{2x}=(m_1+m_2)v_x\\\\0+1150\times 36=(950+1150)v_x\\\\v_x=\frac{41400}{2100}\\\\v_x=19.71\ m/s[/tex]
There is no component of initial velocity for first car in east direction, as it is moving in the north direction. So, [tex]u_{1x}=0[/tex]
Now, conserving momentum along the north direction, we have:
Initial momentum = Final momentum
[tex]m_1u_{1y}+m_2u_{2y}=(m_1+m_2)v_y\\\\950\times 20+0=(950+1150)v_y\\\\v_y=\frac{19000}{2100}\\\\v_y=9.05\ m/s[/tex]
There is no component of initial velocity for second car in north direction, as it is moving in the east direction. So, [tex]u_{2y}=0[/tex]
Therefore, the final velocity is given as:
[tex]v=v_x\vec{i}+v_y\vec{j}\\\\v=19.71\vec{i}+9.05\vec{j}[/tex]
The magnitude of final velocity is given as:
[tex]|\vec{v}|=\sqrt{(v_x)^2+(v_y)^2}\\\\|\vec{v}|=\sqrt{19.71^2+9.05^2}\\\\|\vec{v}|=\sqrt{470.387}=21.69\ m/s[/tex]
The direction is given as:
[tex]\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{9.05}{19.71})=24.66^\circ[/tex]
So, the final velocity has a magnitude of 21.69 m/s and is at 24.66° north of east.
