# Problem 8-75 Inelastic collision - Part 7

Two trucks have a completely inelastic collision. After the collision, they have a common velocity of $11 \;m/s$ at $35^\circ$ south of west. Before the collision, one truck $(\text{mass} \; 2.3\times10^4\; kg)$ has a velocity of $15\; m/s$ at $51^\circ$ south of west. Determine the initial velocity (magnitude and direction) of the other truck $(\text{mass} \; 1.2\times10^4\; kg).$

Accumulated Solution

$v_{1x} = (-15) \cos 51 \\ v_{2x} = v_2 \cos \theta \\ v_x{'} = (-11) \cos 35 \\ m_1v_{1x} + m_2v_{2x} = (m_1 + m_2)v_x{'} \\ v_{2x} = -8.2\; m/s \\ v_{1y} = (-15) \sin 51 \\ v_{2y} = v_2 \cos \theta \\ v_y{'} = (-11) \sin 35 \\ v_{2y} = 3.9 \;m/s$

$(2.3 \times 10^4\; kg)(-15) \sin 51 + (1.2 \times 10^4\; kg)v_{2y} = (3.5 \times 10^4\; kg)(-11) \sin 35 \\ v_{2y} = 3.9\; m/s$

So you have $v_{2x} = -8.2\; m/s$ and $v_{2y} = 3.9 \;m/s$

The minus sign on $v_{2x}$ means that:

(B)  The initial choice of direction of $v_{2x}$ was incorrect.