# Problem 8-75 Inelastic collision - Part 7 - B

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

Correct. The original choice of direction for $v_2$ in the figure (up and to the right) was incorrect but it doesn't matter for we now know it is up $(v_y +)$ to the left $(v_x -)$ as in this figure: The velocity $v_2$ is given by:

(A)   $3.9 + 8.2 = 12.1\; m/s$

(B)   $3.9 - 8.2 = -4.3\; m/s$

(C)   $[3.9^2 + (-8.2)^2]^{1/2} = 9.1 \;m/s$