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