Question

A 1800 kg object moving to the east at 13 m/s, collided with a 200 kg object that was moving to the north at 32 m/s. What was the angle of motion

Answer 1

$\begin{gathered} For\text{ object moving to the east} \\ v1x=13m/s \\ v1y=0\text{ m/s} \\ m1=1800kg \\ For\text{ object moving to the north} \\ v2x=0m/s \\ v2y=32m/s \\ m2=200kg \\ \theta=? \\ vx=\frac{m1v1x}{m1+m2} \\ vx=\frac{(1800kg)(13m/s)}{1800kg+200kg} \\ vx=11.7\text{ m/s} \\ vy=\frac{m2v2y}{m1+m2} \\ vy=\frac{(200kg)(32m/s)}{1800kg+200kg} \\ vy=3.2\text{ m/s} \\ \theta=\tan^{-1}(\frac{vy}{vx}) \\ \theta=\tan^{-1}(\frac{3.2m/s}{11.7m/s}) \\ \theta=15.3\text{\degree} \\ The\text{ angle of motion is 15.3\degree} \end{gathered}$