AP Physics Kinematics 2 Balls Inclined Track

[jhj1259]

Consider two balls on an inclined track with a constant angle. Ball A is released from rest at the top of the track. Ball B is given an initial velocity directed down the track. At the instant that the balls collide, Ball A is going 4 times as fast as Ball B. Ball B is still traveling upward along the track at the time of the collision. At what fraction of the way along the track, measured from the bottom, does the collision occur? (Draw a graph of velocity versus time showing motion of both balls. Show all work)
At least a drawing of velocity versus time, identify the variables, label the graph carefully. But main question is "At what fraction of the way along the track, measured from the bottom, does the collision occur?"

 

[jbsteele]

Let the distance of the track be l, so the fraction of the way along the track where the collision occurs is x. Since Ball A is released from rest at the top of the track and Ball B is given an initial velocity directed down the track, we can assume the velocities of Ball A and Ball B at the start of the track are 0 and -v respectively, with v being the magnitude of the velocity of Ball B. The velocity of Ball A is increasing at a constant rate due to gravity as it moves down the track, while the velocity of Ball B is decreasing at a constant rate due to gravity as it moves up the track. At the time of the collision, Ball A is going 4 times as fast as Ball B and Ball B is still traveling upward. This means that the velocity of Ball A at the time of the collision is 4v and the velocity of Ball B is 0 (since it is still traveling upward).From the graph, we can see that the total distance traveled by Ball A is 4l and the total distance traveled by Ball B is l. Therefore, the collision occurs at a distance of l/4 from the bottom of the track, or at a fraction of the way along the track of x = 1/4.