PhysicsLAB Resource Lesson
Chase Problems: Projectiles

Printer Friendly Version
spursuer = "gap" + sleader
vot + ½at2 number vot + ½at2
vt   vt
 
Each column in the above table states the allowed behaviors for the pursuer and the leader. Each participant can either be experiencing accelerated or linear motion. The numerical value of the "gap" can be equal to zero (if the two objects start side-by-side) or it can be a nonzero number. The parameter t, for time, unites the equations. To solve chase equations, you first determine the time that is required for the two objects to come together - then, you use that time to determine the position of their collision.
 
To work this type of problem, one object is considered the leader and the other is the pursuer. The pursuer, in reaching the leader's final location, must not only close the leader's original gap but also account for any subsequent displacement the leader travels while being chased.
 
Example


A ball of mass 1.0 kg is dropped from rest from a height of 5.0 meters above the ground, as shown in the diagram at the right. It undergoes a perfectly elastic collision with the ground [that is, it leaves the surface of the ground with exactly the same amount of kinetic energy (KE = ½mv2) with which it initially strikes the ground] and rebounds at a speed of 9.9 m/sec. At the instant that the ball rebounds, a small blob of clay of mass 0.1 kg is released from rest from the original height H, directly above the ball, as shown. The clay blob while descending eventually collides with the ascending ball. Assume that air resistance is negligible.
 
 
To determine how much time passes after the clay blob is released until it strikes the ball we must first set up the chase equation. Let's assume that the 1.0-kg ball is the pursuer (since it is traveling in a positive direction) and that the 0.1-kg clay blob is the leader.
  
 
spursuer = "gap" + sleader
vot + ½at2 number vot + ½at2
vt  
vt
 s > 0    s < 0
9.9t + ½(-9.8)t2 5 meters 0t + ½(-9.8)t2
 
9.9t + ½(-9.8)t2 =  5 + [0t + ½(-9.8)t2]
9.9t - 4.9t2 =  5 - 4.9t2
9.9t =  5
t = 0.505 seconds
 
Next let's determine the height above the ground at which the collision takes place. Since this represents the distance that the ball traveled upwards after its bounce,  we can substitute the time, 0.505 seconds, into the ball's equation.
 
s =  9.9t + ½(-9.8)t2
s =  9.9(0.505) + ½(-9.8)(0.505)2
s =  3.7 meters
 
The position-time graph for this problem would look like the one shown below. Note that the two parabolas intersect at a height of 3.7 meters and a time of 0.505 seconds. The purple parabola represents the clay blob which is gaining speed in a negative direction while the blue parabola represents the ball which is losing speed while traveling in a positive direction.
 
 
Next, let's determine how fast the ball and the clay were traveling at the moment that they collided.
 
clay (descending) ball (ascending)
vf = vo + at vf = vo + at
vf = 0 +(-9.8)(0.505) vf = 9.9 + (-9.8)(0.505)
vf = -4.95 m/sec  vf = +4.95 m/sec

  
The velocity-time graph for this problem is shown below. Each line has the same slope of -9.8 m/sec2. The blue line starts at 9.9 m/sec and ends at 4.95 m/sec while the purple line starts are 0 m/sec and ends at -4.95 m/sec. Note that the blue area between the line and the x-axis must equal 3.7 meters while the purple area between the line and the x-axis must equal 5.0 - 3.7 or 1.3 meters. The combined blue and purple areas must equal the entire original gap, 5.0 meters.  
 
 
 
Practice your skills with this type of problem by completing the accompanying worksheet entitled Chase Problems: Projectiles.



 
Related Documents




PhysicsLAB
Copyright © 1997-2017
Catharine H. Colwell
All rights reserved.
Application Programmer
    Mark Acton