Presenter: Tony Wayne, Albemarle High School, firstname.lastname@example.org.
PH.7 The student will investigate and understand properties of fluids. Key concepts include
density and pressure.
Fluids and pressure
Shows that pressure exerts a force and balance of pressure
The bazooka shoots a ping pong ball. Don’t have anyone stand in front of it.
Get a student volunteer to place the pingpong ball in the rear of the bazooka. Have her place her hand over the rear end to seal it. In the front end place a NEW index card. (“New,” means no folds or creases in the card). Turn on the vacuum. After about 8 seconds, on your signal, the student QUICKLY removes her hand. The pingpong ball will accelerate to the front, pushing the card off the front as it exits the barrel. With a 1 hp vacuum it will leave at about 8 m/s.
How the physics is demonstrated
When the student removes her hand. The pressure is unbalanced. The 1 atmosphere of pressure at the rear of the bazooka pushes the ball forward.
Construction and Tips Regarding the Demonstration
The bazooka is made from a piece of strong PVC pipe the same diameter at ping pong ball. The pipe can be found at a hardware store. It costs about $3 (2004 price). Take a pingpong ball with you and look for a thick walled pipe that the ball snugly fits in. Mine is a 5 foot long pipe. Buy a “T” section that fits the pipe. This will go on the front of the bazooka where the ball exists. Wrap pieces of duck tape around the vacuum hose so it makes a snug fit in the “T” section of pipe. It will fit into the section that points 90° from the pipe’s axis. I use the cheapest SEARS Shop Vac. It’s 1 hp and about $30 (2004 price). This does the job and it small enough to store.
Here is a calculation you can do.
The pingpong ball’s diameter is 3.8 cm. It’s area is p(0.019 m)2.
The pingpong ball’s mass is 2.80 grams (0.00280 kg)
Assume the pressure in the bazooka, when the vacuum is turned on is 0.90 atm.
The pressure behind the pingpong ball, when the hand is removed is 1.0 atm
How fast is the ball moving after traveling 1.5 meters?
F = (1.0 atm – 0.90 atm)(1.01x 105 Pa/atm) p(0.019 m)2 = 0.00789 N
Find the acceleration from F=ma
0.00789 N = (0.00280 kg)a
a = 2.81 m/s2
Use kinematics to find the velocity.
V2 = (Vo)2 + 2ax
V = 4.6 m/s
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