By Tom Schuster, Peter Park, and Jason Schachter

1
The Completion of a Circuit Using the Potential
Energy of a Spring Lab

• By Tom Schuster, Peter Park, and Jason Schachter
• In loving memory of
• Todd Greenberg

2
The Setup
3
What We Want

• To determine the spring constant using the
  concepts of conservation of energy, conservation
  of momentum, and projectile motion.
• To determine the resistance of a bulb using Ohms
  Law.

4
The Way it Works

• The compressed spring is released, firing the
  first ball into a collision with the second ball
  of equal mass.
• In order to conserve momentum, the second ball is
  sent accelerating down the track.
• The ball goes over a ramp into projectile motion
  with an initial velocity of both vertical and
  horizontal components.
• The ball drops into a funnel lined with aluminum
  foil that is connected to a circuit, which is
  then completed.
• When the circuit is completed, the bulb becomes
  illuminated.

5
What We Need (part 1)

• Mass of each metal ball (m1, m2) (Triple-beam
  balance)
• Distance between the top of the track and the
  ground (yi) (meter stick)
• Distance between the top of the ramp and the
  ground (yf) (meter stick)
• Horizontal distance between the ramp and the
  bottle (?x) (meter stick)
• Time for the ball to travel from the ramp to the
  bottle (t) (stopwatch)
• Angle of the slope of the ramp (?) (protractor)
• Stretch distance of spring (x) (meter stick)

6
What We Need (part 2)

• Voltage of the battery (Vt) (voltmeter)
• Current of the circuit (it) (ammeter)
• Resistance of the ball (Rball) (resistance meter)

7
The Physics (part 1)

• Vx?x/t
• VrVxcos?
• .5mVi2mgyi .5mVf2mgyf
• (m1m2, Vi20 m/s, Vf10m/s)
• mVi1mVi2mVf1mVf2
• .5kx2 .5mVf2
• k?

8
The Physics (part 2)

• VtitRt
• RtRballRammeterRbulb
• Rbulb?

9
What We Got
10
The Physics Again (part 1)

• Vx(0.74 m)/(.14 s)
• VxVcos(35º)
• .5(0.0671 kg)Vi2(0.0671 kg)(9.8 m/s2)(0.916 m)
  .5(0.0671 kg)Vf2(0.0671 kg)(9.8 m/s2)(0.17 m)
• (m1m2, Vi20 m/s, Vf10m/s)
• (0.0671 kg)Vi1(0.0671 kg)(0 m/s)
• (0.0671 kg)(0 m/s)(0.0671 kg)Vf2
• .5k(0.028 m)2 .5(0.0671 kg)Vf2
• k?

11
The Physics Again (part 2)

• (1.4 V)(0.12 A)Rt
• Rt(4.5 ohms)(5.1 ohms)Rbulb
• Rbulb?

12
The Procedure is Answered!

• Vx.51 m/s
• Vr.62 m/s
• V23.97m/s
• V13.97 m/s
• k 2312.15 N x M
• Rbulb2.1 ohms

13
What Could Have Went Wrong

• Collision between the two balls may not have been
  perfectly elastic, as the surface may have been
  sloped during the collision.
• The ball experienced friction that was not
  accounted for on the track.
• Human error when using the stopwatch.
• -Because the time being determined was so
  small, the possibility of error is high.