Conservation of Momentum Lab
Part 1: Elastic collision
Materials: ramp, meter stick, scales, glass and metal balls
Procedure: take the following measurements and calculate the time of free fall.
Momentum Before Collision:
STEEL SPHERE (kg) : GLASS SPHERE (kg) :
HEIGHT (m): TIME (s): t = Ö (2dy / g)
Release the steel sphere and let it roll down the apparatus and record distance traveled:
TRIAL 1: TRIAL 2: TRIAL 3: AVERAGE:
Calculate the velocity and calculate the momentum.
VELOCITY u1 = = dx / t MOMENTUM = = m1u1
Repeat the procedure, this time with the glass ball being hit by the steel ball.
m1u1 + m2u2 = m1v1 + m2v2
Why is m2u2 zero?
Momentum After Collision
Steel ball / dx= / V1 =
Glass ball / dx= / V2 =
Total system momentum before collision: m1u1
Total system momentum after collision: m1v1 + m2v2
Conclusions:
Is the value for the total momentum of the system of the two balls after the collision within 10% of the momentum of the system before the collision?
You found the total momentum of the steel and glass balls after the collision by adding them together. Why didn't you subtract the values?
Is the momentum of the steel ball greater before or after the collision? Why did it increase/decrease?
Why can you use the free-fall time to calculate the horizontal velocities of the spheres?
Suppose you used another ball instead of the glass ball. After the collision takes place, the new ball falls to the floor, while the steel ball rebounds and travels back up the curved track. Describe at least one property of the new ball that makes this scenario possible, and explain your reasoning.
Part 1: Push Carts
Materials: 2 carts, metric tape, stop watch
Procedure: People of different masses push each other apart
Measure distance traveled and time to stop
Data:
Cart (kg) / Person (kg) / Time (s) / Distance (m) / Accel (m/s2) / Vel (m/s)Calculations:
a = 2.d / t2 vi = a.t = 2.d / t
v1 = a1.t1 = 2.d1 / t1
v2 = a2.t2 = 2.d2 / t2
Momentum:
0 = m1v1 + m2v2
Conclusions:
Is momentum conserved? Explain.
Is kinetic energy conserved? Explain
What would happen if only one person pushed?
Is this an elastic or inelastic collision?
What would happen if the people tossed a heavy object between them?
Why are the people in this picture smiling?
Part 2: Ballistic Pendulum
Momentum Conserved in ALL collisions
Total Energy Conserved in ALL collisions
Momentum of Ball when fired must equal combined Momentum of Pendulum + Ball just after ball hits pendulum
Neglecting friction, the Kinetic Energy of the Pendulum + Ball at the moment of getting attached must get totally converted to the Potential Energy of Pendulum + Ball when it swings to maximum possible height
Mass of ball = m1 = kg, u1 =?
Mass of pendulum = m2 = 0.116 kg, u2 = 0 m/sec
Gain in height = h =
Gain in PE = (m1 + m2).g.h = ½(m1 + m2) v2 or v = (2.g.h)1/2
m1.u1 = (m1 + m2) .v
Velocity of ball u1 =
Kinetic energy conserved? Momentum conserved?
Part 4: Inelastic collisions
Using dynamic carts on a low-friction track and the motion detector, measure the velocity of two carts before and after an inelastic collision. Using the formulas from the chapter, determine whether or not momentum and kinetic energy is conserved.
Trial A: two carts of equal mass
Momentum:
m1u1 + m2u2 = m1v1 + m2v2
Kinetic Energy:
KE1 = ½ m1 v12 KE2 = ½ m2 v22
Trial B: heavy cart strikes light cart
Momentum:
m1u1 + m2u2 = m1v1 + m2v2
Kinetic Energy:
KE1 = ½ m1 v12 KE2 = ½ m2 v22
Trial C: light cart strikes heavy cart
Momentum:
m1u1 + m2u2 = m1v1 + m2v2
Kinetic Energy:
KE1 = ½ m1 v12 KE2 = ½ m2 v22
Conclusions:
How is this activity similar to the way trains are put together for the Metro?
What happens to the kinetic energy that is lost in these collisions?
Compare this exercise to the one you just completed with the ballistic pendulum.