Inelastic Collisions

Purpose:

To analyze the motion of two low friction carts during an inelastic collision and verify that the law of conservation of linear momentum is obeyed.

Equipment:

Computer with logger pro software, lab pro, motion detector, horizontal track, two carts, 500 g masses (3), triple beam balance, bubble level

Introduction:

fig 1

In Fig 1 is the setup of the experiment. If we take the two carts to be an isolated system, then we can conclude momentum will be conserved. If the two carts experience a completely inelastic collision, then the law of conservation of momentum says:

P_i = P_f

m₁v₁ + m₂v₂ = (m₁ + m₂)V

Where v₁ and v₂ are the velocities of each cart before collision and V is the velocity of the combined mass after the collision.

Procedure:

1) Set up the apparatus as shown in fig 1. Use the bubble level to verify that track is level. Measure the mass of each object:

m₁ = 0.5024 kg

m₂ = 0.5134 kg

Connect the motion detector to the computer and open logger pro.

2) Check to make sure the motion detector is working properly by pressing collect then moving the cart nearest the detector back and forth. Does it provide a reasonable graph of position vs time?

Yes, as the cart is moved away from the detector, the functional value increases steadily, and as the cart is moved towards the detector, the functional value decreases. 

Position the carts so that their velcro pads are facing each other.

3) With the second cart (m₂) at rest, give the first cart (m₁) a gentle push away from the detector. Observe the position vs. time graph before and after the collision. Pictured in fig 2 is an educated guess as to what the position vs. time graph should look like.

fig 2

What should these graphs look like? 

The slope of the graph after the collision should be less than the slope of the graph after the collision. Pictured in fig 2 is my prediction of what the graph should look like.

The slope of the position vs. time graphs will give us our velocities directly before and after the collision. To avoid dealing with friction, we will find velocities at the instant before and after the collision. 

Is this a good approximation, why or why not?

Yes, because it will give us the instantaneous velocity directly before and after impact. 

Select a very small range of data directly before the collision and apply a linear fit to the range. Record the slope (velocity) of this line. Repeat for a small range of data points directly after the collision. 

v₁ = 0.4171 m/s

V = 0.2305 m/s

Pictured in fig 3 is the position vs. time graph of one trial run we completed.

fig 3

4) Repeat for two more collisions. Calculate the momentum of the system the instant before and after the collision for each trial and find percent difference.

The folloing calculations are for the first trial in this part of the expirement.

P_i = P_f

m₁v₁ + m₂v₂ = (m₁ + m₂)V

where:

m₁ = 0.5024 kg

m₂ = 0.5134 kg

v₁ = 0.4171 m/s

V = 0.2305 m/s

P_i = (0.5024 kg)( 0.4171 m/s) + (0.5134 kg)(0)

P_i = 0.21 kg m/s

P_f = (0.5024 kg + 0.5134 kg)(0.2305 m/s)

P_f = 0.23 kg m/s

Percent difference:

[(0.23 kg m/s - 0.21 kg m/s)/ 0.23 kg m/s]*100 = 10.5%

Pictured in fig 4 is data table of each trial and the momentum calculations

fig4

5) Add 500 g of mass to cart number 2 and repeat steps 3 and 4. 

m₂ = 1.01 kg

Pictured in fig 5 is the data table for the additional mass on cart two.

fig 5

What do the graphs of velocity vs. time, and acceleration vs. time look like?

Pictured in fig 6 are my predictions of what velocity vs. time and acceleration vs. time should look like.

fig 6

6) remove the mass from cart two, and now add the mass to cart one.

m₁ = 1.00 kg

Pictured in fig 7 is the data collected for the trials with mass added to cart one. 

fig 7

The average of all the percent differences that we had found was 10 % difference from the law of conservation of momentum. 

How well is the law obeyed based on the results of your experiment? 

The law of conservation of momentum was reasonably obeyed. Although we did encounter some larger percent differences, it may have been due to the range of data that was selected.

8) For each of the above nine trials, calculate the kinetic energy of the system before and after the collision. Find the percent kinetic energy lost during each collision. Show sample calculations here:

∆K/K_i * 100 = percent difference

[(K_f - K_i) /K_i ]*100

[(1/2*(m₁ + m₂)*V² – ½* m₁*v₁²)/ ½* m₁*v₁²]*100

[((m₁ + m₂)*V²)/ m₁*v₁²) – 1]*100

[((0.5024 kg + 0.5134 kg)* 0.2305² m/s)/ 0.5024 kg * 0.4171² m/s) – 1]*100 = 38 % difference

Pictured in fig 8 is the data showing the kinetic energy before and after collision, as well as the percent kinetic energy lost.

fig 8

9) Do theoretical calculations for ∆K/K_i * in a perfectly inelastic collision 

        1- a mass (m) colliding with an identical mass (m) initially at rest

Theoretical Calculation 1

        2- a mass (2m) colliding with a mass (m) initially at rest

Theoretical Calculation 2

        3- a mass (m) colliding with a mass (2m) initially at rest

Theoretical Calculation 3

Conclusion:

In completing this lab, one was able to analyze the motion of two low friction carts during an inelastic collision and verify momentum is conserved. I learned that if we can take a system where there are no outside forces acting within the system, the momentum will be conserved. Possible sources of error in this lab would include the following:
       1) Selecting a bad range of data from the position vs. time function, resulting in an inaccurate velocity value. 
       2) The carts may have taken a few milliseconds to actually stick together. 
       3) Friction was present in the system, resulting in a slight loss of momentum.
Further investigations that could be followed after this lab would be to determine the effect of the impulse on the momentum, or maybe even taking friction into account.