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# An Experiment to Determine Values for Velocity and Momentum

• Subcategory: Marketing
• Topic: Value
• Pages: 2
• Words: 1007
• Published: 03 August 2018 Download Print

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In this experiment we determined values for kinetic energy and momentum of a system both before and after elastic and inelastic collisions. Using the values for initial velocity and mass we then calculated values for the final velocities and compared that to our experimental values.

In case of elastic collisions, what would happen if m1 is much, much larger than m2? Conversely, what would happen if m1 is much, much smaller than m2?

If m1 is much larger than m2 the loss of momentum for m1 will be close to 0. For our experiment m1 will have a final velocity that is very close to it’s initial velocity (consistent with the maintained momentum). The final velocity of m2 will be greater than m1, as it experiences the force of m1. If m1 is much smaller than m2 it will maintain most of it’s momentum from impacting m2, but the velocity of m1 will change directions. The final velocity of m2 will be close to 0 as the force of m1 will not be great enough to overcome static friction.

In your lab manual the equations 4.4 and 4.5 give the final velocities of the two objects in terms of the initial velocity and the masses. Now, are the final velocities you found in your trials consistent with these equations?

For trial 1 of elastic collision, v0 = 0.3 m/s, and m1 = 250g and m2 = 250g, the formula for final velocity of m1, (v1) is calculated as:

v1 = v0*((m1-m2)/(m1+m2))

v1 = 0.3*((250-250)/(250+250))

v1 = 0

v2 is calculated using the formula for v2:

v2 = 2*v0*(m1/(m1+m2))

v2 = 2*0.3*(250/500)

v2 = 0.3

The experimental value for v1 and v2 of trial 1 were 0 and 0.29 respectively, these values are consistent with the calculated values shown above.

For trial 2 of elastic collision, v0 = 0.36 m/s, and m1 = 500g and m2 = 250g, the formula for final velocity of m1, (v1) is calculated as:

v1 = v0*((m1-m2)/(m1+m2))

v1 = 0.36*((500-250)/(500+250))

v1 = 0.12

v2 is calculated using the formula for v2:

v2 = 2*v0*(m1/(m1+m2))

v2 = 2*0.36*(500/750)

v2 = 0.48

The experimental value for v1 and v2 of trial 1 were 0.11 and 0.46 respectively, these values are consistent with the calculated values shown above.

For trial 3 of elastic collision, v0 = 0.29 m/s, and m1 = 250g and m2 = 500g, the formula for final velocity of m1, (v1) is calculated as:

v1 = v0*((m1-m2)/(m1+m2))

v1 = 0.29*((250-500)/(250+500))

v1 = -0.10

v2 is calculated using the formula for v2:

v2 = 2*v0*(m1/(m1+m2))

v2 = 2*0.29*(250/750)

v2 = 0.19

The experimental value for v1 and v2 of trial 1 were -0.07 and 0.17 respectively, these values are consistent with the calculated values shown above.

Using the velocities, make a TABLE for the momentum and kinetic energy of each PAScar before and after collision? Calculate the percent difference between TOTAL final and initial momenta and kinetic energy in each trial, and comment on the conservation of momentum and conservation of energy.

 Initial Final % difference Trial p Ke p Ke % difference p % difference Ke Elastic m1=m2 75 22.5 72.5 21.025 3.33% 6.56% Elastic m1>m2 180 64.8 170 58.95 5.56% 9.03% Elastic m1

For the elastic collision trials momentum and kinetic energy are mostly conserved, ideally they would be completely conserved, but our system is not isolated. Other forces are present which prevent us from obtaining perfect results. In the inelastic collision trial momentum is still conserved (mostly) even though kinetic energy is lost, this is because momentum is not converted as kinetic energy is. The impact of the cars in the inelastic trial converts over half of the kinetic energy into thermal energy. For the elastic collision trials we do not observe this loss except for in trial 3 where the force of the magnets is not strong enough to keep the vehicles from impacting.

In the inelastic collision, why do you need to measure only one final velocity?

Because the masses are joined together upon impact which quickly equalizes their velocities. As the joined masses travel along the track yielding a single final velocity.

In the inelastic collision, why do you think the kinetic energy is not conserved?

The energy is converted to another form of energy (thermal) during the collision. Specifically, it is because the two objects are impacting one another that the kinetic energy is converted and not conserved.

What physical law(s) predicts the conservation of momentum? Explain briefly and clearly.

This is predicted by Newton’s 3rd law which talks about action-reaction pairs. The sum of the forces in a closed system will be 0 as each force has a force that is directly counter to it. For momentum this holds true as well because momentum is essentially a product of force.

In the case of inelastic collisions, what would happen if m1 is much, much larger than m2?

Conversely, what would happen if m1 is much, much smaller than m2?

The formula for the final velocity of an inelastic collision is as follows:

v12 = v0*(m1/(m1+m2))

Looking at the formula you can see that as you increase m1 to much larger values, m2 becomes insignificant and m1/(m1+m2) begins to approach 1, this also means that it will approach the final velocity (v12). For our purposes this indicates that the final velocity will nearly equal the initial velocity. If m1 is much smaller than m2 then m1/(m1+m2) will begin to approach 0 as will the final velocity.

Imagine the PAScars m1 and m2 are both on the track, at rest, and with their bumpers touching each other. The mass m1 = 2 m2. A firecracker is placed between the bumpers and explodes, sending the PAScars in opposite directions. What was the initial momentum of the system (before the explosion)? What can you say about the final momentum of the system?

The initial momentum is 0 as we know that p=mv, and the velocity of cars and firecracker was 0. The final momentum of the system will also be 0 if we were to sum each vector of momentum in the system.

### Conclusion

The calculated values for final velocity and momentum had very low percentage error when compared to theoretical values. This is sufficient enough to illustrate the concepts of the lab. Our system for collisions was imperfect in that is was not a closed system and was therefore susceptible to outside forces. Remember: This is just a sample from a fellow student.

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