Investigating Momentum

Investigating Momentum

AIM To determine the momentum of a system before and after an interaction, and to compare these two quantities.

The interaction in this investigation is a mutual force propelling the two trolleys in opposite directions. The force is produced using the spring-loaded propulsion device on one of the trolleys.

METHOD Set up the following apparatus. Compress the spring-loaded propulsion device on the left hand trolley and tie secure the trolleys together using a loop of string. Set up a camera to record the motion of the trolleys when they move apart when the propulsion device is triggered. Do this by burning the string holding the two trolleys together. Ensure that a scale is present in the area to be photographed e.g. a metre rule.

In this investigation, both trolleys are initially stationary and they are accelerated by the release of the spring-loaded propulsion device on one of the trolleys. A camera is used to record a movie that can be analysed using LoggerPro.

Measure and record the masses of the two trolleys.

Trolley on left, mass ______Additional mass on left trolley ______

Trolley on right, mass ______Additional mass on left trolley ______

ANALYSIS Use LoggerPro to analyse the movie to determine the speeds of the trolleys after the interaction. Since the trolleys are initially accelerating as the spring pushes them apart, do not use this part of the record. LoggerPro can be accessed via the Start menu and it is in the Science folder.

From this information, and the measured masses of the two trolleys, find the momentum after the "explosion”. Compare the sum of these values with the total initial momentum and comment on this result in terms of the law of conservation of momentum (you should state this law). Make a conclusion appropriate to the aim of the investigation.

Repeat the analysis using a second movie to establish reliability of your conclusion.

Complete the questions on the following page as a part of this report.

Questions

1. Using Newton's second law, F = ma and the definition of acceleration, determine an alternative expression for this law in terms of momentum.

2. Define the term IMPULSE and state the two equivalent units used for this quantity.

3. Compare the impulse of each trolley.

Set out the answers to the following questions carefully. Write (a) the relevant equation in symbolic form (b) substitute without calculation (c) complete the calculation and (d) include the units in the answer

4. Calculate the momentum of a 60 kg person running at 10 m s-1.

5. How fast would a 1000 kg car travel to have the same momentum a 60 kg person running at 10 m s-1?

6. What is the momentum of a 20 g bullet travelling at 500 m s-1?

7. A 360 g trolley moving at 2.4 m s-1 collides and coalesces with a 200 g trolley travelling in the opposite direction at 1.8 m s-1. What is the resultant velocity of the trolleys?

References: These should be read carefully. Complete the activities in each site.

http://www.physicsclassroom.com/Class/momentum/U4L2b.html

http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/momentum/u4l1a.html

Momentum Conservation

One of the most fundamental laws in physics is the law of momentum conservation. The law of momentum conservation can be stated as follows.

For any interaction occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

The above statement tells us that the total momentum of a collection of objects (a system) is conserved" - that is the total amount of momentum is a constant or unchanging value.

Typical Results

Here are two images from a typical movie of this “explosive” interaction. They are separated by 0.281 s.

Here is the analysis of the movies, done using Logger Pro.

Answers to questions

1. Using Newton's second law, F = ma and the definition of acceleration, determine an alternative expression for this law in terms of momentum.

Newton’s second law: F = ma

Substituting the definition for acceleration, a = ∆v/∆t, into this expression.

F = m(∆v/∆t) = m(v-u)/t = (mv-mu)t

Hence the net force is equal to the rate of change of momentum.

Clarification: the expression (mv-mu) is the final momentum minus the initial momentum – i.e. the change in momentum – and dividing it by the time that it takes gives us what physicists would call “the rate of change of momentum”.

2. Define the term IMPULSE and state the two equivalent units used for this quantity.

Impulse is defined as the change in momentum.

Since F = (mv-mu)t from the above …

Impulse can be calculated from the expression Ft = mv=mu

Hence two equivalent units for impulse can be seen from this equation.

From the left hand side of the equation the units of impulse are newton seconds (N s) and this is equivalent to the unit for momentum (change) on the right hand side of the equation, kilogrammetres per second (kg m s-1)

3. Compare the impulse of each trolley.

Since each trolley was initially at rest, each trolley had an initial momentum of zero.

Hence the change in momentum of each trolley is equal to its final momentum.

Right trolley: change in momentum = 0.93 kg m s-1

Left trolley: change in momentum = -0.94 kg m s-1

NB: if the results were exactly consistent with the law of conservation of momentum, the change in momentum (impulse) of each trolley would be the same magnitude and in opposite directions.

There is a clear connection between Newton’s third law and the law of conservation of momentum.

In this specific situation, Newton’s third law states “The force produced on the right hand trolley by the plunger pushing them apart is equal and opposite to the force produced by the plunger on the left hand trolley”.

Since equal forces act in opposite directions on the trolleys for the same amount of time, the impulse, Ft, must be the same for both.

The law of conservation of momentum is equivalent to Newton’s third law – it is just an alternative way of stating the same physics!

4. Calculate the momentum of a 60 kg person running at 10 m s-1.

p = mv = 60 x 10 = 600 kg m s-1

5. How fast would a 1000 kg car travel to have the same momentum a 60 kg person running at 10 m s-1?

p = mv

600 = 1000v

v = 0.6 m s-1

6. What is the momentum of a 20 g bullet travelling at 500 m s-1?

p = mv = 20 x 10-3 x 500 = 10 kg m s-1

7. A 360 g trolley moving at 2.4 m s-1 collides and coalesces with a 200 g trolley travelling in the opposite direction at 1.8 m s-1. What is the resultant velocity of the trolleys?

The 360 g trolley

p360 = mv = 0.36 x 2.4 = 0.864 kg m s-1

The 200 g trolley

p200 = mv = 0.2 x -1.8 = -0.360 kg m s-1

(the minus sign indicates that the velocity is in the opposite direction)

The total initial momentum = p360 + p200 = 0.864 + (-0.360) = 0.504 kg m s-1

The final momentum, pf = (m360 + m200) x v and this must be equal to the initial momentum.

(m360 + m200) x v = 0.504

(0.36 + 0.2)v = 0.504

0.56v = 0.504

v = 0.9 m s-1