Experiment 8: Conservation of Energyand Linear Momentum

Figure 8.1: Ballistic Pendulum and Equipment

EQUIPMENT

Ballistic Pendulum30 cm RulerDigital BalanceTriple-Beam Balance

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2 Experiment 8: Conservation of Energy and Linear Momentum

Advance Reading

Text: Conservation of Energy, Conservation of Lin-ear Momentum, Mechanical Energy, Kinetic Energy,Gravitational Potential Energy, Elastic Potential En-ergy, Elastic and Inelastic Collisions.

Objective

To determine the velocity of a ball as it leaves the bal-listic pendulum using conservation of linear momen-tum and conservation of energy considerations; to per-form error analysis.

Theory

Energy is always conserved. There are different formsof energy, and one form of energy may be transformedinto another form of energy. Mechanical energy, aform of energy being investigated today, is not alwaysconserved.

The mechanical energies being investigated today arekinetic energy, KE, and gravitational potential energy,PEgrav:

KE =1

2mv2 (8.1)

where a mass, m, has energy due to its speed, v, and

PEgrav = mgh (8.2)

where a mass m has energy as a result of its position(height, h), and g is acceleration due to gravity.

Linear momentum, �p, is always conserved in an iso-lated system. An isolated system is a system in whichall forces acting on the system are considered.

Linear momentum p is given by:

�p = m�v (8.3)

where a mass, m, has a velocity, v. There are threedistinct categories of collisions: elastic, inelastic, andcompletely inelastic.

Elastic collisions result in conservation of both linearmomentum and mechanical energy. Billiard balls areoften used as examples when discussing elastic colli-sions.

Inelastic collisions result in deformation of one or moreof the objects involved in the collision. Although lin-ear momentum is conserved, mechanical energy is not.Car wrecks are examples of inelastic collisions.

Completely inelastic collisions refer to collisions thatresult in objects becoming attached to each other after

the collision (i.e., stuck together). The objects thusmove, after collision, with the same velocity. Thesecollisions, like inelastic collisions, conserve linear mo-mentum but not mechanical energy.

This experiment investigates completely inelastic colli-sions to determine the initial velocity of the ball. Whenthe ballistic pendulum is fired, the ball is caught andheld by the catcher; the two objects, ball and catcher,move together as one object after collision. We assumethat the linear momentum of the system before and af-ter the collision is conserved, and that no energy is lostduring the ball’s flight.

First Process:

Conservation of linear momentum between state 1 andstate 2 (just before and just after a collision) is:

�p1 = m�v1 = �p2 = m�V2 (8.4)

Relevant to this experiment, we consider a collisionbetween a ball of mass m and a catcher of mass M .Before the collision, state 1, the velocity of the catcheris 0.0 m/s. Just after the collision, state 2, the velocityof the ball and the catcher are equal.

�p1 = m�v1 = �p2 = (m+M)�V2 (8.5)

Second Process:

While mechanical energy is not conserved during an in-elastic collision. However, after the collision, the ball-catcher system has KE due to its motion. We assumethat mechanical energy is conserved (i.e., ignore rota-tional energy and energy losses due to friction). KE istransformed into PEgrav as the pendulum arm swingsup to a height h.

The initial mechanical energy is all KE, as the pen-dulum arm is at the lowest possible position. The fi-nal energy is all PEgrav when the pendulum rises andstops, state 3. Therefore, conservation of mechanicalenergy is:

KE =1

2(m+M)V 2

2 = PEgrav = (m+M)gh (8.6)

When a pendulum rotates θ◦, the center of mass, cm,rises an amount h = hf − hi. Refer to Fig. 8.2.

By measuring the change in height, h, of the center ofmass of the pendulum arm, PEgrav can be determined.A mark is scribed on the pendulum arm, just abovethe catcher. Once h is determined, one can calculateV2 using Eq. 8.6, which, in turn, allows calculation ofv1. Note the use of capital V for V2; this is to distin-guish it from the notation used for the initial velocityof the ball when fired from detent 2, v2.

Experiment 8: Conservation of Energy and Linear Momentum 3

Figure 8.2: The change in height of the center of massis h ≡ hf − hi

4 Prelab 8: Conservation of Energy and Linear Momentum

Name:

1. State the Law of Conservation of Linear Momentum. (10 pts)

2. State the Principle of Conservation of Mechanical Energy. (10 pts)

3. Solve the following equations (2 equations with 2 unknowns) for x in terms of: m, g, h. Refer to Appendix A: MathReview if necessary. (10 pts)

6x = 9y

5y2 = mgh

4. Solve the following equations (2 equations with 2 unknowns) for x in terms of: m, M , g, h. (20 pts)

mx = (m+M)y

12 (m+M)y2 = (m+M)gh

(continued on next page)

Prelab 8: Conservation of Energy and Linear Momentum 5

5. Solve Eq. 8.5 and Eq. 8.6 (2 equations with 2 unknowns) for v1 in terms of: m, M , g, h. (20 pts)

mv1 = (m+M)V2 (Eq. 8.5)

12 (m+M)V 2

2 = (m+M)gh (Eq. 8.6)

6. You shoot a ball, m = 50.0 g, into a catcher, M = 200.0 g; the center of mass rises 15.0 cm. Calculate v1. Referto your answer for Question 5. (20 pts)

7. You will fire the spring gun 3 times from the first detent and measure the change in height of the (pendulum +ball) for each shot. Write the equation for the change in height of the first shot. (10 pts)

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Name: Section: Date:

Worksheet - Exp 7: Conservation of Energy and Linear Momentum

Objective:This experiment uses principles of conservation to determine the velocity of a ball as is leaves the ballistic pendulum.

Theory: Energy is always conserved in one form or another. However, there are many forms of energy, and one formof energy may be transformed into another kind of energy. Thus, the mechanical energy studied in this experimentmight not always be conserved. Relevant energies for this experiment are kinetic energy and gravitational potentialenergy:

KE = 12mv2 A mass, m, has energy due to its velocity, v.

PEgrav = mgh An object has energy due to gravity, g, and its elevation, h.

Momentum is always conserved in an isolated system, i.e. - one in which all forces are considered. Linear momentumis given by �p = m�v. Momentum and velocity are both vector quantities, and their direction must be consideredwhen adding the motion of multiple objects.

1. A running back weighing 100 kg runs headlong into a defender weighing 120 kg. The two players are movingat 7 m/s and 5 m/s, respectively. Calculate the total momentum and the total kinetic energy of the system.[Calculations including velocity should reflect direction of travel.]

pi = (2 pts)

KEi = (2 pts)

pf = (2 pts)

KEf = (2 pts)

The two players collide and continue moving tothe right. Calculate the momentum and kineticenergy of the system after the collision. [Hint:Assume the two players are an isolated system.]

2. Which of the above quantities is/are conserved? (4 pts) What kind of collision is this? (3 pts)

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Procedure:3. Measure the mass of the ball (m) and

catcher arm (M). [Unscrew the knurled nut to remove the arm; screw it back when finished.] (4 pts)

m = M =

4. Measure the radius to the center of mass of the ball/pendulum system. The center of mass is the top of theopening of the ball catcher.

r =

5. Fire the ball into the catcher and calculate the change in height of the ball/pendulum system.Δh = r − r cosΘ

6. Record Δh for 3 trials at the first detent (short range). Calculate the average Δh for this detent. Record it inthe table provided. Repeat for detent 2 (medium range) and detent 3 (long range).

Detent #1 Detent #2 Detent #3

Trial

#1

Trial

#2

Trial

#3

Average

Δh

7. As the pendulum arm rises, its energy changes from one form to another. a) Describe this transformation ofenergy. (5 pts) b) How can this process be visually observed? (5 pts)

8. Considering the average Δh during this process, how much gravitational potential energy is being stored in theball + pendulum with each fire?

PE1 = (6 pts) PE2 = (6 pts) PE3 = (6 pts)

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9. Conservation of energy dictates that the total energy remains constant throughout the motion of the pendulum[the force of friction is ignored]. Use this knowledge of energy conservation to determine the velocity of theball + pendulum just after collision. Show work below and record in the table.

θ

KE + PE final

= KE + PE initial

10. Consider the collision of the ball and pendulum that resulted in the motion described above. Are both momentumand energy conserved in this inelastic process? (6 pts) You may wish to refer to the example on the first page.

11. Write an equation relating the value of the conserved quantity (or quantities) before and after the collision. (5 pts)Calculate the initial velocity of the ball before collision, vi. Show work below and record in the table.

Average Δh vb+p After collision vi Firing Velocity

(3 pts ea.) (5 pts ea.) (4 pts ea.)

Detent

#1

Detent

#2

Detent

#3

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12. When fired from the first detent, what percent of the ball’s initial energy remains in the ball/pendulum systemafter the collision? (6 pts) Where did the dissipated energy go? (6 pts)

13. Imagine that the ball, fired from the first detent, transfers 100% of its kinetic energy to the pendulum. How highwould the pendulum swing with this extra energy, given that it is not carrying the mass of the ball? Solve usingenergy equations only. (6 pts)