Physics Including Human Applications 122 Chapter 6 Momentum and Impulse GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational definition: impulse elastic collision impulsive force inelastic collision momentum rocket propulsion Impulse Problems Use the relationship between impulse and change in momentum to solve problems. Conservation of Momentum Explain the principle of conservation of momentum. Collision Problems State the difference in conditions between an elastic impact and an inelastic impact, and use both kinds of conditions to solve problems. Momentum and Energy Problems Apply the principles of conservation of momentum and conservation of energy to solve problems. PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 3, Kinematics, Chapter 4, Forces and Newton's Laws, and Chapter 5, Energy.
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Chapter 6 Momentum and Impulse
GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational definition: impulse elastic collision impulsive force inelastic collision momentum rocket propulsion Impulse Problems Use the relationship between impulse and change in momentum to solve problems. Conservation of Momentum Explain the principle of conservation of momentum. Collision Problems State the difference in conditions between an elastic impact and an inelastic impact, and use both kinds of conditions to solve problems. Momentum and Energy Problems Apply the principles of conservation of momentum and conservation of energy to solve problems. PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 3, Kinematics, Chapter 4, Forces and Newton's Laws, and Chapter 5, Energy.
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Chapter 6 Momentum and Impulse
6.1 Introduction The manager of the major league baseball team, which had just won its tenth straight game, was quoted as saying, "Well, we've really got the momentum now. I think we can win the pennant." After the Judiciary Committee of the House of Representatives had passed three articles of impeachment against President Richard M. Nixon in 1974, a spokesman for the House was reported to have said, "The momentum for impeachment is building up." If you suddenly decided to take a long trip would your friends say that you were acting on an impulse? In fact, if you are noted for making and acting upon decisions quickly you are probably described by others as an impulsive person. What are the connotations of these two words, momentum and impulse? How does a baseball team have momentum after ten victories and not after one victory? How does the action of a representative political body have momentum after three articles? Would it also have momentum after one article? And your sudden decision. Suppose you had waited twice as long to make up your mind. Would you still have been acting on an impulse? The idea of momentum conveyed by the examples above does imply the motion of something and also has a quantitative quality. The idea of impulse conveys the feeling of quickness, of doing something in a short amount of time. How are such ideas related to the use of the definitions of momentum and impulse in physics? Are they similar or contradictory? We hope you will be able to answer that question when you finish this chapter.
6.2 Momentum and Impulse Suppose we have two blocks moving to the right on a frictionless surface in the same line, with block A moving faster than block B see Figure 6.1. Block A will eventually come in contact with block B. While they are in contact, block A is exerting a force FBAon block B, and, in turn, block B is exerting a force FABon block A, but FABis in the opposite direction of FBA.
In accord with Newton's third law of motion, these two forces are equal in magnitude and oppositely directed. One can write FAB= -FBA. These forces are not constant. Both before and after contact these forces are zero. The magnitude of the net forces varies with time by starting at zero at the instant of contact, (t1), increasing to some maximum value, and then decreasing again to zero at the instant the blocks lose contact. A plot of
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the magnitude of FABor FBAas a function of time might be shown as in Figure 6.2. A force of this type which is a function of time is called an impulsive force. Can you name or give examples of impulsive forces? One example is kicking a football see Figure 6.3.
The product of the net force and the time interval over which it acts is called the impulse of the force. This is represented in Figure 6.2as the area under the curve. Because FABand FBAare equal in magnitude and opposite in direction, and the time of contact is the same for each of the bodies, the impulse of FABis equal in magnitude and opposite in direction to the impulse of FBA. If the force is constant, the area under the curve is given by the product of the magnitude of the force times the change in time, F(t2- t1) where t1and t2are the times that contact begins and ends, respectively. The impulse of a constant force is given by F(t2- t1). The dimensions of impulse in terms of M, L, and T are MLT-1. The units of impulse are newton-second (N-sec) in the SI system. Impulse is a vector quantity. You will recall that the unit of force (newton) is equal to mass times acceleration, force = (mass)(length)/(time)2. Therefore, the units of impulse are (force)(time) or (mass)(length)/(time). But (length)/(time) is the definition of velocity. So impulse has the same dimensions as the product of mass times velocity. We call this physical quantity the momentum of a body, momentum of a body = p= mv(6.1) Momentum is a vector quantity whose SI units are kg-m/sec.
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Momentum and impulse are both vector quantities that are given by the product of a scalar quantity and a vector quantity. The direction of impulse is determined by the direction of the net force, and the direction of momentum is given by the direction of the velocity. An impulse may be applied to a body to change its momentum. In fact the impulse applied to a body is equal to its change in momentum: impulse of a force = change in momentum of the body upon which the force acts, impulse = Δp (6.2) EXAMPLES1. A body of mass 1.60 kg is acted upon by an impulsive force as shown in Figure 6.4. What is the impulse of the force? What is the change in momentum of the body and the change in velocity? The impulse of the force is represented by the triangular area. Thus impulse = 1/2 (base of triangle)(height) = 1/2 x 800 x .010 = 4.00 N-sec in the positive direction The change in momentum is equal to the impulse, 4.00 N-sec. Δp = mΔv = 4.00 N-sec 1.60 Δv = 4.00 N-sec Δv = 2.50 m/sec in the positive direction
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2. Short periods of rapid deceleration are characteristic of physical impacts. The current state of knowledge of human tolerances for impacts is not very complete. In fatal collisions a head injury is usually the cause of death. If the head is subjected to an impulse of about 100 N-sec in an abrupt collision, death may ensue. Consider a motorcyclist riding without a helmet to cushion his fall. Assume the mass of his head is 6 kg. If he has an abrupt collision, how slow must he be going to survive? 100 N-sec = Δp = 6 kg Δv Δv = 16.7 m/sec, approximately 37 mph!
3. Given the force-time graph for a sprinter's start as shown in Figure 6.5, we wish to find the change in speed between t= 0.100 sec and t= 0.250 sec for a sprinter with a mass of 75.0 kg. The impulse is the area under the curve ABCDor the area of the closed polygon ABCDEFG. We can find this area by adding together the area of the triangle BCD, the area of the trapezoid ABDE, and the area of the rectangle AEGF. For the area of the triangle BCD, we obtain
impulse1= area1= 1/2 (base)(height) where the base is BDand the height is given by 1900 N - 900 N.
impulse1 = 1/2 (0.130)(1000) I1= 65.0 N-sec For the area of the trapezoid, we obtain impulse2= area2= 1/2 (height)(base1+ base2) = 1/2 (650 N)(0.130 sec + 0.150 sec) I2= 91.0 N - sec For the area of the rectangle AEFG, we obtain impulse3= area3= (base)(height) = (0.150 sec)(150 N) I3= 37.5 N-sec total impulse = (65 + 91.0 + 37.5) N-sec = 193.5 N-sec change in velocity = impulse/mass = 193 N-sec/75.0 kg = 2.58 m/sec
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4. A baseball with mass 142 g (weight, 5.5 oz) has a velocity of 29.4 m/sec when it is struck by a bat. After impact the ball is traveling with a velocity of 34.4 m/sec in a direction just opposite to the original velocity. What is the impulse imparted to the ball by the bat? Let us take the direction of the velocity after impact vfas positive. The change in momentum is mvf- mviwhere viis the initial velocity. impulse of the bat = change in momentum Impulse of the bat = m(vf- vi) but the direction of vi is negative relative to vf, so in terms of the magnitudes of the velocities viand vfimpulse = m(vf + vi) = 0.142 x (34.3 + 29.4) = 0.142 x 63.7 = 9.00 kg-m/sec = 9.00 N-sec
6.3 Conservation of Linear Momentum As we stated earlier from Newton's third law, impulse of the net FAB= -impulse of the net FBA(6.3) Since the change in momentum is equal to the impulse of the force producing the change in momentum, we can rewrite Equation 6.3 in terms of momentum change, change of momentum of body A = - (change of momentum of body B), ΔpA= -ΔpB mAΔvA= - mBΔvB(6.4) mAvAf- mAvAi= - (mBvBf- mBvBi(6.5) where mA= mass of body A, mB= mass of body B, vAi= initial velocity of body A, vAf= final velocity of body A, vBi= initial velocity of body B, and vBf= final velocity of body B. The above Equation 6.5 can be written with all the initial state terms on one side and the final state terms on the other. mAvAi+ mBvBi= mAvAf+ mBvBf(6.6) The left side of the above equation is the total momentum before collision, and the right side is the total momentum after collision, that is, the momentum before collision is equal to the momentum after collision. The relationship is called the principle of conservation of linear momentum.
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The principle applies for any collision. Remember that momentum and impulse are vector quantities. Glancing collisions occur more frequently than head-on collisions, and the situation is complex because of the vector properties of momentum. The total momentum of a system can only be changed by the action of external forces on the system. The internal forces produce opposite and equal changes in momentum that cancel. Thus, we conclude that the total momentum of an isolated system is constant in magnitude and direction. This is equivalent to saying that the vector sum of the momenta before collision is equal to the vector sum of momenta after collision, (Σmivi)before= (Σmtvt)after(6.7) for an isolated system, where Σindicates the process of adding the momentum of the individual bodies. EXAMPLES
1. Suppose body A is moving in the xdirection with velocity vAiand collides with body B which is at rest. Body A is deflected at angle θabove the xaxis and body B is deflected on angle φbelow the xaxis after the collision see Figure 6.6. The xcomponents of momentum are equal before and after the collision. mAvAi= mAvAfcosθ+ mBvBfcosφ(6.8) The initial ycomponent of momentum is zero, so after the collision the ymomentum components must add to equal 0, 0 = mAvAfsinθ - mBvBfsinφ(6.9)
2. A body A of mass 2.00 kg and velocity of 3.00 m/sec collides with a body B of mass
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3.00 kg that is at rest. After collision, A is moving with a velocity of 2.00 m/sec at an angle of 53o above the original direction. What is the velocity of B after the collision? Taking the original direction of vAas the xdirection, we have for the xcomponents of momentum pi= pfTaking the original direction of vAas the xdirection, we have for the xcomponents of momentum 2.00(3.00) = 2.00(2.00)(0.600) + 3.00vBcosφ 3.00vBcosφ= 6.00 - 2.40 = 3.60 kg-m/sec For the ycomponents of momentum, we have 0 = 2.00(2.00)(0.800) - 3.00 vB sinφ3.00 vB sinφ= 3.20 kg-m/sec tanφ= 3.20/3.60 = 0.889 φ= 41.5o Squaring the above equations for vB and adding them together, we get 9.00 vB
6.4 Collisions In any collision the total momentum is always conserved. However, the same is not always true of kinetic energy. If the kinetic energy of the bodies involved in a collision is the same before and after impact, the collision is said to be a perfectly elastic collision. If the bodies stick together after impact, the collision is said to be completely inelastic. These represent the two extreme cases. All intermediate cases are possible and are called partially elastic collisions. We have seen many examples of a partially elastic collision. Consider a collision between two automobiles, see Figure 6.7. Momentum is conserved, but the kinetic energy before the collision is greater than the kinetic energy after the collision. Evidence of this is the change in shape of the vehicles as a result of the collision. These "transformations" require an expenditure of energy which comes from the kinetic energy of the system.
The only known perfectly elastic collisions are those between atomic and nuclear particles, and not all of these collisions are perfectly elastic. A collision within your experience that is nearly perfectly elastic is the one between two billiard balls. For a perfectly elastic collision between particles A and B, one can write mAvAi+ mBvBi= mAvAf+ mBvBfconservation of momentum (6.6) If U1= U2, mgh1= mgh2, in Equation 5.10, one can obtain:
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(½) mAv2Ai+ (½) mBv2
Bi= (½) mAv2Af+ (½) mBv2
Bfconservation of kinetic energy where the subscripts A and B designate the two particles in collision, the subscript i designates the velocity before collision, and the subscript f designates the velocity after collision. These are two independent equations and can be solved in general to find two unknown quantities, such as the two final velocities. Let us consider the special case of a two-particle head-on, elastic collision where one body of mass mBis initially at rest (vBi= 0). We can rearrange these equations to have all mAterms on one side and all mBterms on the other side of the equal sign. (½) mBv2
Bf= (½) mA(v2Af- v2
Bf)conservation of energy (6.10) mBvBf= mA(vAi- vAf) conservation of momentum (6.11) For head-on collisions the velocities are all along the same line of action. We can treat the velocities as positive or negative scalars. Thus we have two equations and two unknowns, and hence a solution is possible. One approach is to eliminate the masses from the two equations by dividing Equation 6.10 by Equation 6.11. The result is given by the following equality between the final velocity of particle B and the sum of the initial and final velocities of particle A, vBf= vAi+ vAf(6.12) This expression for vBf can be substituted into Equation 6.11 to obtain a general relationship between the final and the initial velocities of particle A, vAf = {(mA- mB) /(mA+ mB)} vAi(6.13) This relationship is valid for all two particle head-on collisions when one of the objects is initially at rest. We can use this expression to eliminate uAffrom Equation 6.11 and obtain a relationship between the initial velocity of particle A and the final velocity of particle B, vBf = {2mA/(mA+ mB)} vAi(6.14) EXAMPLE OF A PERFECTLY ELASTIC HEAD-ON COLLISIONA neutron (mass = 1.67 x 10-27kg) collides head on with a nitrogen nucleus at rest (mass = 23.1 x 10-27kg). The initial velocity of the neutron is 1.50 x 107m/sec. Find the final velocity of the neutron, and the final velocity of the nitrogen nucleus. To find the final velocity of the neutron, we can use Equation 6.13: vAf = {(mA- mB)/(mA+ mB)} vAi= (1.67 - 23.1) x 10-27kg/(23.1 + 1.67) x 10-27kg x (1.50 x 107m/sec) vAf = -1.30 x 107m/sec The negative value for vAf in this case, means that the direction of the velocity of the neutron is reversed by the collision. To find the final velocity of the nitrogen nucleus, we can use Equation 6.14: vBf = {2mA/(mA+ mB)} vAi = {2(1.67 X 10-27kg)/(23.1 + 1.67)10-27kg} x (1.50 x 107m/sec) vBf = 2.02 x 106m/sec Let us consider a completely inelastic collision between two objects, A and B. The objects stick together and travel off with a velocity of vf.
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mAvAi+ mBvBi= (mA+ mB))vfconservation of momentum (6.15) (½) mAv2
Ai+ (½) mBv2Bi- (½) (mA+ mB)v2
f= loss of kinetic energy (6.16) EXAMPLEA 2.00-g rifle bullet is fired into a 2.00 kg block of a ballistic pendulum. The pendulum is 4.00 m long, and it rises 3.60 cm as a result of impact (Figure 6.8). Find the speed of the bullet, and the loss of kinetic energy.
Let ma= mass of bullet, mb= mass of block, vai= initial velocity of bullet, vf= velocity of block and bullet, and vbi= initial velocity of the block = 0. Upon impact momentum is conserved so mavai= (ma+ mb)vfAfter impact, energy is conserved so (½) (ma+ mb)vf
2= (ma+ mb) gh Then vf
2 = 2gh. Putting in the numerical values in the last equation, vf
2= 2 x 9.80 x .0360 = 0.704 vf = 0.840 m/sec Then from conservation of momentum, with numerical values, we obtain 2vai = (2000 + 2.00)(0.840) vai = 1001.00 x 0.840 = 841 m/sec loss in KE = (½) mavai
2- (½) (ma+ mb)vf2= (½) (2.00 x 10-3) x (841)2– (½) (0.002)(0.841)2= 707 -
0.71 = 706 J 6.5 Ballistocardiography The conservation of momentum law suggests that as the blood is pumped from the heart the body should recoil. The recording of this recoil is called ballistocardiography. We want to look at the physical basis for such measurements. While measurements are being made, the patient is placed on a "frictionless" platform. Since there are no external forces acting, the total momentum change is zero. We write the following equation for the blood momentum and the momentum for the rest of the body system, mbvb+ mBvB= 0 (6.17) thus mBvB= -mbvb(6.18) where subscript b represents blood and subscript B represents the body. Let Δtrepresent the period of the pump stroke of the heart (time of contraction). We can derive the equation in terms of the distance of body movement. Multiplying both sides
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of Equation 6.12 by the stroke time Δtand recalling that vΔtis just the displacement, mbvbΔt= – mBvBΔt or mbΔxb= – mBΔxB(6.19) where xbis the blood displacement and xB is the body displacement during the time of heart contraction. We have the relationship between the displacement of the body and the displacement of the blood, ΔxB = (mb /mB ) Δxb(6.20) We can estimate the body displacement by using some typical data. The mass of blood per heart stroke is about 60 g, and we will assume a body mass of 50 kg. A reasonable estimate of the movement of the center of mass of the blood for a contraction can be obtained by using the average velocity of blood flow from the heart 50 cm/sec and a typical contraction time of 0.13 sec. This gives a blood displacement of 6.5 cm. Substituting these values into Equation 6.20 we get ΔxB = {60 g/50 x 103g} (-6.5 cm) = -7.8 x 10-3cm This small displacement is detectable with the sensitive displacement transducers available today. A typical ballistocardiogram of displacement versus time is shown in Figure 6.9.
6.6 Rocket Propulsion Rocket propulsion is a practical example of the conservation of momentum. The momentum of the rocket is initially zero. When the stream of exhaust gases start out the back of the rocket, they have a momentum in the backward direction. For the total momentum of the rocket and fuel system to be conserved, the rocket must be given an equal and opposite momentum. To simplify the quantitative calculations, let us construct a two- particle model of a rocket. One particle is determined by the rocket ship, the engine, the remaining fuel and the fuel storage tanks. The other particle is the fuel burned and ejected in a given time period (Figure 6.10).
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We use the conservation of momentum to determine the exit velocity of the fuel ve, the total thrust exerted by the fuel on the rocket ship, T, and the final velocity of the rocket ship vf. The initial momentum of the system is given by (mr+ mf)viwhere vi is the initial velocity of the rocket and the fuel, mris the mass of the rocket, mf is the mass of the fuel. Then the fuel is burned and ejected in a time of tseconds, the ejected fuel will have an ejection velocity vein a direction opposite to vi. The ejected fuel will exert a thrust upon the rocket, which will attain a final velocity vf. The final momentum must be equal to the initial momentum, (mr+ mf)vi= mr vf– mfve(6.21) where the minus sign indicates that veis opposite in direction to vi and vf, but along the same line, so we dropped the vector notation. The change in momentum of the rocket ship is equal to the impulse imparted to the rocket ship by the ejected gases. Since we know the time of fuel ejection, we can calculate the thrust by dividing the impulse by the time, net impulse imparted to the rocket ship = mr (vf - vi) net thrust = impulse/time = mr (vf - vi) / t(6.22) EXAMPLEThe following data are for a three-stage Saturn rocket launched in 1971 for a trip to the moon. The first-stage fuel burn of liquid oxygen and hydrogen lasted for 165 sec. The initial velocity was zero. The initial total mass of rocket and fuel was 2.91 x 106kg. The mass of the rocket and its velocity at the end of the burn were 8.36 x 105kg and 2.32 x 103m/sec. What is the velocity of the ejected fuel, and what is the average net thrust applied to the rocket ship? From Equation 6.21 we can write an equation for the velocity of the ejected fuel: ve = { mr vf- (mr+ mf)vi}/ mfve= {(8.36 x 165kg)(2.32 x 103m/sec) - (2.91 x 106)kg x 0 m/sec}/(2.91 x 106- 8.36 x 105)kg ve= 9.37 x 102m/sec The average thrust can be found from Equation 6.22: thrust = {(8.36 x 105)kg x (2.32 x 103- 0) m/sec}/165 sec = 1.18 x 107N
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SUMMARY Use these equations to evaluate how well you have achieved the goals of this chapter. The answers to these questions are given at the end of this summary with the number of the section where you can find the related content material. Definitions1. The impulse of a force is equal to the change of resulting
a. energy b. momentum c. potential energy d. kinetic energy e. all of these
2. The momentum of a car is defined as ______ and is a vector in the direction of _______. 3. Elastic collisions conserve both _______ and ________. 4. Inelastic collisions are characterized by the objects' _______ and conservation of ________. 5. Jet, or rocket, propulsion is possible because the momentum of the vehicle equals the ________. Impulse Problems6. If a 10.0-N force acts for 1.00 x 10-2sec on a 0.100-kg ball at rest the resulting change in velocity is ________. 7. The average force in a head-on collision with a wall of a 0.100-kg ball moving at 10.0 m/sec is 10.0 N. Find the total contact time for the impact with the wall. Dt= ________. (Assume elastic collision.) Conservation of Momentum8. The condition necessary for conservation of momentum in a given system is that a. energy is conserved b. one body is at rest c. no external force acts d. internal forces equal external forces e. none of these 9. The superposition principle applied to a system of particles leads to the momentum of the system equal to ______. Collision Problems10. The greatest impulse is involved in collisions that are a. perfectly inelastic b. partially elastic c. perfectly elastic d. same for all 11. When a 0.1-kg block moving at a speed of 10.0 m/sec collides perfectly inelastically with a 0.400-kg block at rest, the final speed of the blocks is a. 33.3 m/sec b. 3.33 m/sec c. 2.5 m/sec d. 2 m/sec e. none of these Momentum and Energy Problems12. The energy lost in problem 11 is what percent of initial kinetic energy? 13. If the block at rest in problem 11 is attached to a spring with a force constant of 100 N/m, find the distance the spring is compressed by the inelastic collision. 14. In a head-on inelastic collision between two cars, which of the cars will suffer the
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greater damage -- the heavier or lighter vehicle? (Assume both have same momentum before collision.) Answers1. b (Section 6.2) 2. mv, v(Section 6.2) 3. momentum, energy (Section 6.4) 4. sticking together, momentum (Section 6.4) 5. exhaust momentum (Section 6.6) 6. 1 m/sec (Section 6.2) 7. 2 x 10-1sec (Section 6.2)
8. c (Section 6.3) 9. sum of miviSection 6.3) 10. c (Section 6.4) 11. d (Section 6.4) 12. 80 percent (Section 6.4) 13. 14.1 cm (Section 6.4) 14. lighter car. It has the greatest change
in energy (Section 6.4) ALGORITHMIC PROBLEMSListed below are the important equations from this chapter. The problems following the equations will help you to translate words into equations and to solve single concept problems. Equationsimpulse = Ft = Δ(mv) = m(vf- vi) = Δp = change in momentum (6.2) mAvAi+ mBvBi= mAvAf+ mBvBf(6.6) (Σmivi)before= (Σmtvt)after(6.7) (½) mAv2
Ai+ (½) mBv2Bi- (½) (mA+ mB)v2
f= loss of kinetic energy (6.16) Problems1. A tennis ball makes a perfectly elastic collision with a wall and bounces
straight back with a speed of 10.0 m/sec. The mass of the ball is 100 g and the time of contact with the wall is 1.00 x 10-2sec. Find the force acting on the ball exerted by the wall.
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2. A force of 100 N acts for 1.00 x 10-3sec on a 1.00-kg mass. Find the change in momentum of the mass.
3. A boy (mass = 40.0 kg) runs at a speed of 5.00 m/sec and jumps onto a sled (mass = 10.0 kg). Find the speed of the boy and the sled.
4. a. Find the impulse needed to stop a car (mass = 1000 kg) moving at 5.00 m/sec. b. Find the force needed to bring this car to rest in 2.00 sec. 5. A railroad car (mass = M) moving at 10.0 m/sec overtakes another car (mass = M)
moving in the same direction with a speed of 4.00 m/sec. Find the velocity of the cars after they collide and couple together.
6. Find the fraction of kinetic energy lost in the collision of problem 5. Where did this energy go?
Answers1. 200 N 2. 0.1 kg- m/sec 3. 4.00 m/sec
4. a. 5000 N-sec b. 2500 N 5. 7.00 m/sec 6. 0.155
EXERCISES These exercises are designed to help you apply the ideas of a section to a physical situation. When appropriate, the numerical answer is given at the end of each exercise. Section 6.21. A 5.00-kg rifle fires a 10.0 g bullet with a speed of 600 m/sec. What is the
speed of recoil of the rifle? If the rifle is stopped in 2.00 cm, what is its deceleration and stopping force? [1.20 m/sec, -36.0 m/sec2, 180 N]
2. A 142-g baseball arrives at a bat with a speed of 40.0 m/sec. It is in contact with the bat for 0.020 seconds, and it leaves the bat going in the opposite direction with a speed of 80.0 m/sec. What is the impulse of the bat on the ball? What is the average force of the bat on the ball? [17.0 N-sec, 850 N]
3. A staple gun fires 10 staples into a wall. Each staple has a mass of 0.400 g and a velocity 80.0 m/sec. What is the impulse acting on the wall? [0.320 N-sec]
Section 6.44. Two balls, A (mass 2.00 kg) and B (mass of 2.50 kg), approach each other head-on eith speeds of 6.00 and 10.0 m/sec respectively. Assume an elastic collision. What is the speed of each ball after the collision? [vAf= - 11.8 m/sec, vBf= 4.22 m/sec]
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PROBLEMSEach problem may involve more than one physical concept. The answer is given at the end of the problem. 5. A 60.0-g ball A is traveling west with a speed of 1.00 m/sec and collides with ball B
(mass 60.0 g) at rest. After the collision ball B moves in a direction 30 o west of south. Assume the collision is perfectly elastic. What is the direction of ball A, and the speed of each? [30o north of west, 0.500 m/sec, 0.866 m/sec]
6. An alpha particle of mass 6.66 x 10-27kg has a velocity of 1.00 x 107m/sec when it collides head-on with an oxygen nucleus at rest, and with mass four times that of an alpha particle. What is the velocity of each after collision? [4.00 x 106m/sec, -6.00 x 106m/sec]
7. An elastic head-on collision occurs between a bombarding mass mand a mass Mat rest. Assume the bombarding mass has kinetic energy K. Calculate the kinetic energy of m, Km, and M, KM, after impact for five different cases: m= 0.01M; m= 0.1M; m= M; m= 10M; m= 100M. What conclusions can you reach? See
Table 8. A completely inelastic head-on collision occurs between a bombarding mass mand a target of mass Mat rest. Assume the mass mhas kinetic energy K. Calculate the kinetic energy after impact for cases: m= 0.01M; m= 0.1M; m= M; m= 10M; m= 100 M. What conclusion can you reach? [0.00990K, 0.091K, 0.500K, 0.909K, 0.990K]
9. A 1500-kg automobile is going west on Main Street at 15.0 m/sec when it collides with a 4000-kg truck going south on Pine Street at 8.00 m/sec. They become coupled together in the collision. What is the velocity after impact, and how much kinetic energy is lost in the collision? [7.10 m/sec, 35.1o west of south, 1.6 x 105J]
10. A 10.0-g bullet is fired into a 5.00-kg wood block which is suspended by long cords so that it can swing as a ballistic pendulum. The impact of the bullet raises block's center of mass 5.00 cm. What was the speed of the bullet? How much kinetic energy is lost? What becomes of the lost kinetic energy? [496 m/sec, 1.23 x 103J]
11. Two pendulum balls are pulled apart until both are raised 40.0 cm above equilibrium position. When they are released, they collide head-on at the bottom of their swing. If they stick together, find how high they swing and the energy that goes into heating them up after the collision. The masses of the balls are 300 g and 200 g. [1.60 cm, 1.88 J]
12. Three blocks each of mass mare located on a frictionless track as shown in Figure 6.11. After block A is released it makes an inelastic collision with block B and the two of them in turn collide with and stick to block C. Find the final velocity of the three blocks as they move off together. Find the total energy converted to heat in this entire process. [ (2/3) SQR RT(gh/2), mgh/3] See Diagram
13. A neutron of mass 1.67 x 10-27kg and a velocity of 2.00 x 104m/sec makes a head-on, completely inelastic, collision with a boron nucleus, mass 17.0 x 10-27kg, originally at rest. What is the velocity of the new nucleus and its final kinetic energy? What is the loss of kinetic energy in this collision? [1.79 x 103m/sec, 2.99 x 10-20J, 3.04 x 10-19J]
14. An arrow of mass Mis shot into a block (mass = 19M) on the end of a cord 1.00 m in
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length. The arrow sticks in the block and the combination swings through an arc of 37o. What properties of the system are conserved? Find the speed of the arrow when it hit the block. [momentum, vi= 39.6 m/sec]
15. During the second stage of its launch a Saturn rocket burns 4.52 x 105kg of fuel in 391 sec. The mass of rocket and fuel at the beginning of the second stage is 6.66 x 105kg. The average net thrust was 2.32 x 106N. Use the two particle model to find the velocity of the ejected fuel. What was the final velocity of the rocket if the velocity at the beginning of stage two was 2.73 x 103m/sec [+7.23 x 102m/sec in the same direction as vi, 6.96 x 103m/sec]
16. If the average force acting on a basketball is equal to its weight when it bounces on the floor and returns to 81 percent of its initial height, H, find the time of contact for the ball on the floor. [ 2.68 SQR RT(H/g)]
17. A rubber raft of mass 3Mfloats past a dock at 1 m/sec toward the east. A girl (mass = 2M) runs down the dock and jumps into the raft with a speed of 2 m/sec in the north direction. Find the final velocity of the raft and girl and the fraction of kinetic energy lost in this collision. [vE= 0.6 m/sec, vN= 0.8 m/sec, 54.5 percent KElost]
