WEEK-6 Recitation PHYS 3 FOCUS ON CONCEPTS Section 7. The Impulse Momentum Theorem Mar, 08. Two identical cars are traeling at the same speed. One is heading due east and the other due north, as the drawing shows. Which statement is true regarding the kinetic energies and momenta o the cars? (a) They both hae the same kinetic energies and the same momenta. (b) They hae the same kinetic energies, but dierent momenta. (c) They hae dierent kinetic energies, but the same momenta. (d) They hae dierent kinetic energies and dierent momenta.. Six runners hae the mass (in multiples o m0), speed (in multiples o 0), and direction o trael that are indicated in the table. Which two runners hae identical momenta? (a) B and C (b) A and C (c) C and D (d) A and E (e) D and F Runner Mass Speed Direction o Trael A m 0 0 Due north B m0 0 Due east C m0 0 Due south D m0 0 Due west E m0 0 Due north F m0 0 Due west 7. A particle moes along the +x axis, and the graph shows its momentum p as a unction o time t. In each o the our regions a orce, which may or may not be nearly zero, is applied to the particle. In which region is the magnitude o the orce largest and in which region is it smallest? (a) B largest, D smallest (b) C largest, B smallest (c) A largest, D smallest (d) C largest, A smallest (e) A largest, C smallest Section 7. The Principle o Conseration o Linear Momentum
0. As the text discusses, the conseration o linear momentum is applicable only when the system o objects is an isolated system. Which o the systems listed below are isolated systems?. A ball is dropped rom the top o a building. The system is the ball.. A ball is dropped rom the top o a building. The system is the ball and the earth. 3. A billiard ball collides with a stationary billiard ball on a rictionless pool table. The system is the moing ball. 4. A car slides to a halt in an emergency. The system is the car. 5. A space probe is moing in deep space where graitational and other orces are negligible. The system is the space probe. (a) Only and 5 are isolated systems. (b) Only and 3 are isolated systems. (c) Only 3 and 5 are isolated systems. (d) Only 4 and 5 are isolated systems. (e) Only 5 is an isolated system. Section 7.3 Collisions in One Dimension 3. Two objects are inoled in a completely inelastic one-dimensional collision. The net external orce acting on them is zero. The table lists our possible sets o the initial and inal momenta and kinetic energies o the two objects. Which is the only set that could occur? Initial (Beore Collision) Final (Ater Collision) Momentum Kinetic Energy Momentum Kinetic Energy a. Object : +6 kg m/s 5 J +8 kg m/s 9 J Object : 0 kg m/s 0 J b. Object : +8 kg m/s 5 J +6 kg m/s J Object : kg m/s 7 J c. Object : 3 kg m/s J + kg m/s 4 J Object : +4 kg m/s 6 J d. Object : 0 kg m/s 3 J 8 kg m/s J Object : 8 kg m/s 8 J Section 7.4 Collisions in Two Dimensions 5. Object is moing along the x axis with an initial momentum o +6 kg m/s, where the + sign indicates that it is moing to the right. As the drawing shows, object collides with a second object that is initially at rest. The collision is not head-on, so the objects moe o in dierent directions ater the collision. The net external orce acting on the two-object system is zero. Ater the collision, object has a momentum whose y component is 5 kg m/s. What is the y component o the momentum o object ater the collision? (a) 0 kg m/s (b) +6 kg m/s (c) +5 kg m/s (d) 6 kg m/s (e) The y component o the momentum o object cannot be determined.
Answers o FOC. (b) Kinetic energy, m, is a scalar quantity and is the same or both cars. Momentum, m, is a ector quantity that has a magnitude and a direction. The two cars hae dierent directions, so they hae dierent momenta.. (d) Momentum is a ector quantity that has a magnitude and a direction. The magnitudes (m 0 0 ) and directions (due north) are the same or both runners. 7. (b) According to the impulse-momentum theorem, the net aerage orce is equal to the change in the particle s momentum diided by the time interal. This ratio is greatest in region C. The ratio is equal to zero in region B, since the change in the particle s momentum is zero there. 0. (a) The net external orce acting on the ball/earth system is zero. The graitational orces that the ball and earth exert on each other are internal orces, or orces that the objects within the system exert on each other. The space probe is also an isolated system, since there are no external orces acting on it. 3. (c) Since the net external orce acting on the two objects during the collision is zero, the total linear momentum o the system is consered. In other words, the total linear momentum beore the collision ( 3 kg m/s + 4 kg m/s + kg m/s) equals the total linear momentum ater the collision (+ kg m/s). Furthermore, some kinetic energy is lost during a completely inelastic collision, which is the case here. The inal kinetic energy (4 J) is less that the total initial kinetic energy ( J + 6 J 7 J). 5. (c) The net external orce acting on the two objects during the collision is zero, so the total momentum o the system is consered. In two dimensions this means that the y-component o the initial total momentum (0 kg m/s) is equal to the y-component o the inal total momentum. Since the y-component o the inal momentum o object is 5 kg m/s, then the y-component o the inal momentum o object must be +5 kg m/s [0 kg m/s ( 5 kg m/s) +5 kg m/s].
Problems 4. In a perormance test, each o two cars takes 9.0 s to accelerate rom rest to 7 m/s. Car A has a mass o 400 kg, and car B has a mass o 900 kg. Find the net aerage orce that acts on each car during the test. REASONING The impulse-momentum theorem, as expressed in Equation 7.4, states that the impulse acting on each car is equal to the inal momentum o the car minus its initial momentum: where ΣF is the net aerage orce that acts on the car, and t is the time interal during which the orce acts. SOLUTION We assume that the elocity o each car points in the +x direction. The net aerage orce acting on each car is: Car A ( 400 kg)( 7 m/s) ( 400 kg)( 0 m/s) m m + Σ F 0 + 400 N t 9.0 s Car B ( 900 kg)( 7 m/s) ( 900 kg)( 0 m/s) m m + Σ F 0 + 5700 N t 9.0 s *. A student (m 63 kg) alls reely rom rest and strikes the ground. During the collision with the ground, he comes to rest in a time o 0.040 s. The aerage orce exerted on him by the ground is +8 000 N, where the upward direction is taken to be the positie direction. From what height did the student all? Assume that the only orce acting on him during the collision is that due to the ground. REASONING We will diide this problem into two parts, because the orces acting on the student change abruptly at the instant o impact. In the irst part, the student alls reely rom rest, under the sole inluence o the conseratie graitational orce. Thus, the student s total mechanical energy E is consered up to the instant o impact. We will use the energy conseration principle to determine the student s initial height H in terms o the student s elocity impact at that instant. The second part o the student s motion begins at impact, when the orce F ground due to the ground oerwhelms the graitational orce and brings the student to rest. The orce o the ground is nonconseratie, so instead o the energy conseration principle, we will apply the impulse-momentum theorem Fground t m m0 (Equation 7.4) to analyze the
collision. Because this time interal begins at impact, 0 is the student s impact elocity: 0 impact. SOLUTION We begin with the energy conseration principle 0 0 m + mgh m + mgh (Equation 6.9b) applied to the student s all to the ground. Falling rom rest implies 0 0 m/s, and the student s inal elocity is the impact elocity: impact. Thus, we hae () For the student s collision with the ground, the impulse-momentum theorem gies F t m m (Equation 7.4). The collision brings the student to rest, so we know ground 0 that 0 m/s, and Equation 7.4 becomes Fground t m0. Soling or the impact speed 0, we obtain Fground t 0 impact () m Substituting Equation () into Equation () yields Fground t impact m ( Fground t) ( + 8 000 N)( 0.040 s) H 6.7 m g g gm 9.80 m/s 63 kg ( )( )
9. ssm A lumberjack (mass 98 kg) is standing at rest on one end o a loating log (mass 30 kg) that is also at rest. The lumberjack runs to the other end o the log, attaining a elocity o +3.6 m/s relatie to the shore, and then hops onto an identical loating log that is initially at rest. Neglect any riction and resistance between the logs and the water. (a) What is the elocity o the irst log just beore the lumberjack jumps o? (b) Determine the elocity o the second log i the lumberjack comes to rest on it. SSM REASONING The system consists o the lumberjack and the log. For this system, the sum o the external orces is zero. This is because the weight o the system is balanced by the corresponding normal orce (proided by the buoyant orce o the water) and the water is assumed to be rictionless. The lumberjack and the log, then, constitute an isolated system, and the principle o conseration o linear momentum holds. SOLUTION a. The total linear momentum o the system beore the lumberjack begins to moe is zero, since all parts o the system are at rest. Momentum conseration requires that the total momentum remains zero during the motion o the lumberjack. Here the subscripts "" and "" reer to the irst log and lumberjack, respectiely. Let the direction o motion o the lumberjack be the positie direction. Then, soling or gies m (98 kg)(+3.6 m / s) m 30 kg.5 m / s The minus sign indicates that the irst log recoils as the lumberjack jumps o. b. Now the system is composed o the lumberjack, just beore he lands on the second log, and the second log. Graity acts on the system, but or the short time under consideration while the lumberjack lands, the eects o graity in changing the linear momentum o the system are negligible. Thereore, to a ery good approximation, we can say that the linear momentum o the system is ery nearly consered. In this case, the initial momentum is not zero as it was in part (a); rather the initial momentum o the system is the momentum o the lumberjack just beore he lands on the second log. Thereore, m + m m + m 0 0 Total momentum just ater lumberjack lands Initial momentum In this expression, the subscripts "" and "" now represent the second log and lumberjack, respectiely. Since the second log is initially at rest, 0 0. Furthermore, since the lumberjack and the second log moe with a common elocity,. The statement o momentum conseration then becomes
Soling or, we hae m + m m 0 m 0 (98 kg)(+3.6 m / s) m + m 30 kg + 98 kg +. m / s The positie sign indicates that the system moes in the same direction as the original direction o the lumberjack's motion. *5. ssm By accident, a large plate is dropped and breaks into three pieces. The pieces ly apart parallel to the loor. As the plate alls, its momentum has only a ertical component and no component parallel to the loor. Ater the collision, the component o the total momentum parallel to the loor must remain zero, since the net external orce acting on the plate has no component parallel to the loor. Using the data shown in the drawing, ind the masses o pieces and. REASONING No net external orce acts on the plate parallel to the loor; thereore, the component o the momentum o the plate that is parallel to the loor is consered as the plate breaks and lies apart. Initially, the total momentum parallel to the loor is zero. Ater the collision with the loor, the component o the total momentum parallel to the loor must remain zero. The drawing in the text shows the pieces in the plane parallel to the loor just ater the collision. Clearly, the linear momentum in the plane parallel to the loor has two components; thereore the linear momentum o the plate must be consered in each o these two mutually perpendicular directions. Using the drawing in the text, with the positie directions taken to be up and to the right, we hae x direction m (sin 5.0 ) + m (cos 45.0 ) 0 () y direction m (cos 5.0 ) + m (sin 45.0 ) m 3 3 0 () These equations can be soled simultaneously or the masses m and m.
SOLUTION Using the alues gien in the drawing or the elocities ater the plate breaks, we hae, m ( 3.00 m/s) sin 5.0 + m (.79 m/s) cos 45.0 0 () m ( 3.00 m/s) cos 5.0 + m (.79 m/s) sin 45.0 (.30 kg)( 3.07 m/s) 0 () Subtracting () rom (), and noting that cos 45.0º sin 45.0º, gies m.00 kg. Substituting this alue into either () or () then yields m.00 kg. 34. The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass o 0.05 kg and is moing along the x axis with a elocity o +5.5 m/s. It makes a collision with puck B, which has a mass o 0.050 kg and is initially at rest. The collision is not head-on. Ater the collision, the two pucks ly apart with the angles shown in the drawing. Find the inal speeds o (a) puck A and (b) puck B. REASONING The net external orce acting on the two-puck system is zero (the weight o each ball is balanced by an upward normal orce, and we are ignoring riction due to the layer o air on the hockey table). Thereore, the two pucks constitute an isolated system, and the principle o conseration o linear momentum applies. SOLUTION Conseration o linear momentum requires that the total momentum is the same beore and ater the collision. Since linear momentum is a ector, the x and y components must be consered separately. Using the drawing in the text, momentum conseration in the x direction yields ( cos 65 ) ( cos37 ) m m + m () A 0A A A B B while momentum conseration in the y direction yields A A ( ) m ( ) 0 m sin 65 sin 37 () B B
Soling equation () or B, we ind that B m m A A Substituting equation (3) into Equation () leads to m A 0A a. Soling or A gies B ( sin 65 ) ( sin 37 ) m sin 65 maa( cos 65 ) + sin 37 A A ( ) ( cos37 ) (3) A 0A + 5.5 m/s sin 65 sin 65 cos 65 + cos 65 + tan 37 tan 37 3.4 m/s b. From equation (3), we ind that B ( 0.05 kg)( 3.4 m/s)( sin 65 ) ( 0.050 kg)( sin 37 ).6 m/s 59. mmh The carbon monoxide molecule (CO) consists o a carbon atom and an oxygen atom separated by a distance o.3 0 0 m. The mass mc o the carbon atom is 0.750 times the mass mo o the oxygen atom, or mc 0.750 mo. Determine the location o the center o mass o this molecule relatie to the carbon atom. REASONING AND SOLUTION system as Equation 7.0 gies the center o mass o this two-atom x cm mx m c c c + mx + m I we take the origin at the center o the carbon atom, then x c 0 m, and we hae o o o x cm 0 mx c c + mx o o xo.3 0 m 6.46 0 m m + m m / m + 0.750 m / m + ( ) ( ) c o c o o o