hysics HW #7 Solutions Chapter 7: Focus On Concepts: 2, 6, 0, 3 robles: 8, 7, 2, 22, 32, 53, 56, 57 Focus On Concepts 7-2 (d) Moentu is a ector quantity that has a agnitude and a direction. The agnitudes ( 0 0 ) and directions (due north) are the sae for both runners. Focus On Concepts 7-6 (c) According to the ipulse-oentu theore, the ipulse is equal to the change in the particle s oentu. The agnitude of the oentu change is the sae in regions A and C; therefore, in these regions the particle experiences ipulses of the sae agnitude. The oentu in region B is constant. Therefore, the change in oentu is zero, and so is the ipulse. Focus On Concepts 7-0 (a) The net external force acting on the ball/earth syste is zero. The graitational forces that the ball and earth exert on each other are internal forces, or forces that the objects within the syste exert on each other. The space probe is also an isolated syste, since there are no external forces acting on it. Focus On Concepts 7-3 (c) Since the net external force acting on the two objects during the collision is zero, the total linear oentu of the syste is consered. In other words, the total linear oentu before the collision ( 3 kg /s + 4 kg /s + kg /s) equals the total linear oentu after the collision (+ kg /s). Furtherore, soe kinetic energy is lost during a copletely inelastic collision, which is the case here. The final kinetic energy (4 J) is less that the total initial kinetic energy ( J + 6 J 7 J). roble 7-8 REASONING We will apply the ipulse oentu theore as gien in Equation 7.4 to sole this proble. Fro this theore we know that, for a gien change in oentu, greater forces are associated with shorter tie interals. Therefore, we expect that the force in the stiff-legged case will be greater than in the knees-bent case. SOLUTION a. Assuing that upward is the positie direction, we find fro the ipulseoentu theore that ( 75 kg)( 0 /s) ( 75 kg)( 6.4 /s) 5 f 0 Σ F + 2.4 0 N t 3 2.0 0 s
b. Again using the ipulse-oentu theore, we find that ( 75 kg)( 0 /s) ( 75 kg)( 6.4 /s) 3 f 0 Σ F + 4.8 0 N t 0.0 s c. The net aerage force acting on the an is Σ F FGround + W, where F Ground is the aerage upward force exerted on the an by the ground and W is the downward-acting weight of the an. It follows, then, that FGround ΣF W. Since the weight is W g, we hae Stiff legged F ΣF W Ground Knees bent F ΣF W Ground ( )( ) 5 2 5 + 2.4 0 N 75 kg 9.80 /s + 2.4 0 N ( )( ) 3 2 3 + 4.8 0 N 75 kg 9.80 /s + 5.5 0 N roble 7-7 REASONING AND SOLUTION The collision is an inelastic one, with the total linear oentu being consered: ( + 2 )V The ass 2 of the receier is ( )( ) 5 kg 4.5 /s 5 kg 84 kg V 2.6 /s 2 roble 7-2 REASONING The two-stage rocket constitutes the syste. The forces that act to cause the separation during the explosion are, therefore, forces that are internal to the syste. Since no external forces act on this syste, it is isolated and the principle of conseration of linear oentu applies: + + ) f 2 f2 2 0 after separation before separation where the subscripts "" and "2" refer to the lower and upper stages, respectiely. This expression can be soled for f.
SOLUTION Soling for f gies f ( + ) 2 2 0 f2 2400 kg + 200 kg ( 4900 / s) (200 kg)(5700 / s) 2400 kg +4500 / s Since f is positie, its direction is the sae as the rocket before the explosion. roble 7-22 REASONING During the breakup the linear oentu of the syste is consered, since the force causing the breakup is an internal force. We will assue that the +x axis is along the original line of otion (before the breakup), and the +y axis is perpendicular to this line and points upward. We will apply the conseration of linear oentu twice, once for the oentu coponents along the x axis and again for the oentu coponents along the y axis. SOLUTION The ass of each piece of the rocket after breakup is, and so the ass of the rocket before breakup is 2. Applying the conseration of oentu theore along the original line of otion (the x axis) gies cos 30.0 + cos 60.0 2 or cos 30.0 + cos 60.0 2 2 0 2 0 f, x Applying the conseration of oentu along the y axis gies 0, x sin 30.0 sin 30.0 sin 60.0 2 0 or 2 sin 60.0 f, y 0, y () (2) a. To find the speed of the first piece, we substitute the alue for 2 fro Equation (2) into Equation (). The result is sin 30.0 cos 30.0 + cos 60.0 2 sin 60.0 0 Soling for and setting 0 45.0 /s yields 2 ( 45.0 /s) 0 sin 30.0 sin 30.0 cos 30.0 + cos 60.0 cos 30.0 + cos 60.0 sin 60.0 sin 60.0 2 77.9 /s
b. The speed 2 of the second piece can be found by substituting 77.9 /s into Equation (2): sin 30.0 ( ) 77.9 /s sin 30.0 2 sin 60.0 sin 60.0 45.0 /s roble 7-32 REASONING The weight of each ehicle is balanced by the noral force exerted by the road. Assuing that friction and other resistie forces can be ignored, we will treat the twoehicle syste as an isolated syste and apply the principle of conseration of linear oentu. SOLUTION Using 0, car and 0, SUV to denote the elocities of the ehicles before the collision and applying the principle of conseration of linear oentu, we hae 0 car0, car + SUV 0, SUV after collision before collision Note that the total oentu of both ehicles after the collision is zero, because the collision brings each ehicle to a halt. Soling this result for 0, SUV and taking the direction in which the car oes as the positie direction gies 0, SUV SUV ( 00 kg)( 32 /s) car0, car 4 /s 2500 kg This result is negatie, since the elocity of the sport utility ehicle is opposite to that of the car, which has been chosen to be positie. The speed of the sport utility ehicle is the agnitude of 0, SUV or 4 /s. roble 7-53 REASONING The syste consists of the luberjack and the log. For this syste, the su of the external forces is zero. This is because the weight of the syste is balanced by the corresponding noral force (proided by the buoyant force of the water) and the water is assued to be frictionless. The luberjack and the log, then, constitute an isolated syste, and the principle of conseration of linear oentu holds. SOLUTION a. The total linear oentu of the syste before the luberjack begins to oe is zero, since all parts of the syste are at rest. Moentu conseration requires that the total oentu reains zero during the otion of the luberjack. + 0 f 2 f2 just before the jup Initial oentu
Here the subscripts "" and "2" refer to the first log and luberjack, respectiely. Let the direction of otion of the luberjack be the positie direction. Then, soling for f gies f f2 (98 kg)(+3.6 / s) 2 230 kg.5 / s The inus sign indicates that the first log recoils as the luberjack jups off. b. Now the syste is coposed of the luberjack, just before he lands on the second log, and the second log. Graity acts on the syste, but for the short tie under consideration while the luberjack lands, the effects of graity in changing the linear oentu of the syste are negligible. Therefore, to a ery good approxiation, we can say that the linear oentu of the syste is ery nearly consered. In this case, the initial oentu is not zero as it was in part (a); rather the initial oentu of the syste is the oentu of the luberjack just before he lands on the second log. Therefore, + + f 2 f2 0 2 02 just after luberjack lands Initial oentu In this expression, the subscripts "" and "2" now represent the second log and luberjack, respectiely. Since the second log is initially at rest, 0 0. Furtherore, since the luberjack and the second log oe with a coon elocity,. The stateent f f2 f of oentu conseration then becoes Soling for f, we hae f + f 2 f 2 02 2 02 (98 kg)(+3.6 / s) + 230 kg + 98 kg 2 +. / s The positie sign indicates that the syste oes in the sae direction as the original direction of the luberjack's otion. roble 7-56 REASONING AND SOLUTION According to the ipulse-oentu theore, Equation 7.4, ( ΣF) t f 0, where ΣF is the net aerage force acting on the person. Taking the direction of otion (downward) as the negatie direction and soling for the net aerage force ΣF, we obtain ( ) ( )[ ] f 0 62.0 kg.0 /s ( 5.50 /s) Σ F t.65 s The plus sign indicates that the force acts upward. +65 N
roble 7-57 REASONING AND SOLUTION The oentu is zero before the beat. Conseration of oentu requires that it is also zero after the beat; thus so that 0 p p + b b p ( b / p ) b (0.050 kg/85 kg)(0.25 /s) 4.5 0 /s