Week 6 hoework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign ersions of these probles, arious details hae been changed, so that the answers will coe out differently. The ethod to find the solution is the sae, but you will need to repeat part of the calculation to find out what your answer should hae been. WebAssign Proble 1: Consult Interactie Solution 7.9 at www.wiley.co/college/cutnell for a reiew of proble-soling skills that are inoled in this proble. A strea of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water strea has a elocity of + 16.0 /s, while the exiting water strea has a elocity of 16.0 /s. The ass of water per second that strikes the blade is 30.0 kg/s. Find the agnitude of the aerage force exerted on the water by the blade. REASONING During the tie interal t, a ass of water strikes the turbine blade. The incoing water has a oentu 0 and that of the outgoing water is f. In order to change the oentu of the water, an ipulse ( Σ F t is applied to it by the stationary turbine blade. Now ( Σ F t = F t, since only the force of the blade is assued to act on the water in the horizontal direction. These ariables are related by the ipulse-oentu theore, F t = f 0, which can be soled to find the aerage force F exerted on the water by the blade. SOLUTION Soling the ipulse-oentu theore for the aerage force gies The ratio / ( t so the aerage force is F t t = f 0 = ( f 0 is the ass of water per second that strikes the blade, or 30.0 kg/s, F = ( = ( 30.0 kg/s ( 16.0 /s ( + 16.0 /s = 960 N f 0 t
The agnitude of the aerage force is 960 N. WebAssign Proble 2: A 55-kg swier is standing on a stationary 210-kg.oating raft. The swier then runs off the raft horizontally with a elocity of +4.6 /s relatie to the shore. Find the recoil elocity that the raft would hae if there were no friction and resistance due to the water. REASONING The su of the external forces acting on the swier/raft syste is zero, because the weight of the swier and raft is balanced by a corresponding noral force and friction is negligible. The swier and raft constitute an isolated syste, so the principle of conseration of linear oentu applies. We will use this principle to find the recoil elocity of the raft. SOLUTION As the swier runs off the raft, the total linear oentu of the swier/raft syste is consered: ss + rr = 14243 Total oentu after swier runs off raft { 0 Total oentu before swier starts running where s and s are the ass and final elocity of the swier, and r and r are the ass and final elocity of the raft. Soling for r gies r r ( 55 kg ( + 4.6 /s ss = = = 1.2 /s 210 kg WebAssign Proble 3: The lead feale character in the oie Diaonds Are Foreer is standing at the edge of an offshore oil rig. As she fires a gun, she is drien back oer the edge and into the sea. Suppose the ass of a bullet is 0.010 kg and its elocity is + 720 /s. Her ass (including the gun is 51 kg. (a What recoil elocity does she acquire in response to a single shot fro a stationary position, assuing that no external force keeps her in place? (b Under the sae assuption, what would be her recoil elocity if, instead, she shoots a blank cartridge that ejects a ass of at a elocity of + 720 /s? REASONING For the syste consisting of the feale character, the gun and the bullet, the su of the external forces is zero, because the weight of each object is balanced by a corresponding upward (noral force, and we are ignoring friction. The feale character, the gun and the bullet, then, constitute an isolated syste, and the principle of conseration of linear oentu applies.
SOLUTION a. The total oentu of the syste before the gun is fired is zero, since all parts of the syste are at rest. Moentu conseration requires that the total oentu reains zero after the gun has been fired. + = 0 1 f1 2 f2 Total oentu after gun is fired Total oentu before gun is fired where the subscripts 1 and 2 refer to the woan (plus gun and the bullet, respectiely. Soling for f1, the recoil elocity of the woan (plus gun, gies f1 f2 (0.010 kg(720 / s = 2 = = 51 kg 1 0.14 / s b. Repeating the calculation for the situation in which the woan shoots a blank cartridge, we hae 4 f2 (5.0 10 kg(720 / s = 2 3 = = 7.1 10 / s f1 51 kg 1 In both cases, the inus sign eans that the bullet and the woan oe in opposite directions when the gun is fired. The total oentu of the syste reains zero, because oentu is a ector quantity, and the oenta of the bullet and the woan hae equal agnitudes, but opposite directions. WebAssign Proble 4: Multiple-Concept Exaple 7 presents a odel for soling probles such as this one. A 1055-kg an, stopped at a traffic light, is hit directly in the rear by a 715-kg car traeling with a elocity of +2.25 /s. Assue that the transission of the an is in neutral, the brakes are not being applied, and the collision is elastic. What is the final elocity of (a the car and (b the an? REASONING Since the collision is an elastic collision, both the linear oentu and kinetic energy of the two-ehicle syste are consered. The final elocities of the car and an are gien in ters of the initial elocity of the car by Equations 7.8a and 7.8b. SOLUTION a. The final elocity f1 of the car is gien by Equation 7.8a as = 1 2 f 1 01 + 1 2 where 1 and 2 are, respectiely, the asses of the car and an, and 01 is the initial elocity of the car. Thus,
f1 715 kg 1055 kg = ( + 2.25 /s = 0.432 /s 715 kg + 1055 kg b. The final elocity of the an is gien by Equation 7.8b: 1 f 2 01 1 2 ( 2 2 715 kg = = ( + 2.25 /s = + 1.82 /s + 715 kg + 1055 kg WebAssign Proble 5: Consult Multiple-Concept Exaple 8 for background pertinent to this proble. A 2.50-g bullet, traeling at a speed of 425 /s, strikes the wooden block of a ballistic pendulu, such as that in Figure 7.14. The block has a ass of 215 g. (a Find the speed of the bullet/block cobination iediately after the collision. (b How high does the cobination rise aboe its initial position? REASONING a. During the collision between the bullet and the wooden block, linear oentu is consered, since no net external force acts on the bullet and the block. The weight of each is balanced by the tension in the suspension wire, and the forces that the bullet and block exert on each other are internal forces. This conseration law will allow us to find the speed of the bullet/block syste iediately after the collision. b. Just after the collision, the bullet/block rise up, ultiately reaching a final height h f before coing to a oentary rest. During this phase, the tension in the wire (a nonconseratie force does no work, since it acts perpendicular to the otion. Thus, the work done by nonconseratie forces is zero, and the total echanical energy of the syste is consered. An application of this conseration law will enable us to deterine the height h f. SOLUTION a. The principle of conseration of linear oentu states that the total oentu after the collision is equal to that before the collision. ( + = + bullet block f bullet 0,bullet block 0,block 14 424443 14444244443 Moentu after collision Moentu before collision Soling this equation for the speed f of the bullet/block syste just after the collision gies
f + = + bullet 0,bullet block 0,block bullet block ( 0.00250 kg ( 425 /s + ( 0.215 kg ( 0 /s = = 0.00250 kg + 0.215 kg 4.89 /s b. Just after the collision, the total echanical energy of the syste is all kinetic energy, since we take the zero-leel for the graitational potential energy to be at the initial height of the block. As the bullet/block syste rises, kinetic energy is conerted into potential energy. At the highest point, the total echanical energy is all graitational potential energy. Since the total echanical energy is consered, we hae 1 2 ( + gh = ( + bullet block f 2 bullet block f 144424443 144424443 Total echanical energy at Total echanical energy at the top of the swing, the botto of the swing, all potential all kinetic Soling this expression for the height h f gies h ( 4.89 /s 2 = = = g 9.80 /s 1 2 1 2 f 2 f 2 1.22 WebAssign Proble 6: Multiple-Concept Exaple 9 reiews the concepts that play roles in this proble. The drawing shows a collision between two pucks on an air-hockey table. Puck A has a ass of 0.025 kg and is oing along the x axis with a elocity of + 5.5 /s. It akes a collision with puck B, which has a ass of 0.050 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of (a puck A and (b puck B. REASONING The net external force acting on the two-puck syste is zero (the weight of each ball is balanced by an upward noral force, and we are ignoring friction due
to the layer of air on the hockey table. Therefore, the two pucks constitute an isolated syste, and the principle of conseration of linear oentu applies. SOLUTION Conseration of linear oentu requires that the total oentu is the sae before and after the collision. Since linear oentu is a ector, the x and y coponents ust be consered separately. Using the drawing in the text, oentu conseration in the x direction yields ( cos65 ( cos37 = + (1 A 0A A fa B fb while oentu conseration in the y direction yields A fa Soling equation (2 for fb, we find that ( ( 0 = sin 65 sin 37 (2 B fb fb = A fa B ( sin 65 ( sin 37 (3 Substituting equation (3 into Equation (1 leads to A 0A ( sin 65 AfA = AfA ( cos 65 + sin 37 ( cos37 a. Soling for fa gies fa 0A + 5.5 /s = = = sin 65 sin 65 cos 65 + cos 65 + tan 37 tan 37 3.4 /s b. Fro equation (3, we find that fb ( 0.025 kg ( 3.4 /s ( sin 65 ( 0.050 kg ( sin 37 = = 2.6 /s WebAssign Proble 7: Consider the two oing boxcars in Exaple 5. Deterine the elocity of their center of ass (a before and (b after the collision. (c Should your answer in part (b be less than, greater than, or equal to the coon elocity f of the two coupled cars after the collision? Justify your answer.
REASONING AND SOLUTION The elocity of the center of ass of a syste is gien by Equation 7.11. Using the data and the results obtained in Exaple 5, we obtain the following: a. The elocity of the center of ass of the two-car syste before the collision is 1 01 2 02 ( = c before + + 1 2 3 3 (65 10 kg( + 0.80 /s + (92 10 kg( +1.2 /s = = 3 3 65 10 kg + 92 10 kg +1.0 /s b. The elocity of the center of ass of the two-car syste after the collision is ( 1 f 2 f c after + = = f = + 1 2 +1.0 /s c. The answer in part (b should be the sae as the coon elocity f. Since the cars are coupled together, eery point of the two-car syste, including the center of ass, ust oe with the sae elocity. WebAssign Proble 8: When juping straight down, you can be seriously injured if you land stiff-legged. One way to aoid injury is to bend your knees upon landing to reduce the force of the ipact. A 75-kg an just before contact with the ground has a speed of 6.4 /s. (a In a stiff-legged landing he coes to a halt in 2.0 s. Find the aerage net force that acts on hi during this tie. (b When he bends his knees, he coes to a halt in 0.10 s. Find the aerage net force now. (c During the landing, the force of the ground on the an points upward, while the force due to graity points downward. The aerage net force acting on the an includes both of these forces. Taking into account the directions of the forces, find the force of the ground on the an in parts (a and (b. 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 ipulse-oentu theore that ( 75 kg ( 0 /s ( 75 kg ( 6.4 /s 5 f 0 Σ F = = = + 2.4 10 N t 3 2.0 10 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 10 N t 0.10 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 downwardacting weight of the an. It follows, then, that FGround = Σ F W. Since the weight is W = g, we hae Stiff legged F = Σ F W Ground Knees bent F = Σ F W Ground ( ( 5 2 5 = + 2.4 10 N 75 kg 9.80 /s = + 2.4 10 N ( ( 3 2 3 = + 4.8 10 N 75 kg 9.80 /s = + 5.5 10 N WebAssign Proble 9: To iew an interactie solution to a proble that is siilar to this one, go to www.wiley.co/college/cutnell and select Interactie Solution 7.55. A 0.015-kg bullet is fired straight up at a falling wooden block that has a ass of 1.8 kg. The bullet has a speed of 810 /s when it strikes the block. The block originally was dropped fro rest fro the top of a building and had been falling for a tie t when the collision with the bullet occurred. As a result of the collision, the block (with the bullet in it reerses direction, rises, and coes to a oentary halt at the top of the building. Find the tie t. REASONING We will diide the proble into two parts: (a the otion of the freely falling block after it is dropped fro the building and before it collides with the bullet, and (b the collision of the block with the bullet. During the falling phase we will use an equation of kineatics that describes the elocity of the block as a function of tie (which is unknown. During the collision with the bullet, the external force of graity acts on the syste. This force changes the oentu of the syste by a negligibly sall aount since the collision occurs oer an extreely short tie interal. Thus, to a good approxiation, the su of the external forces acting on the syste during the collision is negligible, so the linear oentu of the syste is consered. The principle of conseration of linear oentu can be used to proide a relation between the oenta of the syste before and after the collision. This relation will enable us to find a alue for the tie it takes for the bullet/block to reach the top of the building.
SOLUTION Falling fro rest ( 0, block = 0 /s, the block attains a final elocity block just before colliding with the bullet. This elocity is gien by Equation 2.4 as = + a t { block 0, block 123 Final elocity of block just before bullet hits it Initial elocity of block at top of building where a is the acceleration due to graity (a = 9.8 /s 2 and t is the tie of fall. The upward direction is assued to be positie. Therefore, the final elocity of the falling block is block = a t (1 During the collision with the bullet, the total linear oentu of the bullet/block syste is consered, so we hae that ( + bullet block f 14 424443 = + bullet bullet block block 144 4244443 Total linear oentu after collision Total linear oentu before collision (2 Here f is the final elocity of the bullet/block syste after the collision, and bullet and block are the initial elocities of the bullet and block just before the collision. We note that the bullet/block syste reerses direction, rises, and coes to a oentary halt at the top of the building. This eans that f, the final elocity of the bullet/block syste after the collision ust hae the sae agnitude as block, the elocity of the falling block just before the bullet hits it. Since the two elocities hae opposite directions, it follows that f = block. Substituting this relation and Equation (1 into Equation (2 gies Soling for the tie, we find that ( + ( at = + ( at bullet block bullet bullet block t ( ( 0.015 kg ( 810 /s bullet bullet = = = a bullet + 2 ( ( + ( + 2 9.80 /s 0.015 kg 2 1.8 kg block 0.34 s Practice conceptual probles: 3. Two objects hae the sae oentu. Do the elocities of these objects necessarily hae (a the sae directions and (b the sae agnitudes? Gie your reasoning in each
case. REASONING AND SOLUTION a. Yes. Moentu is a ector, and the two objects hae the sae oentu. This eans that the direction of each object s oentu is the sae. Moentu is ass ties elocity, and the direction of the oentu is the sae as the direction of the elocity. Thus, the elocity directions ust be the sae. b. No. Moentu is ass ties elocity. The fact that the objects hae the sae oentu eans that the product of the ass and the agnitude of the elocity is the sae for each. Thus, the agnitude of the elocity of one object can be saller, for exaple, as long as the ass of that object is proportionally greater to keep the product of ass and elocity unchanged. 4. (a Can a single object hae kinetic energy but no oentu? (b Can a syste of two or ore objects hae a total kinetic energy that is not zero but a total oentu that is zero? Account for your answers. REASONING AND SOLUTION a. If a single object has kinetic energy, it ust hae a elocity; therefore, it ust hae linear oentu as well. b. In a syste of two or ore objects, the indiidual objects could hae linear oenta that cancel each other. In this case, the linear oentu of the syste would be zero. The kinetic energies of the objects, howeer, are scalar quantities that are always positie; thus, the total kinetic energy of the syste of objects would necessarily be nonzero. Therefore, it is possible for a syste of two or ore objects to hae a total kinetic energy that is not zero but a total oentu that is zero. 9. The drawing shows a garden sprinkler that whirls around a ertical axis. Fro each of the three ars of the sprinkler, water exits through a tapered nozzle. Because of this nozzle, the water leaes each nozzle with a speed that is greater than the speed inside the ar. (a Apply the ipulse oentu theore to deduce the direction of the force applied to the water. (b Then, with the aid of Newton s third law, explain how the water causes the whirling otion. REASONING AND SOLUTION a. Since the water leaes each nozzle with a speed that is greater than the speed inside the ar, the quantity is positie. Fro the ipulse-oentu theore, ( f 0 f 0 Σ F t =, and we can deduce that there is a net positie or outward ipulse. Therefore, a net outward force is exerted on the water. b. Fro Newton's third law, the water ust exert a net force that is equal in agnitude, but negatie and directed toward the nozzle. The nozzle and the ar, in
turn, oe. whirl. Since each ar is free to rotate about the ertical axis, the ar will 18. Where would you expect the center of ass of a doughnut to be located? Why?. REASONING AND SOLUTION Many objects hae a point, a line, or a plane of syetry. If the ass of the syste is uniforly distributed, the center of ass of such an object lies at that point, on that line, or in that plane. The point of syetry of a doughnut is at the geoetric center of the hole. Thus, the center of ass of a doughnut is at the center of the hole.