CHAPTER 7: Linear Moentu Solution Guide to WebAssign Probles 7.1 [1] p v ( 0.08 kg) ( 8.4 s) 0.4 kg s 7. [] Fro Newton s second law, p Ft. For a constant ass object, p v. Equate the two expression for p. Ft v v Ft. If the skier oves to the right, then the speed will decrease, because the friction force is to the left. v Ft ( 0 s) 5 N 65 kg The skier loses 7.7 s of speed. 7.7 s 7.3 [3] Choose the direction fro the batter to the pitcher to be the positive direction. Calculate the average force fro the change in oentu of the ball. p Ft v F v 0.145 kg t % 5.0 s # #39.9 s & 3.00 10 #3 s 7.4 [5] The force on the gas can be found fro its change in oentu. ( ) * 4.40 10 3 N, towards the pitcher F p t v v ( 4.0 10 4 s) ( 1500 kg s) 6.0 10 7 N downward, t t The force on the rocket is the Newton s 3 rd law pair (equal and opposite) to the force on the gas, and so is 6.0 10 7 N upward. 7.5 [8] Consider the otion in one diension, with the positive direction being the direction of otion of the first car. Let A represent the first car, and B represent the second car. Moentu will be conserved in the collision. Note that 0. v ( 9300 kg) ( 15.0 s # 6.0 s ) 6.0 s p final B v # v A A v 1.4 10 4 kg 1
7.6 [1] Consider the otion in one diension with the positive direction being the direction of otion of the bullet. Let A represent the bullet, and B represent he block. Since there is no net force outside of the block-bullet syste (like frictions with the table), the oentu of the block and bullet cobination is conserved. Note that 0. p final # B.0 kg ( 0.03 kg) 30 s # 170 s 0.69 s 7.7 [15] (a) The ipulse is the change in oentu. The direction of travel of the struck ball is the positive direction. p v ( 4.5 10 # kg) ( 45 s # 0).0 kg s The average force is the ipulse divided by the interaction tie. F p.0 kg s t 3.5 10 #3 s 5.8 10 N 7.8 [16] (a) The ipulse given to the nail is the opposite of the ipulse given to the haer. is the change in oentu. Call the direction of the initial velocity of the haer the positive direction. p nail p haer v i v f ( 1 kg) ( 8.5 s) 0 1.0 # 10 kg s The average force is the ipulse divided by the tie of contact. F avg p t 1.0 10 kg s 1.3 10 4 N 8.0 10 #3 s 7.9 [17] The ipulse given the ball is the change in the ball s oentu. Fro the syetry of the proble, the vertical oentu of the ball does not change, and so there is no vertical ipulse. Call the direction AWAY fro the wall the positive direction for oentu perpendicular to the wall. p v # v final initial ( vsin 45º # #vsin 45º ) vsin 45º ( 6.0 10 # k) ( 5 s)sin 45º.1kg s, to the left 7.10 [18] (a) The average force on the car is the ipulse (change in oentu) divided by the tie of interaction. The positive direction is the direction of the car s initial velocity.
# # 1 s & & F p t v % 0 50 k h % 3.6k h ( ( ( 1500 kg) % ( 1.389 ) 10 5 N * 1.4 ) 10 5 N t % 0.15 s ( % ( The deceleration is found fro Newton s nd law. F a a F ( 1.389 # 105 N) 93 s 1500 kg 7.11 [19] Call east the positive direction. (a) p original v original ( 95 kg) ( 4.0 s) 3.8 10 kg s fullback fullback The ipulse on the fullback is the change in the fullback s oentu. # p fullback v final % fullback v final fullback & ( ( 95 kg )( 0 4.0 s) 3.8 ) 10 kg s (c) The ipulse on the tackler is the opposite of the ipulse on the fullback, so 3.8 10 kg s (d) The average force on the tackler is the ipulse on the tackler divided by the tie of interaction. F p t 3.8 10 kg s 5.1 10 N 0.75 s 7.1 [] Let A represent the 0.440-kg ball, and B represent the 0.0-kg ball. We have 3.30 s and 0. Use Eq. 7-7 to obtain a relationship between the velocities. # + Substitute this relationship into the oentu conservation equation for the collision. + # B 0.0 kg 0.660 kg 3.30 s + 3.30 s + 1.10 s 1.10 s ( east ) 4.40 s( east) 7.13 [3] Let A represent the 0.450-kg puck, and let B represent the 0.900-kg puck. The initial direction of puck A is the positive direction. We have 3.00 s and 0. Use Eq. 7-7 to obtain a relationship between the velocities. # + 3
Substitute this relationship into the oentu conservation equation for the collision. + # B #0.450 kg 1.350 kg 3.00 s + 3.00 s # 1.00 s #1.00 s 1.00 s ( west ).00 s( east) 7.14 [4] Let A represent the ball oving at.00 /s, and call that direction the positive direction. Let B represent the ball oving at 3.00 /s in the opposite direction. So.00 s and 3.00 s. Use Eq. 7-7 to obtain a relationship between the velocities. # 5.00 s + Substitute this relationship into the oentu conservation equation for the collision, noting that B. + + #1.00 s + + 5.00 s 5.00 s +.00 s #6.00 s #3.00 s The two balls have exchanged velocities. This will always be true for 1-D elastic collisions of objects of equal ass. + M 7.15 [31] Fro the analysis in Exaple 7-10, the initial projectile speed is given by v gh. Copare the two speeds with the sae asses. v v 1 + M + M gh gh 1 h h 1 h h 1 5..6 v v 1 7.16 [3] Fro the analysis in the Exaple 7-10, we know that v + M h 1 v % g # + M & gh 0.1607 ( 0.16 Fro the diagra we see that 1 9.8 s 0.08 kg + 3.6 kg ( 0.08 kg) 30 s % # & 4
L ( L h) + x x L ( L h) (.8 ) (.8 0.1607 ) 0.94 7.17 [34] Use conservation of oentu in one diension, since the particles will separate and travel in opposite directions. Call the direction of the heavier particle s otion the positive direction. Let A represent the heavier particle, and B represent the lighter particle. We have 1.5 B, and 0. p final 0 # B # 3 The negative sign indicates direction. Since there was no echanical energy before the explosion, the kinetic energy of the particles after the explosion ust equal the energy added. E added K E A + K E B 1 A + 1 B 1 K E B 3 E 5 added 3 5 7500 J 4500 J K Thus K E A 3.0 10 3 J K E B 4.5 10 3 J ( 1.5 B )( 3 ) + 1 B 5 3 1 ( B ) 5 K E 3 B E A E added K E B 7500 J 4500 J 3000 J 7.18 [35] Use conservation of oentu in one diension. Call the direction of the sports car s velocity the positive x direction. Let A represent the sports car, and B represent the SUV. We have v 0 and Solve for. B. p final + 0 ( ) The kinetic energy that the cars have iediately after the collision is lost due to negative work done by friction. The work done by friction can also be calculated using the definition of work. We assue the cars are on a level surface, so that the noral force is equal to the weight. The distance the cars slide forward is x. Equate the two expressions for the work done by friction, solve for, and use that to find. W fr ( KE final KE initial ) affer 0 1 ( ) collision W fr F fr #x cos 180º µ k ( ) g#x g#x 1 µ k + B + B 3.191 s % 3 s µ k g#x µ k g#x 90 kg + 300 kg 90 kg ( 0.80) ( 9.8 s )(.8 ) 5
7.19 [40] Use this diagra for the oenta after the decay. Since there was no oentu before the decay, the three oenta shown ust add to 0 in both the x and y directions. ( p nucleus ) x p neutrino ( p nucleus ) y p electron p nucleus ( p nucleus ) x + ( p nucleus ) y ( p neutrino ) + ( p electron ) + ( 9.30 10 3 kg s) 1.08 10 kg s ( ( p neutrino ) 9.30 10 3 kg s) tan1 ( 5.40 10 3 kg s) 59.9º 5.40 10 3 kg s # tan 1 p nucleus y p tan 1 electron x p nucleus The second nucleus oentu is 150 o fro the oentu of the electron. 7.0 [46] Use Eq. 7-9a, extended to three particles. x CM x A x B + C x C + C 0.44 1.00 kg 0 + ( 1.50 kg ) ( 0.50 ) + ( 1.10 kg )( 0.75 ) 1.00 kg + 1.50 kg + 1.10 kg 7.1 [47] Choose the carbon ato as the origin of coordinates. x CM Cx C + O x O C + O ( 1 u) ( 0) + ( 16 u) 1.13 1010 1 u + 16 u 7. [48] Find the CM relative to the front of the car. 6.5 10 11 fro the C ato. x CM carx car + front x front + back x back car + front + back ( 1050 kg ) (.50 ) + 70.0 kg 1050 kg + 70.0 kg (.80 ) + 3 ( 70.0 kg )( 3.90 ) + 3( 70.0 kg).74 6