Physics 40 HW #4 Chapter 4 Key NEATNESS COUNTS! Solve but do not turn in the following problems from Chapter 4 Knight

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1 Physcs 40 HW #4 Chapter 4 Key NEATNESS COUNTS! Solve but do not turn n the ollowng problems rom Chapter 4 Knght Conceptual Questons: 8, 0, ; 4.8. Anta s approachng ball and movng away rom where ball was thrown, so ball was thrown wth the greater speed. Ths can be determned numercally as well, treatng Anta as a movng reerence rame wth respect to the ground, so vanta vball 5 m/s. For ball, Anta measures 0 m/s v 5 m/s v 5 m/s. For ball, 0 m/s v 5 m/s v 5 m/s. So ball was thrown wth greater speed Zach should throw hs book outward and toward the back o the car. The book has the same ntal velocty as do Zach and the car, so throw or wll cause the book to land beyond the drveway n the same drecton as the car s travelng. 4.. Snce Zach and Yvette are travelng at the same speed they share the same reerence rame, so Zach should throw the book straght to her (throw.)

2 Problems: 0,, 3, 37, 47, 53, , Model: Use the partcle model or the puck. Solve: Snce the vx vs t and vy vs t graphs are straght lnes, the puck s undergong constant acceleraton along the x- and y- axes. The components o the puck s acceleraton are dvx vx 0 m/s ( m/s) ax.0 m/s dt t 0 s 0 s (0 m/s 0 m/s) ay m/s (0 s 0 s) x y The magntude o the acceleraton s a a a. m/s. Assess: The acceleraton s constant, so the computatons above apply to all tmes shown, not just at 5 s. The puck turns around at t 5 s n the x drecton, and constantly accelerates n the y drecton. Travelng 50 m rom the startng pont n 0 s s reasonable. 4.. Model: Assume the partcle model or the ball, and apply the constant-acceleraton knematc equatons o moton n a plane. Vsualze: Solve: (a) We know the velocty v ˆ ˆ (.0.0 j) m/s at t s The ball s at ts hghest pont at t s, so vy 0 m/s. The horzontal velocty s constant n projectle moton, so vx 0 m/s at all tmes. Thus v ˆ 0 m/s at t s We can see that the y-component o velocty changed by v y.0 m/s between t s and t s Because a y s constant, vy changes by.0 m/s n any -s nterval. At t 3 s, v y s.0 m/s less than ts value o 0 at value o.0 m/s at 0 t s Consequently, at t 0 s, t s At t 0 s, v y must have been.0 m/s more than ts v (.0ˆ 4.0 ˆj ) m/s, At t s, v() (.0ˆ.0 ˆj ) m/s At t s, v() (.0ˆ 0.0 ˆj ) m/s At t 3 s, v(3) (.0ˆ.0 ˆj ) m/s (b) Because v s changng at the rate.0 m/s per s, the y-component o acceleraton s y projectle moton, so the value o g on Exdor s g 0 m/s (c) From part (a) the components o v 0 are v0x.0 m/s and v0y 4.0 m/s. Ths means ay.0 m/s. But ay g or v0 y 4.0 m/s tan tan 63 above x v0x.0 m/s Assess: The y-component o the velocty vector decreases rom.0 m/s at m/s. All the other values obtaned above are also reasonable. t s to 0 m/s at t s Ths gves an acceleraton o

3 4.3. Model: The bullet s treated as a partcle and the eect o ar resstance on the moton o the bullet s neglected. Vsualze: Solve: (a) Usng 0 0y 0 y 0 y y v ( t t ) a ( t t ), we obtan t t (.0 0 m) 0 m 0 m ( 9.8 m/s )( 0 s) s s (b) Usng 0 0x 0 x 0 x x v ( t t ) a ( t t ), (50 m) 0 m v ( s 0 s) 0 m v 78 m/s 780 m/s 0x 0x Assess: The bullet alls cm durng a horzontal dsplacement o 50 m. Ths mples a large ntal velocty, and a value o 78 m/s s understandable Model: Assume the spaceshp s a partcle. The acceleraton s constant, so we can use the knematc equatons. Vsualze: We apply the knematc equaton s s v0t a ( t ) n each drecton. t 35 mn 00 s. Solve: 5 x km (9.5 km/s)(00 s) (0.040 km/s )(00 s) s km 5 y km (0 km/s)(00 s) (0 km/s )(00 s) s km 5 z.0 0 km (0 km/s)(00 s) (.00 km/s )(00 s) s km 3 Roundng to two sg gs gves r (70ˆ 400 ˆj 60 kˆ ) 0 km. Assess: The y-component ddn t change because there was no velocty or acceleraton n the y-drecton.

4 4.47. Model: We wll use the partcle model and the constant-acceleraton knematc equatons n a plane. Vsualze: Solve: The x-and y-equatons o the ball are From the y-equaton, B 0B ( 0B) x( B 0B) ( B) x( B 0B) 65 m 0 m ( 0B cos30 ) B 0 m B 0B ( 0B) y( B 0B) ( B) y( B 0B) 0 m 0 m ( 0B sn30 ) B ( ) B x x v t t a t t v t y y v t t a t t v t g t Substtutng ths nto the x-equaton yelds v 0B gtb (sn30 ) gcos30t 65 m sn30 t.77 s For the runner: 0 m tr.50 s 8.0 m/s Thus, the throw s too late by 0.7 s. Assess: The tmes nvolved n runnng the bases are small, and a tme o.5 s s reasonable. B B Model: We dene the x-axs along the drecton o east and the y-axs along the drecton o north.

5 Solve: (a) The kayaker s speed o 3.0 m/s s relatve to the water. Snce he s beng swept toward the east, he needs to pont at angle west o north. Hs velocty wth respect to the water s v (3.0 m/s, west o north) ( 3.0sn m/s) ˆ (3.0cos m/s) ˆj KW We can nd hs velocty wth respect to the earth v KE v KW v WE, wth v ˆ WE (.0 m/s). Thus v (( 30sn 0) m/s) ˆ (3.0cos m/s) ˆj KE In order to go straght north n the earth rame, the kayaker needs ( v ) 0. Ths wll be true x KE.0.0 sn sn Thus he must paddle n a drecton 4 west o north. (b) Hs northward speed s vy 3.0 cos(4.8 ) m/s.36 m/s. The tme to cross s The kayaker takes 45 s to cross. 00 m t 44.7 s.36 m/s Model: Mke and Nancy concde at t 0 s Use subscrpts B, M, N or the ball, Mke, and Nancy respectvely. Solve: (a) Accordng to the Gallean transormaton o velocty v BN v BM v MN. Mke throws the ball wth velocty v ˆ ˆ BM ( m/s)cos63 ( m/s)sn63 j, and v ˆ NM (30 m/s). Thus wth respect to Nancy v v v (cos63 30) ˆ m/s (sn63 ) ˆj m/s ( 0.0 ˆ 9.6 ˆj ) m/s BN BM NM v y 9.6 m/s tan tan 44.4 v 0.0 m/s The drecton o the angle s 44.4 above the x axs (n the second quadrant). (b) Wth respect to Nancy and x x BN BN (0.0 m/s) t y 0 m (9.6 m/s) t (9.8 m/s ) t (9.6 m/s) t (4.9 m/s ) t

6 4.79. Model: Use the partcle model or the arrow and the constant-acceleraton knematc equatons. Vsualze: Solve: Usng v y v0 y ay ( t t0 ), we get Also usng 0 0x 0 x 0 x x v ( t t ) a ( t t ), v 0 m/s gt v gt y y 60 m 60 m 0 m v t 0 m v v Snce vy/ vx tan , usng the components o v 0 gves 0x 0x x t gt (0.054)(60 m) t s (60 m/ t ) (9.8 m/s ) Havng ound t, we can go back to the x-equaton to obtan v0 60 m/0.566 s 06 m/s 0 m/s x Assess: In vew o the act that the arrow took only s to cover a horzontal dstance o 60 m, a speed o 06 m/s or 37 mph or the arrow s understandable.

7 4.80. Model: Use the partcle model or the arrow and the constant-acceleraton knematc equatons. We wll assume that the archer shoots rom.75 m above the slope (about 5 9). Vsualze: Solve: For the y-moton: For the x-moton: Because y/ x tan5 0.68, 0 0y( 0) y( 0).75 m ( 0 sn0 ) y y v t t a t t y v t gt.75 m (50 m/s)sn 0 y t gt 0 0x( 0) ( x 0) 0 m ( 0 cos0 ) 0 m (50 m/s)(cos0 ) x x v t t a t t v t t.75 m (50 m/s)(sn 0 ) t (50 m/s)(cos0 ) t gt 0.68 t 6. s and s (unphyscal) Usng t 6 s n the x- and y-equatons above, we get y 77.0 m and x 87 m. Ths means the dstance down the slope s x y (87 m) ( 77.0 m) 97 m. Assess: Wth an ntal speed o mph (50 m/s) or the arrow, whch s shot rom a 5 slope at an angle o 0 above the horzontal, a horzontal dstance o 87 m and a vertcal dstance o 77.0 m are reasonable numbers.

8 Turn n the ollowng. Show all your work clearly and neatly. Always draw a dagram, sketchng the stuaton and labelng varables dene drectons, etc Box nal answers. These are BRIEF INCOMPLET solutons. See solutons n glass case or complete solutons.. A partcle moves n the xy plane wth a constant acceleraton gven by a 4. 0j ˆ m/s. At t = 0, ts poston and velocty are 0 m and ( j) m/s, respectvely. What s the dstance rom the orgn to the partcle at t =.0 s? What s the velocty o the partcle? Sketch the dsplacement and velocty vector equatons as shown n the dagram 4.5 n the text. Answer: r (6ˆ 8 ˆj ) m, r 0m v ˆj m / s see keys n glass case or sketches.. A ball s thrown rom an upper story wndow o a buldng. The ball s gven an ntal velocty o 5.00 m/s at an angle o 0.0 above the horzontal. It strkes the ground 3.00 s later. a) Sketch the event, labelng EVERYTHING, ncludng the ntal and nal states. b) How ar horzontally rom the base o the buldng does the ball strke the ground? c) Fnd the heght rom whch the ball was thrown. d) Fnd the velocty that the balloon hts the ground wth (both magntude and drecton.) Answer: R 4. m, H 39.0 m, v (8. m / s, 80.0 ) 3. A quarterback throws a ootball straght toward a recever wth an ntal speed o 0.0m/s, at an angle o 35.0 above the horzontal. At that nstant, the recever s 0.0 m n ront (down eld) o the quarterback. Wth what constant speed should the recever run n order to catch the ootball at the level at whch t was thrown? Answer: v 7.86 / m s 5. A water balloon gun s held.40 m above the ground and ponted 5 degrees above the horzontal. It shoots a water balloon that lands on the ground 4.4 m away n the horzontal drecton. a) Sketch the event, labelng everythng, gvens, drectons, ncludng the ntal and nal states. b) Fnd the total tme the balloon travels n the ar. c) The ntal speed o the water balloon. d) Fnd the velocty that the balloon hts the ground wth. Answer: t 0.84 s, v 5.78 m / s, v (7.80 m / s, 47.8 )

9 4. A dve bomber has a velocty o 80 m/s at an angle below the horzontal. When the alttude o the arcrat s.5 km, t releases a bomb, whch subsequently hts a target on the ground. The magntude o the dsplacement rom the pont o release o the bomb to the target s 3.5 km. a) Fnd the dsplacement n the x drecton. b) Fnd the launch angle as shown. c) Fnd the tme t takes the bomb to strke the target. d) Fnd the velocty the bomb strkes the ground wth, both magntude and drecton and express t as an jk vector too! Answer: When the bomb has allen a vertcal dstance.5 km, t has traveled a horzontal dstance x gven by x 3.5 km.5 km.437 km gx y x tan v cos 50 m 437 m tan 9.8 m s 437 m 80 m s cos 50 m 437 m tan 37.9 m tan tan tan tan Select the negatve soluton, snce s below the horzontal. tan 0.66, Answers: x.437 km, 33.5, t 0.43 s, v (347 m / s, 47.8 ) 6. A tran travels due south at 30m/s (relatve to the ground) n a ran that s blown toward the south by the wnd blowng due south. The path o each randrop makes an angle o 70 degrees wth the vertcal, as measured by an observer statonary on the ground. An observer on the tran, however, sees the drops all perectly vertcally. Determne the speed o the randrops. Answer: v 3.9 m/ s RG 7. Tm n hs Corvette accelerates at the rate o (3.50ˆ.70 ˆj ) m / s, whle Jll n her Jaguar accelerates at ˆ ˆ ( j) m / s. They both start rom rest at the orgn o an xy coordnate system. Ater 5.0 s, Fnd Tm s and Jll s poston vectors, relatve to the orgn, expressed as jk vectors. What s the dstance between them? What s Jll s poston, relatve to Tm? Sketch all the vectors, labelng, etc. Answer: d 76.6 m, r ( 75.0ˆ 5.5 ˆj ) m, JT

10 8. A shp s launched rom shore s crossng a lake, headng 30.0 degrees west o north at 30.0 m/s relatve to the water. The velocty o the shp relatve to the shore s 0.0 m/s due north. Fnd the velocty (jk vector and magntude and drecton) o the water relatve to the shore both by graphcal methods and the method o components. Wrte the vector equaton or the relatve moton Express the drecton o the water relatve to due east (the + x-axs.) Calculate the percent derence or the velocty rom each method, whch should be wthn a ew percent o each other. Answer: v (6. m / s,0 ) WS 9. A coast guard shp s travelng at a constant velocty o 4.0 m/s, due east, relatve to the water. On hs radar screen the navgator detects an object that s movng at a constant velocty. The object s located at a dstance o 30 m wth respect to the shp, n a drecton 3.0 degrees south o east. Sx mnutes later, he notes that the object s poston relatve to the shp has changed to 0 m, 57.0 degrees south o west. What are the magntude and drecton o the velocty o the object relatve to the water? Express the drecton as an angle wth respect to due west. Try solvng both by methods o component and by law o snes and cosnes. Whch s easer? Answer: v (3.04 m / s,5. ) OW