PHYSICS 203-NYA-05 MECHANICS

Transcription

1 PHYSICS 03-NYA-05 MECHANICS PROF. S.D. MANOLI PHYSICS & CHEMISTRY CHAMPLAIN - ST. LAWRENCE 790 NÉRÉE-TREMBLAY QUÉBEC, QC GV 4K TELEPHONE: EXT smanol@slc.qc.ca WEBPAGE: SOLUTIONS TO PROBLEM SET # TWO-DIMENSIONAL MOTION ) The Coyote s chasng the Roadrunner. The Coyote s strapped to jet-powered roller skates whch provde hm wth a constant horzontal acceleraton o 5.0 m/s. He starts 70.0 m rom the edge o a cl, racng towards the Roadrunner. O course at the last nstant, the Roadrunner runs away and the Coyote les o the edge o the cl. I the cl s 00.0 m above the base o a canyon, (a) determne where the Coyote wll land? (b) What wll be the magntude and the drecton o the Coyote's velocty just beore he hts the ground? Assume that the jet-powered skates are on all the tme. To calculate the ntal velocty o the Coyote at the edge o the cl v v ax v 0 x 70. 0m a 5. 0m s v m s 70. 0m 00m s v 45. 8m s Ths velocty wll be only horzontal, and t wll be the ntal velocty o the Coyote when he leaves the top o the cl. As the Coyote alls, the moton wll be two-dmensonal and he wll be subject to an acceleraton whch wll have two components; a horzontal component due to hs jet powered skates and a vertcal component due to gravty. Usng a reerence rame wth the orgn at the pont at whch the Coyote leaves the top o the cl, Intal: t 0 r 0 v 45. 8m s Fnal: t t r x 00m j v v x v y j a 5. 0m s 9. 80m s j constant where t s the tme at whch the Coyote wll ht the ground, x s the horzontal dstance that he wll cover when he hts the ground, 00 m s the heght o the cl, and v x and v y are the nal components o the velocty along the x and y drectons, respectvely. Note the nal velocty along x wll not be equal to the ntal velocty along x because there s an acceleraton along the x-drecton. Applyng the knematcal equatons n the horzontal drecton x x x v t a t x. m s t. m s t x. m s t. m s t v v a t v. m s. m s t x x x x Applyng the knematcal equatons n the vertcal drecton y y y v t a t m t. m s t m. m s t v v a t v. m s t y y y y The rst equaton above can be used to solve or the tme t wll take the Coyote to ht the ground and the result can be used to calculate all o the other quanttes Page o 7

2 Hence 00m m. m s t t. s 4. 90m s y v 9. 80m s t v 9. 80m s 4. 5 s 44. 3m s y x x 45. 8m s t 7. 50m s t x 45. 8m s 4. 5 s 7. 50m s 4. 5 s 360m v. m s. m s t v. m s. m s. s. m s x (a) Thereore the Coyote wll land 360m rom the base o the cl. v v v j. m s. m s j. The magntude and drecton o (b) Hs nal velocty wll be gven by the velocty wll be x y v 3. 6m s 44. 3m s m s 44. 3m s tan below the horzontal (nto the ground) 3. 6m s ) A cannon wth a muzzle velocty o 000 m/s s to be used to destroy a target on top o a mountan. The target s 000 m rom the cannon horzontally and 800 m above the ground. At what angle relatve to the ground should the cannon be red? Ths s a straghtorward projectle moton problem. The acceleraton s due to gravty and t s constant. Selectng a reerence rame wth and orgn at the pont at whch the cannon s stuated, Intal: t 0 r 0 v v cos v sn j Fnal: t t r x yj v v x v y j a. m s j 9 80 constant where v s 000 m/s, x s 000 m, y s 800 m and the nal velocty n the x-drecton v x s equal to the ntal velocty n the x-drecton v x. Applyng the knematcal equatons n the horzontal drecton x x x v t a t x v cos t t 000m 000m s cos t t cos. 00s Applyng the knematcal equatons n the vertcal drecton y y y v t a t y v sn t. m s t Solve the vertcal equaton or tsn 800m 000m s sn t 4. 90m s t m m s sn t. m s t m s sn t. m s t m 4. 90m s 800m 3 t sn t st s 000m s 000m s Square the sn and cos equatons then add them t cos. 00s t cos 4. 00s 3 3 t sn s t s t sn s t s Page o 7

3 t cos t sn. s. s t. s. s t. s t cos sn 4. 64s t. 400 s t s t t 4. 64s 0 Usng the quadratc ormula to solve or t s t t 4. 64s 0 b b 4ac t s 4. 64s a s t s or s t. 4 s or 03. 3s s Snce there are two tmes, there wll be two possble ntal angles and these tmes can be used to calculate the ntal angles. 00s t cos. 00s cos s t sn s t s 03. 3s sn s 03s s Almost straght up or sn s t cos. 00s cos s 3 3 t sn s t s. 4s sn s. 4s s sn ) Two cars, A and B, are movng past a guard G as shown n the gure on the rght. Relatve to the guard G, B travels at a constant speed o 0.0 m/s at an angle = 30. Relatve to the guard G, A has accelerated rom rest at m/s at = 60. At some tme, A has a speed o 40.0 m/s. At ths tme, (a) what are the magntude and drecton o the velocty o A relatve to B? (b) What are the magntude and drecton o the acceleraton o A relatve to B? (a) The velocty o A relatve to G at ths nstant s AG v. m s cos sn j.. j m s. Smlarly, the velocty o B relatve to G at ths nstant s BG v. m s cos sn j.. j m s. In general, vob voa vab, the velocty o object O as seen by B s the vector sum o the velocty o the object O as seen by A and B the velocty o A as seen by B. Here, the object o nterest s car A, O A. We requre the velocty o A as seen by B, hence B B. Ths means that A G. Thereore, vab vag vgb where v GB s the velocty o the guard as seen by B whch s the negatve o the velocty o B as seen by the guard. Hence, AB AG BG v v v.. j m s.. j m s.. j m s and the magntude and drecton o v AB are A G Page 3 o 7

4 v. 68m s 4. 6m s 4. 8m s AB tan north o east or 6. east o north (b) Acceleraton s the change n velocty as a uncton o tme. Usng vab vag vbg, v v v t t t AB AG BG aab aag abg but the acceleraton o B wrt G s zero, B s not acceleratng as seen by the guard. Thereore, aab aag, and the acceleraton o A as seen by B s equal to the acceleraton o A as seen by the guard G,.e. a AG m s at 60 north o east. 4) A 00-m wde rver has a unorm low speed o.0 m/s towards the east. A person leaves the south bank o the rver n a boat whch travels wth a speed o 4.00 m/s n the water and wants to reach a pont on the north bank whch s 8.0 m upstream rom a pont drectly opposte hs/her departure pont. (a) In what drecton must the boat head? (b) How long wll t take to cross the rver? The dagram or ths case s shown on the rght. v WS s the velocty o the water wrt the shore, the velocty o the current as seen by someone on the shore. v s the velocty o the boat as seen by someone on the shore. The boat must end up gong n ths drecton to end up 8.0 m upstream, meanng aganst the current. Thereore, the velocty o the boat as seen by someone on the water, v BW, must be as shown n the dagram. Each o the veloctes can be expressed usng the set o axes shown n the dagram, v. 0m s WS v v cos vsnj v v cos v snj BW BW BW Usng the relatonshp between the vectors shown n the dagram, v vbw vws, The horzontal and vertcal components must satsy horzontal: vbw cos. 0m s v cos, vertcal: v sn v sn respectvely. v BW s gven and the angle can be ound rom the dagram BW 00m tan m 0 Hence, there are two equatons n two unknowns, v and. Solvng or cos and sn v cos m s horzontal: cos. 0m s v cos cos v sn vertcal: sn v sn sn Squarng and addng 8.0 m 00.0 m Page 4 o 7

5 v cos m s v sn cos sn v cos m s v sn cos sn v sn v cos m sv cos m s v v. 0m s 0 v m s v 4. 79m s 0 v m s m s m s v m s or 4. 9m s v 3. 45m s Ths s the speed o boat as seen by someone on the shore. (a) The boat n the water must be headed at the angle wrt the shore v cos m s 3. 45m s cos m s cos cos m s sn sn v sn sn At an angle o 53 north o west or 37 west o north. (b) The tme requred to cross the rver can be calculated once the dstance travelled n the water s ound because the speed o the boat n the water s known. Usng the angle wdth o rver w 00m sn dstance travelled: d 50m dstance travelled sn sn53 d 50m 50 m at a speed o 4.00m s t 6. 6s v 5) A block o mass m equal to.50 kg s pushed up a rctonless nclned plane o heght h equal to.65 m by a constant orce o magntude 8.0 N drected as shown n the gure on the rght. I the angle o nclnaton s 30, and the block starts rom rest, how long wll t take the block to reach the top o the nclne? I the heght o the nclned plane s.65 m then the dstance that the block must travel on the nclned plane s d sn m d 3. 30m I the acceleraton o the block s ound then x v t a t can be used to nd the tme t wll take or the block to travel up the nclned plane. The acceleraton o the block can be obtaned rom a Newton s Second Law analyss. Consder the bd or the block on the let. Each orce can be expressed n terms o components wrt the set o axes shown, exerted by the Earth: exerted by the plane: exerted by person: acceleraton: Applyng Newton s Second Law along each drecton a a F mg sn mg cos j g m F P FPj F F cos F sn j app app app h Page 5 o 7

6 y: FP mg cos Fapp sn 0 not necessary n ths case Fapp cos Newton's Second Law: F ma x: Fapp cos mg sn ma a g sn m 8. 0N cos 30 a 9. 80m s sn m s. 50kg Thereore the tme requred or the block to move up the plane s 3. 30m x vt a t 3. 30m 0. 34m s t t. s. 34m s 6) What must be the magntude and drecton o a horzontal orce F suppled to the wedge o mass M shown n the gure on the let n order or the block m to reman statonary wth respect to mass M? Assume all suraces are rctonless and M =.5 kg, m = 3.50 kg and the angle o nclnaton o the wedge s 7. I the appled orce s drected to the rght then the small block wll dentely end up at the bottom o the wedge. Thereore, t must be drected to the let. Once the appled orce s exerted on the wedge, t wll accelerate horzontally to the let. I the block s to reman statonary wrt the wedge t must have the same acceleraton as the wedge, horzontally to the let. Newton's Second Law: Consder the bd or the block shown on the let. Expressng each orce n terms o the selected reerence rame, exerted by Earth: exerted by wedge: acceleraton: F gm mgj FMm F Mm sn FMm cos j a a Applyng Newton s Second law n both drectons F ma y: FMm cos mg 0 x: FMm sn ma FMm sn ma The y equaton can be solved or the magntude o the orce exerted by the wedge on the block and t can then be used to calculate the acceleraton o the system, y: mg FMm cos mg 0 FMm cos x: mg FMm sn ma sn ma cos a g tan 9. 80m s tan m s The bd or the wedge gves exerted by Earth: exerted by loor: exerted by block: exerted by person: acceleraton: F gm app Mgj FF F F j F mm FmM sn FmM cos j F F a a Applyng Newton s Second law n both drectons app m? M Page 6 o 7

7 Newton's Second Law: F ma y: FF Mg FmM cos 0 x: Fapp FmM sn Ma Fapp FmM sn Ma The acceleraton o ths wedge s the same as the acceleraton o the block calculated prevously. Furthermore, The orce exerted by M on m, F Mm, s equal n magntude to the orce exerted by m on M, F mm, because o Newton s Thrd Law. From the bd or m mg FMm FmM a g tan m s cos F F sn Ma app mm mg Fapp sn Mg tan M m g tan. 5kg 3. 50kg 4. 99m s 79. 8N cos Page 7 o 7