Physics 141H Homework Set #4 Solutions

Transcription

1 Phsics 4H Hmewrk Set #4 Slutins Chapter 4: Multiple-chice: 4) The maimum net frce will result when the tw frces are in the same directin. If s, the net frce will hae manitude F + F. The minimum net frce will ccur if the tw frces are in ppsite directins. In this case, the net frce will hae manitude F - F. Fr an ther relatie directin f the tw frces, the manitude f the net frce will be in between these alues. Thus, (C) is the crrect answer. 9) a) Here s what we knw: the rane is 40 miles, and the shell is launched with an initial anle f 45 (which ies the lnest rane fr a ien initial elcit). We als knw frm smmetr that the shell reaches its maimum heiht 0 miles frm the un (i.e., halfwa t the taret). The heiht at this pint is ien b t sinφ t. We als knw that at the maimum heiht, the cmpnent f elcit is 0. This lets us find t: sinφ t 0 sinφ t Pluin this back int ur equatin fr, we find that ( sinφ ). Nw we cmpare this heiht t the hrizntal distance traeled in the same time. This is csφ sinφ ien b t csφ. Fr the special case f φ 45, sinφ csφ, s we find that: sin φ In ther wrds, the maimum heiht is ½ f the hrizntal distance traeled t the maimum, r abut 0 miles in this case. (E)

2 ( ) sinφ b) Abe we fund that the maimum heiht is ien b. We knw φ ο is 45, s sin φ. Since we fund the maimum t be 0 miles, we hae: ( )( ) 4 3ft/s 0mi ( )( ) 3ft/s 5,800ft 600ft/s Chice (B) is the best apprimatin. 3) The parachute adds a dra frce in the ppsite directin f rait. Assumin this frce is reater in manitude than rait (which had better be true r else the parachutist will keep acceleratin dwnward!), the parachutist will under upward acceleratin immediatel after the chute pens. Nte thuh, that his elcit will still be dwnward. Eentuall the parachutist s dwnward elcit will decrease t er near the terminal elcit, as which pint the dra frce f the air n the parachute will eactl balance rait. Frm then n, the parachutist will drift t Earth at a cnstant (hpefull safe!) elcit. Chice (A) best represents this mtin. Questins: 7) a) N b) Let s assume the rpe were hrizntal. Then the free-bd diaram fr the weiht wuld be: T T m We see that rait acts in the directin, but bth tensins hae nl cmpnents. Therefre the weiht wuld accelerate dwnwards. T aid this, the tensin in the rpe needs t hae a cmpnent, which is nl pssible if the rpe is nt eactl hrizntal. ) Since the cmpnent f the elcit is cnstant, the ttal manitude f elcit is reatest when the cmpnent has the reatest abslute alue, and least when the cmpnent has the least abslute alue. The speed in is reatest just after the particle is launched (assumin it was launched upwards!) and just befre it hits the rund. It reaches its minimum at the maimum heiht.

3 5) a) This is a trick questin! Since all three prjectiles rise t the same heiht, the must all hae the same fliht time. (Remember that the mtin in is independent f that in.) b) A trick aain! All f them hae the same initial elcit in the directin. c) Since all three particles sta in the air fr an equal amunt f time, the elcit is cnstant thruhut the fliht, and particle c es the farthest in befre landin, particle c must hae the reatest hrizntal elcit cmpnent. d) B the same tken, particle a has the smallest hrizntal elcit. Since the ertical elcit cmpnent is the same fr all three particles, this als implies that particle a has the smallest speed. 8) Nt at all. Cnsider the mtin f the particle in and as a functin f time (assume 0 at t 0): t,, t t The secnd equatin tells us that the heiht s. time raph is a parabla. Slin the upper equatin fr t and pluin that int the lwer ne ies us as a functin f : t,,,, Nw we see that the heiht s. distance raph is als a parabla bth are cnsequences f the same phsics.

4 Eercises: 7) a) The free-bd diaram fr the blck is: N P θ m Takin the cmpnents f each ectr, we can find the net frce in the and directins: F P csθ F N + Psinθ m We assume that the flr is strn enuh that the blck can t fall thruh. Therefre F can nt be less than 0, and the flr will pride a nrmal frce lare enuh t uarantee this. In ther wrds, if m is reater than the nrmal frce will make up the difference; if m is less than Psinθ there will be n nrmal frce and the blck will eperience a net frce with a psitie cmpnent. In this prblem, Psinθ N sin5 5.N, while m 5.k 9.8m/s 50N. Since m is reater than Psinθ, a nrmal frce f 46.9N will be applied b the flr, and F will equal 0. S, the net frce n the blck is F P cs θi, and the acceleratin is: F m P csθ m a i.m/s i b) The blck will lift ff the flr when the nrmal frce is 0. As discussed abe, this will happen when Psinθ m, s ( 5.k)( 9.8m/s ) m P 8N sinθ sin 5 c) Just befre the blck lifts ff the flr the net frce will still be nl in the directin, and the acceleratin will be:

5 P csθ 8N cs 5 m 5.k a i i.0m/s i 9) Let s start with the free-bd diaram fr the car: T 7 N m 8 With the aes chsen as shwn, the car mes nl in the directin. The net frce is thus: F T cs 7 m sin8 Since we want the maimum distance the car can in 7.5s, we need t appl the maimum pssible frce. S set T t the breakin strenth f the rpe, 4.6kN. With this we find: F 4.6kN cs 7 00k 9.8m/s sin8 4.kN 3.6kN 500N This means that the car s acceleratin is it traels in the first 7.5s is: F 500N 0.4m/s a, s that the distance m 00k at.5m 7) The heiht is ien b t t,. Takin the ais hrizntal, this becmes t t time and set it equal t 0 (phsicall, this just means that the elcit is 0 at the maimum heiht): sinφ. T find the maimum alue, we take the deriatie with respect t

6 d sinφ t 0 dt sinφ t Usin this time in ur equatin fr, we find that: sinφ sinφ ma sinφ ( sinφ ) ) Fr simplicit, we take 0 at the mment the ball is released. The basket is three feet abe the release pint and 3 ft awa hrizntall, s we want 3ft when 3ft: t cs55 3ft t t sin 55 We can sle the first equatin fr t and plu int the secnd: 3ft 3ft t cs55 sin 55 3ft 3ft 3ft 3ft 3ft 9ft 3ft 3ft 6ft 3ft 3ft 59ft /s ( 3ft ) 3ft/s ( 3ft ) 3ft/s Prblems: 3) a) Start with the free-bd diaram fr the rcket:

7 T 58 m The net frce n the rcket while the enine is firin is: F cs58 6.kN cs58 3.4kN T F T m sin kN 3030k 9.8m/s.kN Therefre, a a 3.4kN 0.7m/s 3030k.kN 7.3m/s 3030k Since the rcket was launched at rest frm 0, its heiht when the rcket burns ut is ien b: a 7.3m/s ( 48s ) t 8.4km b) After the enine burns ut, the rcket is a prjectile. At the mment the rcket burns ut: 8.4km a 0.7m/s ( 48s ) t.3km a t 7.3m/s 48s 350m/s a t 0.7m/s 48s 53m/s We can nw find the time between the enine burnut and the rcket returnin t 0:

8 Applin the quadratic frmula, we find: t t +, 0 ± 4 t,, ( ) 350m/s ± 350m/s m/s 8.4km t 9.8m/s 90.4s r -9.0s We nl care abut the psitie-time slutin (e.. the time after the enine burns ut). We can nw find the hrizntal distance: +, t.3km + 53m/s 90.4s 58.7km 6) a) We hae alread shwn (Eercise 7) that a prjectile reaches a maimum heiht f: ma ( sinφ ) It reaches this pint halfwa thruh its trip; hence the alue fr which ma is reached is half the rane, r (usin Equatin 4-4 frm the tet): ma R sin φ sinφ csφ Therefre, tanθ ( sinφ ) ma ma tan φ sinφ sinφ csφ csφ

9 b) If φ 45, tanφ, s: θ tan 7 9) Fr the ftball t clear the crssbar, must be reater than 3.44m when 50m. This time is ien b: The psitin at that time is:, t t csφ 50m 50m t csφ, t t t sinφ t 3.44m 50m 50m sinφ 3.44m csφ csφ sinφ 50m 50m 3.44m csφ csφ sinφ 50m csφ 5m/s csφ sinφ 0m 50m 3.44m csφ cs φ 50m 9.8m/s 3.44m -3.44m cs φ+50m sinφ csφ 0m 0 cs φ The abe equatin will be true wheneer the numeratr n the left-hand side is 0: -3.44m cs φ+50m sinφcsφ 0m 0 50m cs φ csφ 3.44m cs φ + 0m Squarin bth sides ies: ( ) 500m cs φ cs φ.8m cs φ + 38m cs φ + 400m 4 4 5cs φ 36cs φ

10 Nte that ur units (m ) hae cancelled and can be drpped. This makes sense, since the anle needed desn t depend n hw we measure distance! We can appl the quadratic equatin t sle fr cs φ : cs ( )( ) 36 ± φ 0. r csφ 0.47 r 0.85 φ 6 r 3 ) The manitude f the displacement ectr is ien b: r ( t) r( t) ( t) + ( t) T determine if r is increasin r decreasin, we cmpute: d d + dr dt dt dt S the cnditin fr r alwas increasin is + 0. Thus we want t find the anle f launch such that the abe quantit equals 0 at ne pint durin the fliht. Pluin in what we knw abut,,, and fr prjectile mtin, this becmes: cs cs + sin sin t cs φ tsin φ t sinφ + t t ( sin φ cs φ ) + t sinφ t ( t φ )( φ ) t φ t ( φ t) We can cancel ne t eerwhere (and use a tri identit) t btain: t sinφ t

11 This is a quadratic equatin in t. In eneral, an equatin f the frm a + b + c 0 will hae a slutin when b 4ac 0 (as can be seen frm the quadratic frmula). Applin that t ur prblem, we want the case when we just barel hae a slutin, b 4ac 0, r: 9 sin φ Nw we see that all the s and s cancel ut, which is d news ur answer wn t depend n what planet we re n, nr n hw fast the bject is launched. After the cancellatins, we hae: Chapter 5: 9 sin φ sin φ 9 sinφ 0.94 φ 70 Multiple chice: ) The tensin in each rpe must be 0N t supprt the weiht n each side, and the sprin scale measures the frce applied b the sprin. T see what this is, make a freebd diaram f the scale s hk: T F S Since we knw that the scale is nt acceleratin, the sprin frce F S must be just lare enuh t cancel the tensin, r 0N (C). 5) Let s make a free-bd diaram fr a little piece f the rpe: T m The weiht bein supprted is that f the little bit f rpe, plus all the rpe belw it. At the bttm f the rpe, this weiht is er small (0 in the limit that the little bit f rpe

12 we cnsider is infinitel small), s the tensin is 0. At the tp, the full weiht f the rpe must be supprted, s the tensin is 0.98N. (D) Eercises: 3) a) The free-bd diaram fr the lamp is: T m If the eleatr is deceleratin, the lamp must eperience a net upwards frce: net ( ) F T m j maj T m ma m( a + ) T T 89N m 7.3k a +.4m/s + 9.8m/s b) Trick questin aain! The net acceleratin was.4m/s upwards in part (a), and we alread knw that the tensin is 89 N. 5) a) We start with the free-bd diarams fr the sandba and man: Sandba: Man: T T m S m M

13 Nte that T is the same in bth diarams (the rpe can nl eert ne tensin). Als, all the mtin is in the directin, which is cnenient. We can find the acceleratin f bth bjects: a a F net,s S ms F net,m M mm T ms j m S T mm j m M Since the strin can t be stretched, bth the man and the sandba hae the same manitude f acceleratin but in different directins. S we knw that: T ms mm T m m S m T m m m m m T ( ) T m + m m m M M S M S M S M S S M ( )( )( ) m 74k 0k 9.8m/s SmM T 870N m + m 74k + 0k With this, we can find the man s acceleratin: T mm am j m This is dwnward, as we suspected. M S M 870N 0k 9.8m/s j.9m/s j 0k We can nw find ut hw ln it takes the man t fall m (startin frm rest): His elcit upn landin is: at (.9m/s ) m 4m t 3.6s.9m/s t m

14 at.9m/s 3.6s 6.8m/s b) If the man attempts t climb up the rpe, he will pull it er the pulle faster. This means that the acceleratin f the sandba must increase, which in turn implies that the tensin in the rpe must increase. Since the tensin acts aainst rait, increasin it will reduce the man s acceleratin.