Easy. r p 2 f : r p 2i. r p 1i. r p 1 f. m blood g kg. P8.2 (a) The momentum is p = mv, so v = p/m and the kinetic energy is

Transcription

1 Chapte 8 Homewok Solutions Easy P8. Assume the velocity of the blood is constant ove the 0.60 s. Then the patient s body and pallet will have a constant velocity of m m/ s 0.60 s in the opposite diection. Momentum consevation gives p i p 2i p f p 2 f : 0 m blood m s 54.0 kg m/ s m blood kg 40.5 g P8.2 (a) The momentum is p = mv, so v = p/m and the kinetic enegy is K 2 mv 2 2 m p m 2 p2 2m K 2 mv 2 implies v 2K m so p mv m 2K m 2mK. v 3.00î P8.3 We ae given m = 3.00 kg, 4.00ĵ m/ s (a) p mv 9.00î 2.0ĵ kg m/ s Thus, p x 9.00 kg m/ s and p y 2.0 kg m/ s. p p 2 x p 2 y 9.00 kg m/ s kg m/ s kg m/ s tan p y p x tan

2 P8.8 (a) I F avg t, whee I is the impulse the man must delive to the child: o I F avg t p child m child v f v i F avg m child v f v i t F avg m child v f v i t N 2.0 kg0 60 mi/ h 0.0 s m/ s mi/ h F avg N lb N 720 lb (c) The man s claim is nonsense. He would not be able to exet a foce of this magnitude on the child. In eality, the violent foces duing the collision would tea the child fom his ams. These devices ae essential fo the safety of small childen. P8.5 (a) We wite the law of consevation of momentum as mv i 3mv 2i 4mv f o 4.00 m/ s m/ s v f 4 K f K i 2 4mv 2 f 2 mv i 2 2 3m 2.50 m/ s v 2 2i kg[4(2.50 m/ s) 2 (4.00 m/ s) 2 3(2.00 m/ s) 2 ] J P8.6 (a) The intenal foces exeted by the acto do change the total momentum of the system the fou cas and the movie acto. Consevation of momentum ANS. FIG. P8.6 gives not of

3 (c) 4mv i 3m 2.00 m s m 4.00 m s v i W acto K f K i 6.00 m s 4.00 m s 4 2 3m2.00 m s2 m 4.00 m s W acto kg m s 2 4m2.50 m s m s kj The event consideed hee is the time evesal of the pefectly inelastic collision in the pevious poblem. The same momentum consevation equation descibes both pocesses. P8.2 (a) Fom the text s analysis of a one-dimensional elastic collision with an oiginally stationay taget, the x component of the neuton s velocity changes fom v i to v f = ( 2)v i /3 = v i /3. The x component of the taget nucleus velocity is v 2f = 2v i /3. The neuton stated with kinetic enegy (½)m v i 2. The taget nucleus ends up with kinetic enegy (½) (2 m )( 2v i /3) 2 Then the faction tansfeed is 2m (2v / 2 i 3)2 m v i Because the collision is elastic, the othe 7.6% of the oiginal enegy stays with the neuton. The cabon is functioning as a modeato in the eacto, slowing down neutons to make them moe likely to poduce eactions in the fuel. K n (0.76)( J) J and K C (0.284)( J) J

4 P8.32 (a) The vecto expession fo consevation of momentum, p i p f :, gives p xi p xf and p yi p yf mv i mv cos mv cos [] 0 mv sin mv sin [2] Fom [2], sin sin so ANS. FIG. P8.32 Futhemoe, enegy consevation fo the system of two potons equies so 2 mv 2 i 2 mv 2 2 mv 2 v v i 2 Hence, [] gives v i 2v i cos 2 with 45.0 and 45.0 P8.39 (a) m v CM iv i M m v m 2v 2 M 5.00 kg [(2.00 kg)(2.00î m s 3.00ĵ m s) (3.00 kg)(.00î m s 6.00ĵ m s)] v CM.40î 2.40ĵ m s p Mv CM (5.00 kg)(.40î 2.40ĵ) m s (7.00î 2.0ĵ) kg m s

5 Medium P8.6 (a) The gil-plank system is isolated, so hoizontal momentum is conseved. We measue momentum elative to the ice: p gi p pi p gf p pf. The motion is in one dimension, so we can wite v gi î v gp î v pi î v gi v gp v pi whee v gi denotes the velocity of the gil elative to the ice, v gp the velocity of the gil elative to the plank, and v pi the velocity of the plank elative to the ice. The momentum equation becomes 0 m g v gi î m p v pi î 0 m g v gi m p v pi 0 m g v gp v pi m p v pi 0 m g v gp m g m p m g v pi v gp m g m p v pi Using ou esult above, we find that v gi v gp v pi v gp v gi m g m p v gp m g v gp m g m p m g m p m g m p m g m g m p v gp v gi m v m v m v g gp p gp g gp m g m p m p v gi v gp m g m p

6 P8.9 p Ft p y mv yf v yi m(v cos 60.0) mv cos p x m(v sin 60.0 v sin 60.0) 2mv sin (3.00 kg)(0.0 m/ s)(0.866) 52.0 kg m/ s F avg p x t 52.0 kg m/ s s 260 N The foce is 260 N, nomal to the wall. P8. (a) The impulse deliveed to the ball is equal to the aea unde the F-t gaph. have a tiangle and so to get its aea we multiply half its height times its width: We I Fdt = aea unde cuve I s F 3.5 N s 9.00 kn s N 3.5 N s ANS. FIG. P8. P8.7 The collision between the clay and the wooden block is completely inelastic. Momentum is conseved by the collision. Find the elation between the speed of the clay (C) just befoe impact and the speed of the clay+block (CB) just afte impact: p Bi p Ci p Bf p Cf m B v Bi m C v Ci m B v Bf m C v Cf M 0 mv C mv CB Mv CB m M v C m M m v CB v CB Now use consevation of enegy in the pesence of fiction foces to find the elation between the speed v CB just afte impact and the distance the block slides befoe stopping:

7 K E int 0: 0 2 (m M )v 2 CB fd 0 and fd nd (m M )gd 2 (m M )v 2 CB (m M )gd v CB 2gd Combining ou esults, we have v C (m M ) 2gd m (2.0 g 00 g) 2.0 g v C 9.2 m/ s 2(0.650) 9.80 m/ s 2 (7.50 m) P8.9 The mechanical enegy of the isolated block-eath system is conseved as the block of mass m slides down the tack. we find v, the speed of m at B befoe collision: K i + U i = K f + U f 2 m v m gh v 2(9.80 m/ s 2 )(5.00 m) 9.90 m/ s ANS. FIG. P8.9 Fist Now we use the text s analysis of one-dimensional elastic collisions to find v f, the speed of m at B just afte collision. v f m m 2 v m m m/ s 3.30 m/ s Now the 5-kg block bounces back up to its highest point afte collision accoding to m gh max 2 m 3.30 m/ s 3.30 m/ s 2 h max m/ s m

8 P8.24 We assume equal fiing speeds v and equal foces F equied fo the two bullets to push wood fibes apat. These equal foces act backwad on the two bullets. Fo the fist, K i E mech K f gives kgv 2 F m 0 [] Fo the second, p i p f gives kg v f kgv.04 kg Again, K i E mech K f gives Substituting fo v f, v.04 kgv f kgv 2 Fd.04 kg 2 v 2 f kgv 2 Fd.04 kg kgv.04 kg Fd kgv kg 2 v 2.04 kg 2 Substituting fo v fom [] gives Fd F m d 7.94 cm kg.04 kg P8.29 By consevation of momentum fo the system of the two billiad balls (with all masses equal), in the x and y diections sepaately,

9 5.00 m/ s 0 (4.33 m/ s)cos 30.0 v 2 fx v 2 fx.25 m s 0 (4.33 m/ s)sin 30.0 v 2 fy v 2 fy 2.6 m/ s v 2 f 2.50 m/ s at 60.0 Note that we did not need to explicitly use the fact that the collision is pefectly elastic. ANS. FIG. P We could analyze the object as nine squaes, each epesented by an equal-mass paticle at its cente. But we will have less witing to do if we think of the sheet as composed of thee sections, and conside the mass of each section to be at the geometic cente of that section. Define the mass pe unit aea to be σ, and numbe the ectangles as ANS. FIG. P8.34 shown. We can then calculate the mass and identify the cente of mass of each section. m I = (30.0 cm)(0.0 cm) with CM I = (5.0 cm, 5.00 cm) m II = (0.0 cm)(20.0 cm) with CM II = (5.00 cm, 20.0 cm) m III = (0.0 cm)(0.0 cm) with CM III = (5.0 cm, 25.0 cm) The oveall cente of mass is at a point defined by the vecto equation: CM m ii m i Substituting the appopiate values, CM is calculated to be: CM Calculating, 300 cm cm 2 00 cm 2 {[(300)(5.0î 5.00ĵ) (200)(5.00î 20.0ĵ) (00)(5.0î 25.0ĵ)] cm 3 } 4 500î 500ĵ 000î 4 000ĵ 500î 2 500ĵ CM 600 and evaluating, CM (.7î 3.3ĵ) cm cm

10 P8.52 Using consevation of momentum fom just befoe to just afte the impact of the bullet with the block: mv i = (M+ m)v f o v i M m m v f [] The speed of the block and embedded bullet just afte impact may found using kinematic equations: d = v f t and h 2 gt 2 ANS. FIG. P8.52 be Thus, t 2h g and v f d t d g 2h gd2 2h Substituting into [] fom above gives v i M m m gd 2 2h. Had P8.55 (a) The initial momentum of the system is zeo, which emains constant thoughout the motion. Theefoe, when m leaves the wedge, we must have o m 2 v wedge + m v block = 0 (3.00 kg)v wedge + (0.500 kg)(+4.00 m/s) = 0 ANS. FIG. P8.55 so v wedge = m/ s Using consevation of enegy fo the block-wedge-eath system as the block slides down the smooth (fictionless) wedge, we have

11 K block U system K i wedge K i block U system K f wedge f o 0 m gh 0 2 m 4.00 m/ s m m/ s 2 which gives h m P8.60 (a) Use consevation of the hoizontal component of momentum fo the system of the shell, the cannon, and the caiage, fom just befoe to just afte the cannon fiing: o p xf p xi m shell v shell cos 45.0 m cannon v ecoil kg 25 m/ scos kgv ecoil 0 v ecoil 3.54 m/ s ANS. FIG. P8.60 Use consevation of enegy fo the system of the cannon, the caiage, and the sping fom ight afte the cannon is fied to the instant when the cannon comes to est. K f U gf U sf K i U gi U si kx 2 max 2 mv 2 ecoil 0 0 x max mv 2 ecoil k kg3.54 m/ s N/ m.77 m (c) F s, max kx max F s, max N m.77 m N (d) No. The ail exets a vetical extenal foce (the nomal foce) on the cannon and pevents it fom ecoiling vetically. Momentum is not conseved in the vetical diection. The sping does not have time to stetch duing the cannon fiing. Thus, no extenal hoizontal foce is exeted on the system (cannon, caiage, and shell) fom just befoe to

12 just afte fiing. Momentum of this system is conseved in the hoizontal diection duing this inteval.