Momentum and Collisions

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1 SOLUTIONS TO PROBLES Section 8. P8. m 3.00 kg, (a) omentum and Collisions Linea omentum and Its Consevation v ( 3.00î 4.00ĵ ) m s p mv ( 9.00î.0ĵ ) kg m s Thus, p x 9.00 kg m s and p y.0 kg m s. p p x + p y ( 9.00) + (.0) 5.0 kg m s p tan y ( ' tan.33 p x 307 P8.4 (a) The momentum is p mv, so v p m and the kinetic enegy is K mv m p m p m. K mv implies v K m, so p mv m K m mk. Section 8. Impulse and omentum *P8.6 Fom the impulse-momentum theoem, F (t) p mv f mv i, the aveage foce equied to hold onto the child is F m v f v i t ( kg) 0 60 mi h s 0 m s.37 mi h ' ( 6.44 ) 03 N. Theefoe, the magnitude of the needed etading foce is N cannot exet a foce of this magnitude and a safety device should be used., o 400 lb. A peson 7

2 8 omentum and Collisions P8.7 (a) I Fdt aea unde cuve I ( s) ( N) 3.5 N s 3.5 N s F 9.00 kn s (c) Fom the gaph, we see that F max 8.0 kn FIG. P8.7 P8.9 p Ft m v cos 60.0 mv cos mv sin 60.0 ( 0.0 m s) ( 0.866) p y m v fy v iy p x m v sin 60.0 v sin kg 5.0 kg m s F ave p x t 5.0 kg m s 0.00 s 60 N FIG. P8.9 Section 8.3 Collisions P8.3 (a) mv i + 3mv i 4mv f whee m kg v f (.00 ).50 m s 4 K f K i ( 4m) v f mv i + 3m v i ' (.50 ( 04 )( ) 3.75 ( 0 4 J P8.4 (a) The intenal foces exeted by the acto do not change the total momentum of the system of the fou cas and the movie acto ( 4m) v i ( 3m) (.00 m s) + m( 4.00 m s) v i 6.00 m s m s 4.50 m s FIG. P8.4 W acto K f K i 3m W acto kg (.00 m s) + m( 4.00 m s) ( ) m s 37.5 kj 4 m.50 m s (c) 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.

3 Chapte 8 9 P8.6 v, speed of m at B befoe collision. m v m gh v 9.80 ( 5.00) 9.90 m s v f, speed of m at B just afte collision. FIG. P8.6 v f m m v m + m 3 ( 9.90) m s 3.30 m s At the highest point (afte collision) m gh max m ( 3.30) h max 3.30 m s 9.80 m s m P8.9 (a) Accoding to the Example in the chapte text, the faction of total kinetic enegy tansfeed to the modeato is f 4m m m + m whee m is the modeato nucleus and in this case, m m 48 f 4m m 3m 0.84 o of the neuton enegy is tansfeed to the cabon nucleus. K C ( 0.84) ( J) J K n ( 0.76) ( J) J

4 0 omentum and Collisions P8.0 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 ( kg) v F ( m) 0 Fo the second, p i p f ( kg) v (.04 kg) v f Again, K i + E mech K f : v f ( ) v.04 v Fd kg (.04 kg ) v f Substituting fo v f, v Fd kg (.04 kg ) v.04 ' ( Substituting fo v, Fd ( ) v Fd F m ( ) ' ( v d 7.94 cm P8. At impact, momentum of the clay-block system is conseved, so: mv ( m + m ) v Afte impact, the change in kinetic enegy of the clay-block-suface system is equal to the incease in intenal enegy: m + m 0. kg v f f d µ ( m + m ) gd v 9.80 m s kg 7.50 m v 95.6 m s v 9.77 m s ( kg) v ( 0. kg) ( 9.77 m s) v 9. m s FIG. P8. Section 8.4 Two-Dimensional Collisions P8.7 By consevation of momentum fo the system of the two billiad balls (with all masses equal), 5.00 m s + 0 ( 4.33 m s)cos v fx v fx.5 m s 0 ( 4.33 m s)sin v fy v fy.6 m s v f.50 m s at 60.0 FIG. P8.7 Note that we did not need to use the fact that the collision is pefectly elastic.

5 Chapte 8 P8.9 m vi + m vi ( m + m ) v f : 3.00( 5.00) î 6.00ĵ 5.00 v E m v + m v + m 3 v 3 v ( 3.00î.0ĵ ) m s ( ) + ( )( ) + ( )( ) E E J '( Section 8.5 The Cente of ass P8.33 Let A epesent the aea of the bottom ow of squaes, A the middle squae, and A 3 the top pai. A A + A + A A A A 300 cm, A 00 cm, A 3 00 cm, A 600 cm A A A A 3 A 3 A 300 cm 600 cm 00 cm 600 cm 6 00 cm 600 cm 3 x C x + x + x 3 3 x C.7 cm y C 5.0 cm ( 5.00 cm ) + ( 5.0 cm ) y C 3.3 cm cm( 6 ) cm( 3 ) 5.0 cm 3.3 cm FIG. P8.33

6 omentum and Collisions Section 8.6 P8.37 (a) otion of a System of Paticles m v C ivi m v + m v (.00 kg).00î m s 3.00ĵ m s 5.00 kg kg v C (.40î +.40ĵ ) m s.00î m s ĵ m s p v C ( 5.00 kg) (.40î +.40ĵ ) m s ( 7.00î +.0ĵ ) kg m s Additional Poblems * P8.48 Using consevation of momentum fom just befoe to just afte the impact of the bullet with the block: mv i ( + m) v f o v i + m m v f. () The speed of the block and embedded bullet just afte impact may be found using kinematic equations: Thus, t d v f t and h gt. h g and v f d t d g h gd h. FIG. P8.48 Substituting into () fom above gives v i + m m gd h.

7 Chapte 8 3 P8.5 (a) The initial momentum of the system is zeo, which emains constant thoughout the motion. Theefoe, when m leaves the wedge, we must have m v wedge + m v block 0 o ( 3.00 kg) v wedge + ( kg) ( m s) 0 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 K block + U system + K i wedge K i block + U system + K f wedge FIG. P8.5 f o [ 0 + m gh] + 0 m ( 4.00 ) m ( '0.667) which gives h 0.95 m. P8.59 A pictue one second late diffes by showing five exta kilogams of sand moving on the belt. (a) p x t 5.00 kg m s.00 s 3.75 N The only hoizontal foce on the sand is belt fiction, so fom p xi + f t p xf this is f p x t 3.75 N (c) The belt is in equilibium: F x ma x : +F ext f 0 and F ext 3.75 N (d) (e) (f) W F cos 3.75 N( m)cos0.8 J (m) v 5.00 kg m s.4 J Fiction between sand and belt convets half of the input wok into exta intenal enegy.

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

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