Physics 2A Chapter 9 HW Solutions


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1 Phscs A Chapter 9 HW Solutons Chapter 9 Conceptual Queston:, 4, 8, 13 Problems: 3, 8, 1, 15, 3, 40, 51, 6 Q9.. Reason: We can nd the change n momentum o the objects b computng the mpulse on them and usng the equaton Δ p = J. Snce the start at rest wth zero momentum, we can wrte mv = p =Δ p = J = FΔt. So the nal veloct o ether object equals the mpulse on the object dvded b ts mass. For the rst object, ths wll be For the second object, the veloct s gven b ( v = (1 N(.0 s / m = (4 N s/ m 1 ( v = (15 N(3.0 s /( m = (.5 N s/ m The veloct o the rst object s seen to be hgher, because no matter what the value o m s, the numerator s greater n the rst expresson than n the second expresson. Assess: The second object was subject to a greater orce exerted or a greater tme. But t ended wth less veloct because t was more massve. All three actors are mportant n determnng the nal speed. Q9.5. Reason: The sum o the momenta o the three peces must be the zero vector. Snce the rst pece s travelng east, ts momentum wll have the orm ( p1,0, where p 1 s a postve number. Snce the second pece s travelng north, ts momentum wll have the orm (0, p, where p s a postve number. I a thrd momentum s to be added to these and the result s to be ( 0,0, then the thrd momentum must be ( p1, p. Snce ts eastwest and northsouth components are both negatve, the momentum o the thrd pece must pont south west and so the veloct must be south west. The answer s D. Assess: It makes sense that the thrd pece would need to travel southwest. It needs a western component o momentum to cancel the eastern component o the rst pece and t needs a southern component to cancel the northern component o the second pece. Q9.8. Reason: The mpulse needed to stop a car wthout crumple zones wll be the same as or a car wth crumple zones snce the change n momentum s same. The average orce needed to stop the car s the mpulse dvded b the tme the car takes to stop. A car desgned wth crumple zones wll take longer to stop than a st car, so the orce exerted on the car durng a collson s smaller or a car wth crumple zones. Assess: See Secton 9. or a dscusson o brdge abutments, whch are used or mpactlessenng and work on the same prncple. Q9.13. Reason: I ou do not move backward when passng the basketball t s because the ballou sstem s not solated: There s a net external orce on the sstem to keep ou rom movng backward that changes the momentum o the sstem. I the ballou sstem s solated (sa ou are on rctonless ce, then ou do move backward when ou pass the ball. Assess: I the rcton orce o the loor on ou keeps ou rom movng backward (relatve to the loor, then the law o conservaton o momentum doesn t appl because the sstem sn t solated. But ou could then nclude the loor, buldng, and the earth n the sstem so t (the sstem s solated; then momentum o the sstem s conserved and that means the earth does recol ever so slghtl when ou pass the basketball.
2 P9.3. Prepare: From Equatons 9.5, 9., and 9.8, Newton s second law can be protabl rewrtten as Δp Favg = Δ t In act, ths s much closer to what Newton actuall wrote than F = ma. Solve: Ths allows us to nd the orce on the snowball. B Newton s thrd law we know that the snowball exerts a orce o equal magntude on the wall. Δp mv mv m( v v (0.1 kg(0 m/s 7.5 m/s Favg = = = = = 6.0 N Δt Δt Δt 0.15 s where the negatve sgn ndcates that the orce on the snowball s opposte ts orgnal momentum. So the orce on the wall s also 6.0 N. Assess: Ths s not a large orce, but the snowball has low mass, a moderate speed, and the collson tme s arl long. P9.8. Prepare: Model the tenns ball as a partcle, and ts nteracton wth the wall as a collson. The orce ncreases to F max durng the rst two ms, stas at F max or two ms, and then decreases to zero durng the last two ms. Please reer to Fgure P 9.8. The graph shows that F x s postve, so the orce acts to the rght. Solve: Usng the mpulsemomentum theorem p = p + J, x x x (0.06 kg(3 m/s = (0.06 kg( 3 m/s +area under orce graph Now, 1 1 area under orce curve = F (0.00 s + F (0.00 s + F (0.00 s = (0.004 s F (0.06 kg(3 m/s + (0.06 kg(3 m/s Fmax = = 960 N s max max max max
3 P9.1. Prepare: We can use Equatons 9.6 and 9.8 to calculate the mpulse and Equaton 9. to calculate the average orce. Solve: (a Consder the ball to be movng along the postve xaxs beore t s ht. Then ( v =+ 15.0m/s. Ater the collson, the veloct o the ball s ( v = 0.0 m/s. The mpulse on the ball s gven b Equaton 9.8: x x x x J = ( p ( p = (0.145 kg( 0.0 m/s (0.145 kg(15.0 m/s = 5.08 kg m/s (b Gven the mpulse and duraton o the collson, Equaton 9. gves the average orce on the ball. J x 5.08 kg m/s ( Fx avg = = = 3400 N Δt s x Assess: Note that mpulse and momentum are vectors. The mpulse and orce are drected n the negatve x drecton. P9.15. Prepare: Ths s a problem wth no external orces so we can use the law o conservaton o momentum. Solve: The total momentum beore the bullet hts the block equals the total momentum ater the bullet passes through the block so we can wrte mb( vb + mbl( vbl = mb( vb + mbl( vbl 3 3 ( kg(500 m/s + (.7 kg(0 m/s = ( kg(0 m/s + (.7 kg(v. We can solve or the nal veloct o the block: ( v = 0.31 m/s. bl bl Assess: Ths s reasonable snce the block s about one thousand tmes more massve than the bullet and ts change n speed s about one thousand tmes less. P9.3. Prepare: Even though ths s an nelastc collson, momentum s stll conserved durng the short collson we choose the sstem to be sptball plus carton. Let SB stand or the sptball, CTN the carton, and BOTH be the combned object ater mpact (we assume the sptball stcks to the carton. We are gven m SB = kg, m CTN = 0.00 kg, and (v BOTHx = 0.30 m/s. Solve: ( Px = ( Px ( psbx + ( pctnx = ( pbothx msb( vsbx + mctn ( vctnx = ( msb + mctn( vbothx We want to know ( vsbx so we solve or t. Also recall that ( v CTNx = 0m/ sso the last term n the ollowng numerator drops out. ( msb + mctn( vbothx mctn( vctnx ( kg kg(0.30 m/s ( vsbx = = =.3 m/s msb kg Assess: The answer o.3 m/s s certanl wthn the capablt o an expert sptballer.
4 P9.40. Prepare: There are three parts to the moton n ths problem. The ball alls to the ground, the ball has a collson wth the ground, and nall the ball travels upward under the nluence o gravt. The loor exerts a orce on the ball that creates an mpulse on the ball. The change n momentum o the ball can be calculated wth Equaton 9.8 once the mpulse s known. The veloct o the ball as t leaves the loor can be calculated rom the momentum o the ball ater the collson wth the loor. Solve: Addtonal sgncant gures wll be kept n ntermedate results. The mpulse due to a orce s the area under the orceversustme curve, rom Equaton 9.1. The curve on the graph s a trangle. An trangle has an area that s equal to one hal the length o the base multpled b the heght o the trangle. The heght o the trangle s 1000 N and the base has a length equal to 4 ms. The mpulse s then J 3 (1000 N(4 10 s = = kg m/s The change n momentum o the ball s equal to the mpulse, rom Equaton 9.8. We can nd the veloct o the ball as t leaves the loor knowng ts ntal momentum. The veloct o the ball just beore t hts the loor can be calculated wth the knematc equaton ( v = ( v + a Δ The ntal veloct o the ball s zero, and the acceleraton o the ball s the acceleraton due to gravt. Solvng or the magntude o the veloct o the ball just beore t ht the ground, v = gδ = (9.80m/s (.0m = 6.6m/s The magntude o the momentum o the ball just beore t hts the ground s mv = (0. kg (6.6 m/s = 1.5 kg m/s. Ths s the ntal momentum o the ball just beore the collson, ( p = 1.5 kg m/s. A negatve sgn has been nserted snce the momentum s n the downward drecton. Usng Equaton 9.8 to nd the momentum o the ball rght ater the collson, ( p = J + ( p = kg m/s+( 1.5 kg m/s = 0.75 kg m/s We can now nd the veloct o the ball rght ater t bounces up rom the loor, ( p ( v = = 3.75 m/s m Usng ths veloct as the ntal veloct or the upward part o the moton and usng the same knematcs equaton as above, we nd the ball rebounds to a heght o ( v (3.75 m/s Δ = = = m a ( 9.80m/s Assess: Ths s a ver specal loor. P9.54. Prepare: We arbtrarl pck the drecton the lnebacker was runnng as the postve xdrecton snce he was mentoned rst. I, ater we solve the conservaton o momentum equaton or v, the answer s postve, then we know the lnebacker ends up movng orward; v s negatve, then the quarterback ends up movng orward. We wll use subscrpts l or lnebacker and q or quarterback. v Known m1 = 110 Kg mq = 8 Kg ( vlx =.0m/s ( v = 3.0m/s qx Fnd We emplo the conservaton o momentum. Snce the collson s nelastc (the lnebacker grabs and holds onto the quarterback the nal momentum wll be P = ( m + m v, where v s the answer we seek. l q
5 Solve: P = P ( m + m v = m( v m ( v l q l lx q qx v m( v + m ( v (110 kg(.0 m/s + (8 kg( 3.0 m/s l lx q qx = = = ml + mq 110 kg + 8 kg 0.14 m/s The answer s negatve; ths ndcates that the quarterback ends up movng orward ater the ht. (The lnebacker gets knocked backward. Assess: I all we want to know s the sgn o the answer then we do not reall need to compute or dvde b the denomnator the total mass wll certanl be postve and wll not aect the sgn o the answer. So we could have done a smple mental calculaton o the numerator (8 3 > 110 to gure out whch ootball plaers wns. P9.6. Prepare: Model the two blocks (A and B and the bullet (L as partcles. Ths s a twopart problem. Frst, we have a collson between the bullet and the rst block (A. Momentum s conserved snce no external orce acts on the sstem (bullet + block A. The second part o the problem nvolves a perectl nelastc collson between the bullet and block B. Momentum s agan conserved or ths sstem (bullet + block B. Solve: For the rst collson the equaton (p x = (p x s ml( vlx 1+ ma( vax 1 = ml( vlx 0 + ma( vax 0 (0.01kg( v + (0.500 kg(6 m/s = (0.01kg(400 m/s + 0 kg m/s ( v = 100 m/s Lx 1 Lx 1 The bullet emerges rom the rst block at 100 m/s. For the second collson the equaton (p x = (p x s ( ml mb( vx ml( v Lx 1 + = (0.01kg kg( v x = (0.01kg(100 m/s ( v x =.0m/s Assess: Ths problem nvloves repeated applcaton o the law o conservaton o momentum. Also note that the actual value o m or the separaton between the blocks s not necessar or our calculatons.
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