Problem 6G - Hays High School

Givens3. m 1 = m 2v 1, i = 5.0 m/s to the right=+5.0 m/sv 1, f = 2.0 m/s to the left=−2.0 m/sv 2, f = 5.0 m/s to the right=+5.0 m/sSolutionsMomentum conservationm 1 v 1, i + m 2 v 2, i = m 1 v 1, f + m 2 v 2, fv 2, i =m 1 v 1, f + m 2 v 2, f − m 1 v 1, i⎯⎯⎯m2= v 1, f + v 2, f − v 1, iv 2, i =−2.0 m/s + 5.0 m/s − 5.0 m/s =−2.0 m/sv 2, i = 2.0 m/s to the leftConservation of kinetic energy (check)1⎯2 ⎯ m i v 2 1, i + ⎯ 1 2 ⎯ m 2 v 2 2, i = ⎯ 1 2 ⎯ m 1 v 2 1, f + ⎯ 1 2 ⎯ 2m 2 v 2, fv 1,i 2 + v 2,i 2 = v 1, f 2 + v 2, f2(5.0 m/s) 2 + (−2.0 m/s) 2 = (−2.0 m/s) 2 + (5.0 m/s) 225 m 2 /s 2 + 4.0 m 2 /s 2 = 4.0 m 2 /s 2 + 25 m 2 /s 229 m 2 /s 2 = 29 m 2 /s 2II4. m 1 = 45.0 gv 1, i = 273 km/h to the right=+273 km/hv 2, i = 0 km/hv 1, f = 91 km/h to the left=−91 km/hv 2, f = 182 km/h to the right=+182 km/hMomentum conservationm 1 v 1, i + m 2 v 2, i = m 1 v 1, f + m 2 v 2, fm 2 =⎯ m 1v1,f − m1v1, i (45.0 g)(−91 km/h) − (45.0 g)(273 km/h)⎯ = ⎯⎯⎯⎯⎯v2,i − v2,f0 km/h − 182 km/h−4.1 10 3 g •km/h − 12.3 10 3 g •km/h −16.4 10 3 g •km/hm 2 = ⎯⎯⎯⎯⎯ = ⎯⎯⎯−182 km/h−182 km/hm 2 = 90.1 gConservation of kinetic energy (check)1⎯2 ⎯ m 1 v 2 1,i + ⎯ 1 2 ⎯ m 2 v 2 2,i = ⎯ 1 2 ⎯ m 1 v 2 1, f + ⎯ 1 2 ⎯ m 2 v 2, f1⎯2 ⎯ (45.0 g)(273 10 3 m/h) 2 (1 h/3600 s) 2 + ⎯ 1 2 ⎯ (90.1 g)(0 m/s) 2= ⎯ 1 2 ⎯ (45.0 g)(−91 × 10 3 m/h) 2 (1 h/3600 s) 2 + ⎯ 1 2 ⎯ (90.1 g)(182 × 10 3 m/h) 2 (1 h/3600 s) 2129 J + 0 J = 14 J + 115 J129 J = 129 J5. v 1,i = 185 km/h to the right185 km/hv 2,i = 0 km/hv i,f = 80.0 km/h to the left=−80.0 km/hm 1 = 5.70 10 –2 kgMomentum conservationm 1 v 1, i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,f ⎯m 1⎯m2v 1,i − ⎯ m 1⎯m2v 1,f = v 2,f –v 2,i ⎯m 1⎯m2[185 km/h − (−80.0 km/h)] = v 2,f − 0 km/hv 2,f = ⎯ m 1⎯m2(265 km/h) to the rightConservation of kinetic energyKE i = ⎯ 1 2 ⎯ m 1 v 2 1,i + ⎯ 1 2 ⎯ m 2 v 2 2,i = ⎯ 1 2 ⎯ 2m 1 v 1,iKE f = ⎯ 1 2 ⎯ m 1 v 1, f 2 + ⎯ 1 2 ⎯ m 2 v 2,f21⎯2 ⎯ m 1 v 2 1,i = ⎯ 1 2 ⎯ m 1 v 2 1,f + ⎯ 1 2 ⎯ 2m 2 v 2,f ⎯m 1⎯m2(v 1,i ) 2 = ⎯ m 1⎯m2(v 1,f ) 2 2+ v 2,fCopyright © by Holt, Rinehart and Winston. All rights reserved. ⎯m 1m2⎯ (185 km/h) 2 = ⎯ m m1⎯2 (−80.0 km/h) 2 2+ v 2,fII Ch. 6–12Holt Physics Solution Manual

GivensSolutions ⎯m ⎯m 1 2 (3 .4 2 × 10 4 km 2 / h 2 ) − ⎯m ⎯m 1 2 (6 .4 0 × 10 3 km 2 / h 2 ) = v 2,fv 2,f = ⎯ m ⎯m 1 2 (2 .7 8 × 10 4 k m2/h 2 ) = ⎯m ⎯m 1 2 167 km/h Equating the two results for v 2,f yields the ratio of m 1 to m 2 . ⎯m 1m2⎯ (265 km/h) = ⎯ m m 1 2265 km/h = ⎯ m ⎯m 2 1 (167 km/h)⎯ (167 km/h)m2⎯⎯ =m1⎯ 2 65km/h 2⎯167km/h = 2.52m 2 = (2.52) m 1 (2.52)(5.70 10 –2 kg)m 2 = 0.144 kg6. m 1 = 4.00 × 10 5 kgm 2 = 1.60 × 10 5 kgv 1, i = 32.0 km/h to the rightv 2, i = 36.0 km/h to the rightv 1, f = 35.5 km/h to the rightMomentum conservationm 1 v 1, i + m 2 v 2, i = m 1 v 1, f + m 2 v 2, fv 2, f =v 2, f =m 1 v 1, i + m 2 v 2, i − m 1 v 1, f⎯⎯⎯m2(4.00 × 10 5 kg)(32.0 km/h) + (1.60 × 10 5 kg)(36.0 km/h) − (4.00 × 10 5 kg)(35.5 km/h)⎯⎯⎯⎯⎯⎯⎯⎯⎯1.60 × 10 5 kgII1.28 × 10 7 kg•km/h + 5.76 × 10 6 kg•km/h − 1.42 × 10 7 kg•km/hv 2, f = ⎯⎯⎯⎯⎯⎯⎯1.60 × 10 5 kgv 2, f = ⎯ 4.4 6× 10kg•km/h⎯51.60× 10kgv 2, f =28 km/h to the rightCopyright © by Holt, Rinehart and Winston. All rights reserved.Conservation of kinetic energy (check)1⎯2 ⎯ m 1 v 2 1, i + ⎯ 1 2 ⎯ m 2 v 2 2, i = ⎯ 1 2 ⎯ m 1 v 2 1, f + ⎯ 1 2 ⎯ 2m 2 v 2, f1⎯2 ⎯ (4.00 × 10 5 kg)(32.0 × 10 3 m/h) 2 (1 h/3600 s) 2 + ⎯ 1 2 ⎯ (1.60 × 10 5 kg)(36.0 × 10 3 m/h) 2(1 h/3600 s) 2 = ⎯ 1 2 ⎯ (4.00 × 10 5 kg)(35.5 × 10 3 m/h) 2 (1 h/3600 s) 2 + ⎯ 1 2 ⎯ (1.60 × 10 5 kg)(28 10 3 m/h) 2 (1 h/3600 s) 21.58 × 10 7 J + 8.00 × 10 6 J = 1.94 × 10 7 J + 4.8 × 10 6 J2.38 × 10 7 J = 2.42 × 10 7 JThe slight difference arises from rounding.Section Two—Problem Workbook Solutions II Ch. 6–13

Givens7. m 1 = 5.50 × 10 5 kgm 2 = 2.30 × 10 5 kgv 1, i = 5.00 m/s to the right=+5.00 m/sv 2, i = 5.00 m/s to the left=−5.00 m/sv 2, f = 9.10 m/s to the right=+9.10 m/sSolutionsMomentum conservationm 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,fv 1,f =v 1,f =v 1,f =m 1 v 1,i + m 2 v 2,i − m 2 v 2,f⎯⎯⎯m1(5.50 × 10 5 kg)(5.00 m/s) + (2.30 × 10 5 kg)(−5.00 m/s) − (9.10 m/s)⎯⎯⎯⎯⎯⎯⎯5.50 × 10 5 kg2.75 10 6 kg•m/s – 1.15 10 6 kg•m/s –2.09 × 10 6 kg•m/s⎯⎯⎯⎯⎯⎯5.50 × 10 5 kg= −0.89 m/s rightv 1,f =0.89 m/s leftConservation of kinetic energy (check)1⎯2 ⎯ m 1 v 2 1, i + ⎯ 1 2 ⎯ m 2 v 2 2, i = ⎯ 1 2 ⎯ m 1 v 2 1, f + ⎯ 1 2 ⎯ 2m 2 v 2, f1⎯2 ⎯ (5.50 × 10 5 kg)(5.00 m/s) 2 + ⎯ 1 2 ⎯ (2.30 × 10 5 kg)(−5.00 m/s) 2 = ⎯ 1 2 ⎯ (5.50 × 10 5 kg)(−0.89 m/s) 2 + ⎯ 1 2 ⎯ (2.30 × 10 5 kg)(9.10 × m/s) 2II6.88 × 10 6 J + 2.88 × 10 6 J = 2.2 × 10 5 J + 9.52 × 10 6 J9.76 × 10 6 J = 9.74 × 10 6 JThe slight difference arises from rounding.Copyright © by Holt, Rinehart and Winston. All rights reserved.II Ch. 6–14Holt Physics Solution Manual