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Caroline Jackson Manentum fooblenis Caroline Jackson mamentum Prebicms I-30-2021 ||-30-2021 Ocont • Mamcatum before cellision =0.3 Kg m/s mamentum afler collisien = (0.075 kg) 9l Given : • mass of first ball, m.=25 g • Initial velocity of ficst baii, U: 12 m/S •Mass of secend ball, m2 = Sog Inito vebcity of secondball, Uz=0 lwe know thit collision iseither Ciasdic or inelast ic, the mementum is amays censerved. Camservatien of memenkm- mamentum before Collision is equai te the momentum after the Collision. • Imomentum befere collis ion = m, y tmz U -0 lwhere, m. mass of first bailL, M, =25g= 0.02S kg 0:3kgm/s=(0.0Zs Ka). v=4 ms The velacity of each ball after the Collisianis 4 mis momentum ofter Collisien= Lutting V=4 mis mementum AFer collision = (o.075 ka)4 mıs) mamentum afier Cellision = 0,3 Ka m/s • U, - Initiai velacity of first ball, U = 12 m/s • m2 = mass of second ball, Ma = Soa=0.050 kg •U2 = laitial veiasity df seund buil Uz =0 Put all these into eauasion D •momentum before Collis ian = (o.025 kg) (12 mis)+ (0.050 kg) (0) - momentum before the Collision = 0.3 kg m/s Because the Cellision between the first and sesund ball is inelostic, the twe mass behaves as u Single mass. Momentum after Collision = (m,+m;)v lWhere v is velocity afer collision Putting mi=0.025 Kg and mz =0.050 ky momentum ofer Coliston = (0.025 kg + 0.050 Kg) V Momentum OFAer collision= (0.075 ky)v Applying Conservition of mumentum- momentum before Collisien= momentum aAer colision

Caroline Jackson Manentum fooblenis Caroline Jackson mamentum Prebicms I-30-2021 ||-30-2021 Ocont • Mamcatum before cellision =0.3 Kg m/s mamentum afler collisien = (0.075 kg) 9l Given : • mass of first ball, m.=25 g • Initial velocity of ficst baii, U: 12 m/S •Mass of secend ball, m2 = Sog Inito vebcity of secondball, Uz=0 lwe know thit collision iseither Ciasdic or inelast ic, the mementum is amays censerved. Camservatien of memenkm- mamentum before Collision is equai te the momentum after the Collision. • Imomentum befere collis ion = m, y tmz U -0 lwhere, m. mass of first bailL, M, =25g= 0.02S kg 0:3kgm/s=(0.0Zs Ka). v=4 ms The velacity of each ball after the Collisianis 4 mis momentum ofter Collisien= Lutting V=4 mis mementum AFer collision = (o.075 ka)4 mıs) mamentum afier Cellision = 0,3 Ka m/s • U, - Initiai velacity of first ball, U = 12 m/s • m2 = mass of second ball, Ma = Soa=0.050 kg •U2 = laitial veiasity df seund buil Uz =0 Put all these into eauasion D •momentum before Collis ian = (o.025 kg) (12 mis)+ (0.050 kg) (0) - momentum before the Collision = 0.3 kg m/s Because the Cellision between the first and sesund ball is inelostic, the twe mass behaves as u Single mass. Momentum after Collision = (m,+m;)v lWhere v is velocity afer collision Putting mi=0.025 Kg and mz =0.050 ky momentum ofer Coliston = (0.025 kg + 0.050 Kg) V Momentum OFAer collision= (0.075 ky)v Applying Conservition of mumentum- momentum before Collisien= momentum aAer colision

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Use the results of problem 4 to show that energy is NOT conserved in an inelastic collision. 

Solution to problem 4 in picture.

Caroline Jackson Manentum fooblenis Caroline Jackson mamentum Prebicms I-30-2021 ||-30-2021 Ocont • Mamcatum before cellision =0.3 Kg m/s mamentum afler collisien = (0.075 kg) 9l Given : • mass of first ball, m.=25 g • Initial velocity of ficst baii, U: 12 m/S •Mass of secend ball, m2 = Sog Inito vebcity of secondball, Uz=0 lwe know thit collision iseither Ciasdic or inelast ic, the mementum is amays censerved. Camservatien of memenkm- mamentum before Collision is equai te the momentum after the Collision. • Imomentum befere collis ion = m, y tmz U -0 lwhere, m. mass of first bailL, M, =25g= 0.02S kg 0:3kgm/s=(0.0Zs Ka). v=4 ms The velacity of each ball after the Collisianis 4 mis momentum ofter Collisien= Lutting V=4 mis mementum AFer collision = (o.075 ka)4 mıs) mamentum afier Cellision = 0,3 Ka m/s • U, - Initiai velacity of first ball, U = 12 m/s • m2 = mass of second ball, Ma = Soa=0.050 kg •U2 = laitial veiasity df seund buil Uz =0 Put all these into eauasion D •momentum before Collis ian = (o.025 kg) (12 mis)+ (0.050 kg) (0) - momentum before the Collision = 0.3 kg m/s Because the Cellision between the first and sesund ball is inelostic, the twe mass behaves as u Single mass. Momentum after Collision = (m,+m;)v lWhere v is velocity afer collision Putting mi=0.025 Kg and mz =0.050 ky momentum ofer Coliston = (0.025 kg + 0.050 Kg) V Momentum OFAer collision= (0.075 ky)v Applying Conservition of mumentum- momentum before Collisien= momentum aAer colision
Transcribed Image Text:Caroline Jackson Manentum fooblenis Caroline Jackson mamentum Prebicms I-30-2021 ||-30-2021 Ocont • Mamcatum before cellision =0.3 Kg m/s mamentum afler collisien = (0.075 kg) 9l Given : • mass of first ball, m.=25 g • Initial velocity of ficst baii, U: 12 m/S •Mass of secend ball, m2 = Sog Inito vebcity of secondball, Uz=0 lwe know thit collision iseither Ciasdic or inelast ic, the mementum is amays censerved. Camservatien of memenkm- mamentum before Collision is equai te the momentum after the Collision. • Imomentum befere collis ion = m, y tmz U -0 lwhere, m. mass of first bailL, M, =25g= 0.02S kg 0:3kgm/s=(0.0Zs Ka). v=4 ms The velacity of each ball after the Collisianis 4 mis momentum ofter Collisien= Lutting V=4 mis mementum AFer collision = (o.075 ka)4 mıs) mamentum afier Cellision = 0,3 Ka m/s • U, - Initiai velacity of first ball, U = 12 m/s • m2 = mass of second ball, Ma = Soa=0.050 kg •U2 = laitial veiasity df seund buil Uz =0 Put all these into eauasion D •momentum before Collis ian = (o.025 kg) (12 mis)+ (0.050 kg) (0) - momentum before the Collision = 0.3 kg m/s Because the Cellision between the first and sesund ball is inelostic, the twe mass behaves as u Single mass. Momentum after Collision = (m,+m;)v lWhere v is velocity afer collision Putting mi=0.025 Kg and mz =0.050 ky momentum ofer Coliston = (0.025 kg + 0.050 Kg) V Momentum OFAer collision= (0.075 ky)v Applying Conservition of mumentum- momentum before Collisien= momentum aAer colision
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