Block 1, of mass m1m1m_1 = 6.30 kgkg , moves along a frictionless air track with speed v1v1v_1 = 17.0 m/sm/s . It collides with block 2, of mass m2m2m_2 = 37.0 kgkg , which was initially at rest. The blocks stick together after the collision.Find the magnitude pipip_i of the total initial momentum of the two-block system.Express your answer numerically. Find vfvfv_f, the magnitude of the final velocity of the two-block system.Express your answer numerically. What is the change ΔK=Kfinal−KinitialΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision?Express your answer numerically in joules. I Review I ConstantsBlock 1, of mass mi =6.30 kg , moves along aPart Africtionless air track with speed v1 = 17.0 m/s. It collideswith block 2, of mass m2 = 37.0 kg , which was initiallyat rest. The blocks stick together after the collision.(Figure 1)Find the magnitude p; of the total initial momentum of the two-block system.Express your answer numerically.• View Available Hint(s)Figure1 of 1<>?Before collision:Pi =kg · m/sm2VI2SubmitPart BAfter collision:Find vf ,the magnitude of the final velocity of the two-block system.Express your answer numerically.1View Available Hint(s)P PearsonConvright O 2021 Dearson Education Inc All rights reserved I Terms of LIse IDrivacy Dolicy IPermissions I Contact LIs|2. I Review I ConstantsBlock 1, of mass mi =6.30 kg , moves along aUf =m/sfrictionless air track with speed v1 = 17.0 m/s. It collideswith block 2, of mass m2 = 37.0 kg , which was initiallyat rest. The blocks stick together after the collision.(Figure 1)SubmitPart CFigure1 of 1<>What is the change AK = Kfinal – Kinitial in the two-block system's kinetic energy due to the collision?Express your answer numerically in joules.Before collision:• View Available Hint(s)m2VI2HV ΑΣφ?AK =JAfter collision:Submit1Provide FeedbackNext >P PearsonConvright O 2021 Dearson Education Inc All rights reserved I Terms of LIse IDrivacy Dolicy IPermissions I Contact LIs|2.

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Description: Block 1, of mass m1m1m_1 = 6.30 kgkg , moves along a frictionless air track with speed v1v1v_1 = 17.0 m/sm/s . It collides with block 2, of mass m2m2m_2 = 37.0 kgkg , which was initially at rest. The blocks stick together after the collision.Find the

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Block 1, of mass m1m1m_1 = 6.30 kgkg , moves along a frictionless air track with speed v1v1v_1 = 17.0 m/sm/s . It collides with block 2, of mass m2m2m_2 = 37.0 kgkg , which was initially at rest. The blocks stick together after the collision. Find the magnitude pipip_i of the total initial momentum of the two-block system. Express your answer numerically. Find vfvfv_f, the magnitude of the final velocity of the two-block system. Express your answer numerically. What is the change ΔK=Kfinal−KinitialΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision? Express your answer numerically in joules.

Block 1, of mass m1m1m_1 = 6.30 kgkg , moves along a frictionless air track with speed v1v1v_1 = 17.0 m/sm/s . It collides with block 2, of mass m2m2m_2 = 37.0 kgkg , which was initially at rest. The blocks stick together after the collision. Find the magnitude pipip_i of the total initial momentum of the two-block system. Express your answer numerically. Find vfvfv_f, the magnitude of the final velocity of the two-block system. Express your answer numerically. What is the change ΔK=Kfinal−KinitialΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision? Express your answer numerically in joules.

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter9: Momentum And Its Conservation

Section9.1: Impulse And Momentum

Problem 1PP

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