MTop View ofhorizontal tableDVomA block (which you should treat like a point particle) has mass m = 0.28 kg and slides with initial speed vo. It hits andsticks a distance D = 0.578 from the center of a horizontal rod of length L = 1.7 m and mass M = 0.46 kg, which is initiallyat rest and oriented perpendicular to the path of the block. Everything is on top of a horizontal frictionless table,and the rod has a fixed frictionless pivot through its center that allows it to rotate freely but keeps its center frommoving.After the block hits the rod they rotate together with a constant angular velocity of w = 9.1 rad/sec. What was theinitial speed of the block?vo = 11.49 m/svo = 6.23 m/s%3Dvo = 5.26 m/svo = 7.77 m/svo = 5.72 m/sWhich of the following statements best describes the external forces and torques acting on the rod-block systemduring the collision?The total external torque around the pivot is not zero, but the total external force on the system is zero.Both the total external torque around the pivot and the total external force on the system are zero.The total external torque around the pivot is zero, but the total external force on the system is not zero.

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Description: MTop View ofhorizontal tableDVomA block (which you should treat like a point particle) has mass m = 0.28 kg and slides with initial speed vo. It hits andsticks a distance D = 0.578 from the center of a horizontal rod of length L = 1.7 m and mass M =

Given that: m=0.28 kgD=0.578 mL=1.7 mM=0.46 kgωo=0 rad/sω=9.1 rad/sMoment of inertia of rod about the pivot point:I=112ML2Moment of inertia of rod+block about the pivot point:I'=112ML2+mD2Angular momentum before impact:L=Iωo+mvoDAngular momentum after impact:L'=I'ωApplying conservation of angular momentumL=L'Iωo+mvoD=I'ω(112ML2×0)+mvoD=(112ML2+mD2)ωmvoD=(112ML2+mD2)ωvo=(112ML2+mD2)ωmDvo=(112×0.46×1.72+0.28×0.5782)×9.10.28×0.578vo=11.49 m/sDuring the collision, force is totally internal, no net external force is acting on the system. Since the angular velocity is constant, then conserve angular momentum. Therefore the net external torque has to be zero. Hence Option B is the correct answer.

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A block (which you should treat like a point particle) has mass m = 0.28 kg and slides with initial speed vo. It hits and sticks a distance D = 0.578 from the center of a horizontal rod of length L = 1.7 m and mass M = 0.46 kg, which is initially at rest and oriented perpendicular to the path of the block. Everything is on top of a horizontal frictionless table, and the rod has a fixed frictionless pivot through its center that allows it to rotate freely but keeps its center from moving.

A block (which you should treat like a point particle) has mass m = 0.28 kg and slides with initial speed vo. It hits and sticks a distance D = 0.578 from the center of a horizontal rod of length L = 1.7 m and mass M = 0.46 kg, which is initially at rest and oriented perpendicular to the path of the block. Everything is on top of a horizontal frictionless table, and the rod has a fixed frictionless pivot through its center that allows it to rotate freely but keeps its center from moving.

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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