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A $0.50-\mathrm{kg}$ ball that is traveling at $6.0 \mathrm{m} / \mathrm{s}$ collides head-on witha $1.00-\mathrm{kg}$ ball moving in the opposite direction at a speed of 12.0 m/s. The 0.50-kg ball bounces backward at 14 m/s after the collision. Find the speed of the second ball after the collision.
$v_{2 f}=-2.0 \mathrm{m} / \mathrm{s}$
Physics 101 Mechanics
Chapter 9
Momentum and Its Conservation
Section 2
Conservation of Momentum
Kinetic Energy
Energy Conservation
Moment, Impulse, and Collisions
Cornell University
Rutgers, The State University of New Jersey
University of Washington
McMaster University
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we have two balls will just say m someone equal in 0.50 kilograms. I'm so to Equalling the 1.0 kilograms so easier. Sub scripts on dso weaken Apply the conservation of momentum M sub one visa born initial plus M sub two visa to initial equaling EMS of one these of one final plus m sub two multiplied by These are two final, so we can then solve for the visa to final. This would be equal to M sub one initial or other times of one visa one initial plus m sub two visa to initial minus I'm someone be someone final divided by himself to and so we can solve and say that then visa to final is Equalling 0.50 kilograms multiplied by 6.0 muse per second plus 1.0 kilograms multiplied by negative 12.0 meters per second minus 0.50 kilograms. Most applied by negative 14.0 meters per second, all divided by 1.0 kilogram. And we find that then he's up to final is Equalling negative 2.0 meters per second and here here the negative sign indicates that the direction of the second ball after the collision is opposite to the initial direction and again the the magnitude of the velocity of the second ball. 2.0 meters per second. That is the end of the solution. Thank you for watching.
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