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(II) A $22-\mathrm{g}$ bullet traveling 210 $\mathrm{m} / \mathrm{s}$ penetrates a 2.0 -kg block of wood and emerges going 150 $\mathrm{m} / \mathrm{s}$ . If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

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0.66 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 9

Linear Momentum

Motion Along a Straight Line

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Simon Fraser University

Hope College

University of Winnipeg

Lectures

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her initial Matt initial momentum will just be the initial momentum of the bullet. So, uh, initial mass times initial velocity, uh, off the bullet V I will be equal to final mass and B f times final velocity and V B s, uh, plus massive would times velocity of wood. Who? Mentor of the wood? Uh, after after the bullet penetrates it. Uh, no. Uh, the bullet is not losing any mass. O n b i is and b f uh, and so they're both equal to m sabi. Okay. Ah, and so that's the same. Therefore, velocity of the wood. But what we want is just em. Beauty times The initial velocity of the bullet minus the final velocity of the bullet over massive would so plugging in values. We have 22 grams for the mass of the bullets. So that's 220.0 to to kilograms because we divided by 1000 times, uh, initial velocity to 10 meters per second. Miners final velocity 1 15 years for a second over massive the wood which is two kilograms, and this gives us 0.66 meters per second for the velocity of the wood

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