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A A 4.25 g bullet traveling horizontally with a velocity ofmagnitude 375 $\mathrm{m} / \mathrm{s}$ is fired into a wooden block with mass$1.12 \mathrm{kg},$ initially at rest on a level frictionless surface. The bullet passes through the block and emerges with its speedreduced to 122 $\mathrm{m} / \mathrm{s}$ . How fast is the block moving just after thebullet emerges from it?

0.960 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 8

Momentum

Physics Basics

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

Rutgers, The State University of New Jersey

University of Winnipeg

McMaster University

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in this question, we're told that a bullet with a mass of 4.25 grams or 0.4 to 5 kilograms is traveling with initial speed of 375 meters per second before colliding with a block with a mass of 1.12 kilograms, which is initially at rest. Um, so the mass of the bullet is M little B. The initial speed is, um V. It'll be I pass of the block is big B and THEAN initial speed of the bloc. Which zero is, um V Big B I. After exiting the bloc on the other side, the bullet has a speed of 122 meters per second and were asked to find the final speed of the block, which is v big B F just as the bullet exits it. So this is a conservation of momentum problem. So, um, conservation of momentum tells us that the initial momentum of the entire system will be equal to the final momentum. So since Onley, the um bullet is moving in the initial the initial situation here before it collides, um, the initial momentum is simply equal to the mass of the bullet times the initial speed of the bullet whereas, um, at the end of thea bullets motion. So after it passes through the block both the block and the blood, her moving. So here will have the mass of the bullet times the final velocity of the bullet. Plus, um, the mass of the block, Um, times the final velocity of the block. And these are going to add together because they'll both be moving in the same direction. So, um, we want to solve for, um v big BF So we're going to subtract the first term from both sides and switch the sides. So the mass of the block times the final philosophy of the block is equal to the mass of the bullet, which we're gonna factor out, giving us the massive A bullet times in brackets, the velocity of the the initial velocity of the bullet minus the final velocity of the bullet. Um, now we just need to divide both sides by, um, the mass of the block. So the final velocity of the block is equal to the mass of the bullet over the mass of the block times the initial velocity of the bullet minus the final velocity of the bullet. So substituting in some Constance that were given were given all of these constants. So it's just zero 0.4 to 5 kilograms over. Um, the mess it'll block is 1.12 kilograms and then in brackets here we have, um the initial speed of the bullet is 375 meters per second, while the final speed is 122 years per second. Um, and plugging all of this into a calculator, we finds that the final speed of the block is 0.960 meters per second, and that is our answer for this question.

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