A bullet of mass 0.018 kg traveling horizontally at a high speed of 200 m/s embeds itself in a block of mass 4 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? Vf = 7 x m/s (b) Calculate the total translational kinetic energy before and after the collision. Ktrans/= 180 Ktrans=260 X J x J

A bullet of mass 0.018 kg traveling horizontally at a high speed of 200 m/s embeds itself in a block of mass 4 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? Vf = 7 x m/s (b) Calculate the total translational kinetic energy before and after the collision. Ktrans/= 180 Ktrans=260 X J x J

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Description: A bullet of mass 0.018 kg traveling horizontally at a high speed of 200 m/s embeds itself in a block of mass 4 kg that is sitting at rest on a nearly frictionless surface.(a) What is the speed of the block after the bullet embeds itself in the block

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter8: Momentum And Collisions

Section: Chapter Questions

Problem 1OQ

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Question

A bullet of mass 0.018 kg traveling horizontally at a high speed of 200 m/s embeds itself in a block of mass 4 kg that is sitting at rest on a nearly frictionless surface.
(a) What is the speed of the block after the bullet embeds itself in the block?
Vf = 7
X m/s
(b) Calculate the total translational kinetic energy before and after the collision.
Ktrans,/ = 180
X J
Ktrans,f = 260
X J
(c) Compare the two results and explain why there is a difference.
O The Energy Principle isn't valid for an inelastic collision.
The internal energy of the block-bullet system has increased.
O Some of the momentum is lost in an inelastic collision.

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