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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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BuyFindarrow_forward

College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

A 10.0-g bullet is fired into, and embeds itself in, a 2.00-kg block attached to a spring with a force constant of 19.6 N/m and having negligible mass. How far is the spring compressed if the bullet has a speed of 300. m/s just before it strikes the block and the block sliders on a frictionless surface? Note: You must use conservation of momentum in this problem because of the inelastic collision between the bullet and block.

To determine
The velocity of the block and the bullet just after the collision and the compression of the spring when the bullet comes to rest.

Explanation

Given information: The mass of the bullet is 10 g, mass of the block is 2.00 Kg. The force constant of the spring is 19.6Nm-1 . The speed of the bullet just before the collision is 300ms-1 .

Explanation:

The collision between the bullet and the block is inelastic. Apply law of conservation of momentum to just before and after the collision.

mv=(m+M)V

  • m is the mass of the bullet
  • v is the speed of the bullet before collision
  • M is the mass of the block
  • V is the speed of bullet and block after collision

On re-arranging,

V=mvm+M (1)

The velocity of block and bullet after collision is mvm+M .

Apply the conservation of energy from after the collision to till the bullet comes to rest.

KEi+PEsi=KEf+PEsf

  • KEi is the kinetic energy just after collision
  • PEsi is the potential energy due to the spring just after collision
  • KEf is the kinetic energy when the bullet is stopped
  • PEsf is the potential energy when the bullet is stopped

PEsi=KEf=0

KEi=12(m+M)V2

PEsf=12kx2

  • k is the force constant of the spring
  • x is the compression of the spring

Hence

12(m+M)V2=12k

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