Bullet and a Block: Solve the Problem

In summary, a user is seeking help with a physics problem involving the collision of a bullet and a block. They have attempted to solve it using equations for energy and momentum, but are having difficulty with the momentum equation. After some discussion and clarification, they are able to solve the problem.
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  • #3
Okay... i equated .5kx^2 and .5mv^2 for the block, and got the velocity, v. Then i used mv=MV (conservation of momentum) to find the final velocity of the bullet as it emerges out of the block. But that is not the correct answer.
Help me.
 
  • #4
directdelta said:
Okay... i equated .5kx^2 and .5mv^2 for the block, and got the velocity, v. Then i used mv=MV (conservation of momentum) to find the final velocity of the bullet as it emerges out of the block. But that is not the correct answer.
Help me.
Let's be clear on M vs m and V vs v. Your energy calculation involves the spring and the block, so i assume m is mass of block and v is its velocity when the spring starts to compress. The momentum conservation problem can be assumed complete before the spring compreses.

Look at the momentum problem again. You have a bullet moving with known velocity toward a stationary block. After the collision you know the velocity (from the energy calculation) of the block and you need the final velocity of the bullet. Your momentum equation does not correspond to this situation. See if you can fix it.
 
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  • #5
directdelta said:
http://blog.360.yahoo.com/blog-rkirjZg1crSQi6eQnmqL4njg_w--?cq=1

Please try to solve this problem. Thanks.
Interesting problem.

Since the bullet is traveling at 500 m/sec, it passes through the block before the block really begins to move. So we can treat this as if the block acquires an instantaneous kinetic energy which is then transferred into the spring. We know what the energy transferred to the block is [itex]E = kx^2/2[/itex] where v0 is the speed of the block after collision. So you can work out the speed of the block after the bullet passes through. I think that is what you have done.

Since momentum has to be conserved,

[tex]mv_{f-bullet} + Mv_{block} = mv_{i-bullet}[/tex]

That may be where you are having some difficulty.

AM
 
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  • #6
Thank You. I finally solved it. siggghhhh
 

1. What is the "Bullet and a Block" problem?

The "Bullet and a Block" problem is a physics problem that involves a bullet being fired into a block of wood. The goal is to calculate the velocity of the bullet after it has penetrated a certain distance into the block.

2. What are the main factors that affect the solution to the "Bullet and a Block" problem?

The main factors that affect the solution to this problem are the initial velocity of the bullet, the mass and density of the bullet and block, and the distance the bullet penetrates into the block.

3. How is the "Bullet and a Block" problem solved?

The problem is typically solved using the principles of conservation of momentum and energy. This involves setting up equations using the initial and final velocities and the masses of the bullet and block, and then solving for the unknown variables.

4. What are some real-world applications of the "Bullet and a Block" problem?

The "Bullet and a Block" problem has many real-world applications, such as in forensic science for determining the trajectory and speed of a bullet, in ballistics for designing bulletproof materials, and in engineering for understanding the impact of projectiles on structures.

5. What are some possible extensions or variations of the "Bullet and a Block" problem?

Some possible extensions or variations of the problem include adding friction or air resistance, considering the block to be made of different materials, or changing the initial conditions such as the angle of the bullet's trajectory or the shape of the block.

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