Solving the Bullet & Spring Momentum Problem

In summary: ERICAN MATHEMATICSIn summary, the block initially at rest is connected to a spring that has a force constant of 600 N/m. The block moves 1.6 cm to the right after the bullet passes through it. If the bullet exits the block at a speed of 300 m/s, then the block has lost 0.5 m/s of its initial momentum.
  • #1
Gammage
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Homework Statement


A bullet of mass 4 g moving with an initial speed of 300 m/s is fired into and passes through ta block of mass 5kg. The block, initially at rest on a frictionless horizontal surface, is connected to a spring of force constant 600 N/m. If the block moves a distance of 1.6 cm to the right after bullet passed through it, find the speed v at which the bullet emerges from the block.

Homework Equations


KE = 1/2 m v^2
PEspring = 1/2 k x^2
KEinitial + PEinitial = KEfinal + PEfinal
p = mv
m1v1 +m2v2 = m'1v'1 + m'2v'2

The Attempt at a Solution


I tried KEbulletinitial = PEspring = KEbulletfinal by .5mv1^2 = .5kx^2 + .5mv2^2. I guess its not a conservation of energy problem. I was thinking conservation of momentum but wasn't sure how to find the momentum of the spring. I know the fspring = kx. Any hints?
 
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  • #2
The question says the block moves 1.6cm AFTER the bullet passed through it.

I think what the question means is this:

* The bullet hits the block and goes through it in a very short time.
* As that happens, some momentum is transferred from the bullet to the block.
* When the bullet leaves the block, the bullet and the block both have unknown velocities.
* The block is then slowed down by the spring and its maximum movement is 1.6cm.

You are right that the impact part of the problem does not conserve energy, but you can use conservation of energy after the impact.
 
  • #3
As AlephZero indicates, this is a conservation of momentum problem. It is similar to a ballistic pendulum, which is analysed http://hyperphysics.phy-astr.gsu.edu/hbase/balpen.html#c1" except that the bullet does not embed itself in the block. What you need to determine is the momentum of the block after the bullet passes through which will give you the loss of momentum of the bullet.

AM
 
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1. What is the Bullet & Spring Momentum Problem?

The Bullet & Spring Momentum Problem is a physics problem that involves a bullet being fired into a spring and calculating the resulting momentum of both objects.

2. How is the problem solved?

The problem can be solved using the principle of conservation of momentum, where the total momentum before and after the collision must be equal. This can be expressed mathematically as: mbulletvbullet + mspringvspring = (mbullet + mspring)vf, where m represents mass and v represents velocity.

3. What are the assumptions made in solving this problem?

Some common assumptions made in solving the Bullet & Spring Momentum Problem include the neglect of air resistance, the assumption that the spring is an ideal linear spring with no damping, and the assumption that the collision is perfectly elastic, meaning there is no loss of kinetic energy.

4. How does this problem relate to real-world situations?

The Bullet & Spring Momentum Problem can be applied to real-world situations, such as understanding the physics behind a bullet being fired from a gun or a car colliding with a stationary object. It can also be used in engineering applications, such as designing car safety features or measuring the impact of a collision.

5. What are some potential sources of error in solving this problem?

Some potential sources of error in solving the Bullet & Spring Momentum Problem include measurement errors in velocity and mass, as well as the aforementioned assumptions not being completely accurate in real-world scenarios. Other factors such as the surface on which the objects collide and the angle at which the bullet is fired can also introduce errors in the calculations.

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