Solving the Bullet & Spring Momentum Problem

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
Gammage
Messages
14
Reaction score
0

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?
 
on Phys.org
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.
 
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
 
Last edited by a moderator: