# Bullet through block and spring

## Homework Statement

A bullet of mass 9 g moving with initial speed 300 m/s passes through a block of mass 4 kg, initially at rest on a frictionless, horizontal surface. The block is connected to a spring of force constant 904 N/m. If the block moves 2 cm to the right after the bullet passed through it, find the speed at which the bullet emerges from the block and the magnitude of the energy lost in the collision.

## Homework Equations

Conservation of momentum.

## The Attempt at a Solution

Well initial mometum is just (.009)(300), and it must equal (.009)(Vf)=(4)(Vblk)

to solve for the block velocity immediately after the bullet passes, I used Vf = sqrt(2ad). I said acceleration was (904)(.02)/(4)...(may be where I went wrong). I took that velocity and plugged it back into the original momentum equation and got velocity of the bullet to be 111 m/s, which was wrong.

First draw a basic idea of how this looks; a good force body diagram or a diagram on how this works. Drawing out what happens can get you far when conceptualizing what happens and going about problems.

After you do this, we can start working on the problem :).

But here's a few things to think about:

There's a few things to deal with this such as momentum [bullet through the block] and energy [spring energy].

Last edited:
Doc Al
Mentor
I used conservation of energy (initial KE of bullet = final KE of bullet + spring potential), but I got almost the exact same speed for the bullet. Not even .1 m/s less.
Kinetic energy is not conserved during the collision. This problem involves both conservation of momentum and conservation of energy--separately. (See the links that Kurdt provided.)

Kurdt
Staff Emeritus
Gold Member
If you show the work you have done we can determine where you are going wrong.

EDIT: Doc Al beat me

Kinetic energy is not conserved during the collision. This problem involves both conservation of momentum and conservation of energy--separately. (See the links that Kurdt provided.)
To go along with this.

The reason why you can't use conservation of energy is due to the fact that there maybe thermal energy caused by this or sound and other various forms of energy. Energy is not conserved. Conservation of energy only works with perfectly elastic or a complete inelastic collision.

As I stated in my post there are things at work here =). Spring energy being one and the momentum.

Hootenanny
Staff Emeritus
Gold Member
Energy is not conserved.  :surprised

I get that momentum is conserved during the collision, and mechanical energy after. But why would conservation of energy not work if I have just one unknown in the equation?

.5(.009)(300^2) = .5(.009)(v^2) + .5(904)(.02^2) is my energy conservation equation.

Hootenanny
Staff Emeritus