# Ballistic Spring System Problem

## Homework Statement

A 10.1 g bullet is fired horizontally into a
44.1 g wooden block that is initially at rest on
a rough horizontal surface and connected to a
massless spring of constant 97.8 N/m.
If the bullet-block system compresses the
spring by 0.441 m, what was the speed of
the bullet just as it enters the block? The
acceleration of gravity is 9.8 m/s
2
. Assume
that the coefficient of kinetic friction between
the block surface is 0.554.

## Homework Equations

The equation our teacher gave us to solve it is: 1/2 mv^2= 1/2kx^2 + Fd
but I've used the equation and it hasn't worked for me.[/B]

## The Attempt at a Solution

Using the equation above. Is the equation right?[/B]

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Orodruin
Staff Emeritus
Homework Helper
Gold Member
This depends on how you are using it. Can you show your work please?

Well to start off I plugged in everything that I had from the information given:
1/2(.0101)v^2= 1/2(97.8)(.441)^2 + F(.441)

Then I found F by using F= u(coefficient of friction) x Fn(Normal Force)
so F= (.554)Fn which I found normal force by using Fn= (m+M)g
Fn= (.0101 + .0441)(9.8)= .53116
So when you plug that back in you get
F=(.554)(.53116)
1/2(.0101)v^2= 1/2(97.8)(.441)^2 + F(.441)
After i work this equation out solving for v I always get a consistent answer but it's not right.

haruspex
Homework Helper
Gold Member
I plugged in everything that I had from the information given:
1/2(.0101)v^2= 1/2(97.8)(.441)^2 + F(.441)
There's a step before this.
Consider the process in two stages:
- merging of bullet and block
- subsequent movement of combined bullet and block
Both stages involve loss of mechanical energy, the first in impact, the second in friction.
So the first thing to find out is the speed with which the combined bullet and block start moving.