Ballistic Spring System Problem

AI Thread Summary
The discussion centers on calculating the speed of a bullet just before it enters a wooden block connected to a spring. The equation provided, 1/2 mv^2 = 1/2 kx^2 + Fd, is questioned for its application, particularly in accounting for friction. Participants emphasize the need to analyze the problem in two stages: the collision and the subsequent motion of the bullet-block system. The normal force and friction are calculated to determine the force acting against the motion. A consistent answer is being reached, but it is not yielding the correct result, indicating a potential oversight in the energy loss considerations during the impact and movement.
Branson Holt
Messages
2
Reaction score
0

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.
Answer in units of m/s.

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]
 
Physics news on Phys.org
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.
 
Branson Holt said:
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top