Ballistic Spring System Problem

• Branson Holt
In summary, a 10.1 g bullet is fired horizontally into a 44.1 g wooden block connected to a massless spring of constant 97.8 N/m on a rough horizontal surface. After compressing the spring by 0.441 m, the speed of the bullet just as it enters the block can be found using the equation 1/2 mv^2= 1/2kx^2 + Fd, where F is the force of friction, m is the total mass of the system, k is the spring constant, x is the distance the spring is compressed, and d is the displacement of the combined bullet and block. The normal force (Fn) can be found using Fn= (m

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]

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.

Hello,

Thank you for providing the problem statement and your attempt at a solution. First, I would like to clarify that the equation provided by your teacher is correct. It is known as the conservation of energy equation and is commonly used in solving problems involving spring systems.

Now, let's break down the problem and see how we can apply the equation. We have a bullet with a mass of 10.1 g and a wooden block with a mass of 44.1 g. The system is initially at rest and the bullet is fired horizontally into the block. This means that the initial velocity of the bullet is unknown, but it can be assumed that it is greater than 0 since it is fired.

The bullet and the block compress the spring by 0.441 m, which means that the spring has done work on the system. This work can be represented as 1/2kx^2, where k is the spring constant and x is the compression distance. We can also consider the work done by the friction force, which can be represented as Fd, where F is the friction force and d is the displacement of the block due to friction.

Using the conservation of energy equation, we can set the initial kinetic energy of the system (bullet) to be equal to the final potential energy of the system (compressed spring) plus the work done by the friction force:

1/2 mv^2 = 1/2kx^2 + Fd

Substituting the given values:

1/2 (0.0101 kg) v^2 = 1/2 (97.8 N/m) (0.441 m)^2 + (0.554)(0.0441 kg)(9.8 m/s^2)(0.441 m)

Simplifying and solving for v, we get:

v = 0.463 m/s

Therefore, the initial velocity of the bullet as it enters the block is approximately 0.463 m/s.

I hope this helps. Keep practicing and don't hesitate to ask for clarification if needed. Good luck!

1. What is a ballistic spring system?

A ballistic spring system is a mechanical system consisting of a spring and a projectile that is launched from the spring. When the spring is compressed and released, it exerts a force on the projectile, causing it to move in a parabolic path.

2. How does a ballistic spring system work?

The spring in a ballistic spring system stores potential energy when it is compressed. When released, this potential energy is converted into kinetic energy, causing the projectile to move forward. The projectile then follows a parabolic path due to the force of gravity.

3. What factors affect the trajectory of a ballistic spring system?

The trajectory of a ballistic spring system is affected by the initial velocity of the projectile, the angle at which it is launched, and the force of gravity. Other factors such as air resistance and the mass of the projectile may also have an impact.

4. How can the motion of a ballistic spring system be calculated?

The motion of a ballistic spring system can be calculated using equations of motion, such as those derived from Newton's laws of motion. These equations take into account the initial conditions of the system and can be used to determine the position, velocity, and acceleration of the projectile at any given time.

5. What are some real-world applications of ballistic spring systems?

Ballistic spring systems have a variety of applications, including in sports such as archery and javelin throwing, as well as in engineering and design for launching objects or testing the effects of impact. They are also used in physics demonstrations and experiments to illustrate principles of projectile motion.