Conservation of Momentum and/or Energy

[holmeskaei]

Homework Statement

You have been asked to design a “ballistic spring system” to measure the speed of bullets. A bullet of mass m is fired into a block of mass M. The block, with the embedded bullet, then slides across a frictionless table and collides with a horizontal spring whose spring constant is k. The opposite end of the spring is anchored to a wall. The spring’s maximum compression d is measured.

Find an expression for the bullet’s speed v_{\rm B} in terms of m, M, k, and d.

Homework Equations

Conservation of momentum m1v1f+m2m2F=m1v1i+m2v2i
Conservation of Energy Ki+Ugi(spring)=Kf+Ugf(spring)

The Attempt at a Solution

Conservation of Momentum rearrangement: (m1+m2)vf=m1v1i

Conservation of energy rearrangement:
1/2mvi^2+0=0+1/2k(delta s)^2
1/2mvi^2=1/2k(delta s)^2
However, I am stuck here. I tried to solve it for vf and got something that is not correct. I know they fit together somehow.

 

[LowlyPion]

Looks like the Vf of your inelastic momentum conservation, becomes the Vi of your spring energy equation.

 

[nlightner]

As a scientist, it is important to approach problems like this with a clear understanding of the fundamental principles of conservation of momentum and energy. In this case, we are dealing with a system where the initial momentum and energy of the bullet and block are equal to the final momentum and energy after the collision with the spring.

To find the expression for the bullet's speed, we can start by applying the conservation of momentum equation:

(m + M)vf = mvi

We can then rearrange this equation to solve for vf:

vf = mvi / (m + M)

Next, we can use the conservation of energy equation to relate the kinetic energy of the bullet before and after the collision with the potential energy stored in the compressed spring:

1/2mvi^2 = 1/2k(delta s)^2

Solving for vi, we get:

vi = sqrt(k/m)(delta s)

Finally, we can substitute this value for vi into our expression for vf:

vf = m*sqrt(k/m)(delta s) / (m + M)

This gives us the final expression for the bullet's speed in terms of the given variables. It is important to note that this expression assumes a perfectly elastic collision, meaning that there is no energy lost to other forms (such as heat or sound) during the collision. In a real-world scenario, there may be some energy lost and the final speed of the bullet may be slightly lower than what is calculated using this expression.