Simple Harmonic Motion and Elasticity

[Cheddar]

Homework Statement

A 0.01 kg bullet is fired horizontally into a 2.50 kg wooden block attached to one end of a massless, horizontal spring (spring constant = 845 N/m). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.200 m.
What is the speed of the bullet?

Homework Equations

I only wish I knew...

The Attempt at a Solution

I'd love to try this problem, since it is for extra credit, if anyone has any suggestions...

 

[Cheddar]

Wow, sorry if you read all that, but I didn't realize the teacher changed the EC question. We aren't doing that one anymore luckily.

 

[tomcue]

I would approach this problem by first identifying the relevant principles and equations that apply to the situation. In this case, we are dealing with simple harmonic motion and elasticity. Simple harmonic motion is a type of motion where the restoring force is directly proportional to the displacement from equilibrium, and the motion follows a sinusoidal pattern. Elasticity refers to the ability of a material to deform and return to its original shape when a force is applied.

To solve this problem, we can use the conservation of momentum and energy principles. The momentum of the bullet before the collision is equal to the momentum of the block/bullet system after the collision. We can also use the equation for the potential energy stored in a spring, which is given by U = 1/2kx^2, where k is the spring constant and x is the displacement from equilibrium.

Using these principles and equations, we can set up the following equations:

1) Conservation of momentum: mbulletvbullet = (mbullet + mblock)vfinal
2) Conservation of energy: 1/2mbulletvbullet^2 = 1/2kx^2

We can rearrange equation 1 to solve for the final velocity of the block/bullet system, and then substitute it into equation 2. This will give us an equation with only one unknown variable, the initial velocity of the bullet (vbullet). Solving for vbullet, we get:

vbullet = √[(kx^2)/(mbullet + mblock)]

Plugging in the given values, we get vbullet = 1.76 m/s.

In conclusion, the speed of the bullet is 1.76 m/s.