Conservation of Momentum, Energy

[SuperCass]

Homework Statement

A bullet with mass 20 grams and velocity 100 m/s collides with a wooden block of mass 2 kg. The wooden block is initially at rest, and is connected to a spring with k = 800 N/m. The other end of the spring is attached to an immovable wall. What is the maximum compression of the spring?

Homework Equations

p=mv
K=.5mv^2
Spring potential energy = .5kx^2

The Attempt at a Solution

I tried doing M(bullet)V(initial) = [M(bullet) + M(block)][Vfinal], and then I used that final velocity in the energy equation .5mv^2 = .5kx^2 to find x. I keep getting the answer to be .49 meters, but this isn't correct.

What am I doing wrong?
Thanks!

 

[phyzguy]

I think you have the concepts correct and are just doing the numbers wrong. With these inputs, 0.5mv^2~=1J = 0.5 kx^2, so x^2 ~= 1/400 m^2. How are you getting 0.49m?

 

[SuperCass]

Whoops, I mean I keep getting .049 meters.

But still, doing what you said isn't giving me the right answer!

 

[SuperCass]

Okay, I got it, never mind!

I was doing it correctly, I just didn't put in an exact enough answer!

It ended up being .0497!

 

[rdl]

I would like to commend you for using the conservation of momentum and energy equations to approach this problem. However, it seems like you may have made a slight error in your calculations. Let's take a closer look at the problem and see if we can find the correct solution.

First, let's consider the initial momentum of the system. Before the collision, the bullet has a momentum of p = mv = (0.02 kg)(100 m/s) = 2 kg m/s. Since the wooden block is initially at rest, the total initial momentum of the system is also 2 kg m/s.

Next, let's consider the final momentum of the system. After the collision, the bullet and the block will move together with a final velocity, let's call it vf. Using the conservation of momentum equation, we can set up the following equation:

2 kg m/s = (0.02 kg + 2 kg)(vf)

Solving for vf, we get vf = 0.01 m/s. This means that the bullet and the block will move together with a final velocity of 0.01 m/s.

Now, let's look at the energy equation. The initial kinetic energy of the system is given by K = 0.5mv^2 = 0.5(0.02 kg)(100 m/s)^2 = 100 J. After the collision, this kinetic energy is converted into potential energy stored in the spring. Using the spring potential energy equation, we can set up the following equation:

100 J = 0.5(800 N/m)(x^2)

Solving for x, we get x = 0.05 m. This means that the spring will compress by 0.05 meters, or 5 cm.

So, the maximum compression of the spring is actually 5 cm, not 0.49 meters as you initially calculated. It's possible that you may have made a mistake in your calculations or in setting up the equations. I would recommend going back and double-checking your work to see where the error may have occurred.

In conclusion, as a scientist, it's important to carefully check and review our calculations and equations to ensure accuracy and avoid errors. It's also helpful to break down the problem into smaller parts and consider the physical principles involved, such as conservation of momentum and energy, to guide our approach. Good luck with your future problem-solving endeavors!