Two Blocks Inelastic Collision with Spring

[radiantvibe]

Homework Statement

A solid block of mass m₂ = 8.1 kg, at rest on a horizontal frictionless surface, is connected to a relaxed spring. The other end of the spring is fixed, and the spring constant is k = 230 N/m. Another solid block of mass m₁ = 4.0 kg and speed v₁ = 5.1 m/s collides with the 8.1 kg block. The blocks stick together, and compress the spring.

Homework Equations

The Attempt at a Solution

I'm still trying to grasp the concept of energy, as well as laTeX, but I have so far is:

W_initial=W_final

$\frac{1}{2} m_1 v_1,i^2 = \frac{1}{2} (m_1+m_2) v_f^2 - \frac{1}{2} kx^2$

where $$v_f= \frac{m_1 v_1i + m_2 v_2i}{m_1+m_2}$$

Solving for v_f, I got 3.41 m/s, assuming v_2i equals 0 m/s, however, I feel like this is where I'm going wrong.

 

[Vatsal Sanjay]

Why is there no term in your energy equation accounting for loss during collision in terms of heat and sound (which is nothing but vibrations for surrounding air molecules) energies?
Hint: you won't be able to solve this using energy equation. Think of something else. Another approach. And after you are done with the solution find out the value of the term I mentioned above.

 

[AlephNumbers]

What exactly does the question ask you to solve for? It would appear you are trying to solve for the velocity of the blocks at the time that the spring has reached its maximum compression. Your work seems correct so far.

 

[radiantvibe]

What exactly does the question ask you to solve for? It would appear you are trying to solve for the velocity of the blocks at the time that the spring has reached its maximum compression. Your work seems correct so far.

Sorry, I forgot to include what I was solving for! I am solving for the maximum distance that the spring compresses. I figured out that I had to use the impulse-momentum equation to find the velocity when both blocks are stuck together. Afterwards, I plugged my new velocity in the work-energy theorem with the spring expression $\frac{1}{2} kx^2 = \frac{1}{2} (m_1+m_2)v^2_f$ and solved for x to get my final answer.

 

[haruspex]

Homework Statement

A solid block of mass m₂ = 8.1 kg, at rest on a horizontal frictionless surface, is connected to a relaxed spring. The other end of the spring is fixed, and the spring constant is k = 230 N/m. Another solid block of mass m₁ = 4.0 kg and speed v₁ = 5.1 m/s collides with the 8.1 kg block. The blocks stick together, and compress the spring.

Homework Equations

The Attempt at a Solution

I'm still trying to grasp the concept of energy, as well as laTeX, but I have so far is:

W_initial=W_final

$\frac{1}{2} m_1 v_1,i^2 = \frac{1}{2} (m_1+m_2) v_f^2 - \frac{1}{2} kx^2$

where $$v_f= \frac{m_1 v_1i + m_2 v_2i}{m_1+m_2}$$

Solving for v_f, I got 3.41 m/s, assuming v_2i equals 0 m/s, however, I feel like this is where I'm going wrong.

Your post is rather confusing because of numerous typos. It certainly confused Vatsal.
Haven't you swapped the two masses in your calculation of "v_f"?

 

[Vatsal Sanjay]

ya it did confuse me. I just saw an energy conservation equation whose left hand side part made no sense.

 

[SammyS]

I'm still trying to grasp the concept of energy, as well as laTeX, but I have so far is:

$\frac{1}{2} m_1 v_1,i^2 = \frac{1}{2} (m_1+m_2) v_f^2 - \frac{1}{2} kx^2$

where $$v_f= \frac{m_1 v_1i + m_2 v_2i}{m_1+m_2}$$

For the LaTeX :

Any more than a single symbol in a subscript or superscript requires you to enclose the the several symbols in braces, { } . Liberal use of braces (they don't show up) can help in many other cases too.

You had: $\text{$\frac{1}{2} m_1 v_1,i^2 = \frac{1}{2} (m_1+m_2) v_f^2 - \frac{1}{2} kx^2$}$

Instead do the following $\text{$ \frac{1}{2} m_1 v_{1,i}^{\ 2} = \frac{1}{2} (m_1+m_2) v_f^{\ 2} - \frac{1}{2} kx^2 $}$

To get $\frac{1}{2} m_1 v_{1,i}^{\ 2} = \frac{1}{2} (m_1+m_2) v_f^{\ 2} - \frac{1}{2} kx^2$

$\text{$v_f= \frac{m_1 v_{1,i} + m_2 v_{2,i}}{m_1+m_2}$}$ gives
$$v_f= \frac{m_1 v_{1,i} + m_2 v_{2,i}}{m_1+m_2}$$