Perfectly inelastic collision in a spring situation?

[erik-the-red]

Question

A rifle bullet with mass 8.00 g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring. The impact compresses the spring 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cm.

1. Find the magnitude of the block's velocity just after impact.

2. What was the initial speed of the bullet?

My entire strategy for solving this problem is that it is a perfectly inelastic collision. I reason this because the bullet embeds itself in the block, even though this is in a spring situation.

So, $$(m_a)(v_a,1) + (m_b)(v_b,1) = (m_a + m_b)(v_2)$$.

Plugging in the givens results in $$(.008)(v_a,1) = v_2$$.

Then, I considered the energies of the system as a whole. Spring force is conservative.

I reason that at the point at which the spring is at maximum compression of .15 m, the final velocity of the block is zero. So,

$$(1/2)(m_a)(v_a,1)^2 = (1/2)(300)(.15)^2$$

I get 300 as the spring constant because $$F=kx$$.

Solving for $$v_a,1$$, I got 29.0 m/s. I then multiplied 29.0 by .008 to get the velocity of the system, $$v_2$$.

But, my answer was not correct.

Was I wrong to use the perfectly inelastic collision equation?

 

[erik-the-red]

I recognized my initial error. I should have done $$(1/2)(m_a + m_b)(v_2)^2$$.

 

[quicknote]

Your approach of using the perfectly inelastic collision equation is correct. However, there may be some errors in your calculations. Here are some suggestions to help you solve this problem:

1. Make sure you are using the correct units for all values. In this case, the masses should be in kilograms and the distances in meters.

2. Double check your substitution of values in the equations. For example, in your calculation for the final velocity of the block, you have used a mass of 300 instead of 0.992 kg.

3. In your equation for the conservation of energy, make sure you are using the correct value for the spring constant. In this case, it should be 300 N/m, not 300.

4. Lastly, when solving for the velocity of the system, make sure you are using the correct values in the equation. In this case, it should be (0.008)(v_a,1) = v_2, not (0.008)(v_a,1) = 29.0.

By following these suggestions, you should be able to correctly solve for the magnitude of the block's velocity and the initial speed of the bullet in this perfectly inelastic collision in a spring situation.