# Perfectly inelastic collision in a spring situation?

1. Oct 10, 2005

### 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?

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2. Oct 10, 2005

### erik-the-red

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