# [Solved] Collision, spring and impulse

## Homework Statement

An 8-g bullet is shot into a 4.0-kg block, at rest on a frictionless horizontal surface. The bullet remains lodged in the block. The block moves into a spring and compresses it by 3.0 cm. The force constant of the spring is 1500 N/m. Find the initial speed of the bullet, and the impulse of the block (including the bullet), due to the spring, during the entire time interval in which block and spring are in contact.

The correct answers are 290cm/s for the initial speed of the bullet and 4.7 N.s for the impulse of the block.

## Homework Equations

Perfectly inelastic collision
m1v1i + m2v2i = (m1 + m2)vf

Energy conservation
(1/2)(m)(v^2) = (1/2)(k)(x^2)

## The Attempt at a Solution

With these simultaneous equations,
8/1000 vbullet_initial == (4 + (8/1000))vfinal
(1/2) (4 + (8/1000)) (vfinal^2) == (1/2) (1500) (0.03^2)
I found the speed of the bullet and the speed of the combined masses after impact.
vbullet_initial -> 290.764, vfinal -> 0.580367

I know how to find vbullet_initial, but the problem comes with the second part. If the impulse is change in momentum, m * deltav, I get (4 + (8/1000))(0.58) which is 2.32 N.s. I calculated this as if the change in velocity will result in the block with the embedded bullet being stopped by the spring. Could someone please explain to me why I only get half of the correct value? I get 2.32N.s, and the correct answer is 4.7N.s (2*2.32 rounded off).

[Solved]
Once the block comes to v->0 due to the spring, it pushes it back the other way and conservation of energy makes the impulse = mv -- mv = 2mv.

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