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A 0.0270 kg bullet moving horizontally at 400 m/s embeds itself into an initially stationary 0.500...

A 0.0270 kg bullet moving horizontally at 400 m/s embeds itself into an initially stationary 0.500 kg block.

(a) What is their velocity just after the collision? m/s (

b) The bullet-embedded block slides 8.0 m on a horizontal surface with a 0.30 kinetic coefficient of friction. Now what is its velocity? m/s

(c) The bullet-embedded block now strikes and sticks to a stationary 2.00 kg block. How far does this combination travel before stopping? m

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Answer #1

Part a)

For perfectly inelastic collisions, the final velocity is given by the following equation:

Part b)

In the case of friction, the final velocity of the bullet-block system is:

where

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