# A 0.0230 kg bullet moving horizontally at 500 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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A 0.0230 kg bullet moving horizontally at 500 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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Step 1

Given:

Mass of the bullet = 0.0230 kg

Speed of the bullet = 500 m/s

Step 2

Calculating the final velocity of the bullet-embedded block just after collision: help_outlineImage TranscriptioncloseInitial momentum of the bullet, p, = (0.0230kg) x(500m/s) 11.5kgm/s Initial momentum of the block, p2= 0 As the bullet embeds itself into the block, Final momentum of the bullet-block system, p, (0.0230kg +0.5kg) xv Where 'v is the final velocity of the bullet-embedded block system According to momentum of conversion, p, = p, +p2 Pr (0.0230kg+0.5kg) x v= 11.5kgm/s+0 11.5kgm s (0.0230kg+0.5kg = 21.99m/s 22m/s V= v 22m/s fullscreen
Step 3

Calculating the velocity of the bullet-emb... help_outlineImage TranscriptioncloseInitial velocity of bullet-embedded block system 22 m/s Distance travelled 8 m -g = -0.30 x (9.8m / s2-2.94m /s2 Acceleration Let's assume final velocity after 8 m is v' Using third equation of motion, (v') = u2 + 2as Substituting the values, (v') = (22m/s) +2x(-2.94m/s)x (8m) v' 21m/s fullscreen

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