A 0.0240 kg bullet moving horizontally at 400 m/s embeds itself into an initiall

Question

A 0.0240 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

Please post all answers in decimals not 10*power answers)

Explanation / Answer

a) v=m1u1/m1+m2=18.32 m/sec

b) F=ma=friction

friction=mue*N=mue*mg

a=mue*g

from kinematic equation

v^2=2as

a=v^2/2s

v=sqrt(2*mue*s*g)=6.858 m/sec

c) v=m1u1/total mass=3.8 m/sec

s=v^2/2*mue*g=2.46 m

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