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A 0.0280 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

Question

A 0.0280 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

check_circleAnswer
Step 1

a)

From conservation of momentum, the required velocity just after collision is,

my+my(m+m,)v
[(0.028)(400)]+[(0.5))]-(0.028+0.5)v
v=21.21 m/s
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Image Transcriptionclose

my+my(m+m,)v [(0.028)(400)]+[(0.5))]-(0.028+0.5)v v=21.21 m/s

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Step 2

b)

The acceleration of the system,

(m+m)a=(m+m)g
a-(0.3)(0.8)
=2.94 m/s2
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Image Transcriptionclose

(m+m)a=(m+m)g a-(0.3)(0.8) =2.94 m/s2

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Step 3

Now, from kinematics, the r...

v2-v-2x
v21.21)-[2(2.94)(8)
20.07 m/s
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Image Transcriptionclose

v2-v-2x v21.21)-[2(2.94)(8) 20.07 m/s

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