Block 1 of mass 200 kg slides over a frictionless surface with a velocity of 0.25 cm/s and strikes block 2 of mass 100 kg sliding to the left at 0.75 cm/s. What is the final velocity of each block if the collision isA) Perfectly elastic?B) Perfectly inelastic?C) If there is an external force of 4000 kg*cm/s^2 to the right for 0.008 s during the perfectly inelastic

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Asked Oct 15, 2019
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Block 1 of mass 200 kg slides over a frictionless surface with a velocity of 0.25 cm/s and strikes block 2 of mass 100 kg sliding to the left at 0.75 cm/s. What is the final velocit
y of each block if the collision is
A) Perfectly elastic?
B) Perfectly inelastic?
C) If there is an external force of 4000 kg*cm/s^2 to the right for 0.008 s during the perfectly inelastic
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Expert Answer

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Step 1

(A)

The final velocity of the each block strikes with each other if before the collision is perfectly elastic is,

Substitute the values,

(т — т,)и, + 2т,
и
т+т,
(200kg-100kg)(0.25cm/s)+2(100kg) (-0.75cm/s)
и
200kg +100kg
3-0.42 сп/s
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Image Transcriptionclose

(т — т,)и, + 2т, и т+т, (200kg-100kg)(0.25cm/s)+2(100kg) (-0.75cm/s) и 200kg +100kg 3-0.42 сп/s

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Step 2

The final velocity of the each block strikes with each other if after the collision is perfectly elastic is,

Substitute the values,

(т, - т)и, +2тц
т +т,
(100kg - 200kg)(-0.75сmis) + 2(200kg)(025сms)
200kg +100kg
— 0.58ст/s
help_outline

Image Transcriptionclose

(т, - т)и, +2тц т +т, (100kg - 200kg)(-0.75сmis) + 2(200kg)(025сms) 200kg +100kg — 0.58ст/s

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Step 3

(B)

According to the conservation of momentum, the final velocity of each block strikes with ...

m4mu(m+m,)v
y=
(200kg)(0.25am/s)+(100kg)(-0.75cm/s)
200kg +100kg
y=
=-0.083cm/s
help_outline

Image Transcriptionclose

m4mu(m+m,)v y= (200kg)(0.25am/s)+(100kg)(-0.75cm/s) 200kg +100kg y= =-0.083cm/s

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