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  • Can someone explain A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitutio

A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

  • Option 1)

    0.8

  • Option 2)

    0.25

  • Option 3)

    0.5

  • Option 4)

    0.4

 

Answers (1)

best_answer

As we have learned

Coffecient of Restitution (e) -

e= frac{v_{2}-v_{1}}{u_{2}-u_{1}}

- wherein

Ratio of relative velocity after collision to relative velocity before collision.

 

 Let velocity of 4m mass is v after collision

then mv=\left ( 4m \right )v

\Rightarrow V=\frac{V}{4}

e=\frac{velocity of separation}{velocity of apporoach}=\frac{\frac{v}{4}}{v}=\frac{1}{4}

e=0.25

 

 

 


Option 1)

0.8

This is incorrect

Option 2)

0.25

This is correct

Option 3)

0.5

This is incorrect

Option 4)

0.4

This is incorrect

Posted by

Avinash

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