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5.work,energy,power and collision — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysics5.work,energy,power and collision1 Mark
MCQ
A moving block having mass $m,$ collides with another stationary block having mass $4\,m$ . 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
  • A
    $0.5$
  • ✓
    $0.25$
  • C
    $0.4$
  • D
    $0.8$
✓

Answer

Correct option: B.
$0.25$
b
 Let final velocity of the block of    
  mass $ 4 m = v'$   
  Initial velocity of block of mass $ 4 m = 0 $   
  Final velocity of block of mass $ m = 0$   
  According to law of conservation of   
  linear momentum 

$\begin{gathered}
  mv + 4m \times 0 = 4mv' + 0 \Rightarrow v' = v/4 \hfill \\
  Coefficient\,of\,restitution, \hfill \\
  e = \frac{{\operatorname{Re} letive\,velocity\,of\,separation}}{{\operatorname{Re} letive\,velocity\,of\,separation}} \hfill \\
  \,\,\, = \,\frac{{v/4}}{v} = 0.25 \hfill \\ 
\end{gathered} $

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