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NEET · Physics · STD 11 - 5. work,energy,power and collision

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

  1. A \(0.5\)
  2. B \(0.25\)
  3. C \(0.4\)
  4. D \(0.8\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(0.25\)

Step-by-step Solution

Detailed explanation

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