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
- A \(0.5\)
- B \(0.25\)
- C \(0.4\)
- D \(0.8\)
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 \(\begin{gathered}
mv + 4m \times 0 = 4mv' + 0 \Rightarrow v' = v/4 \hfill \\
Coefficient\,of\,restitution, \hfill \\
e = \frac{{\operatorname{Re} letive\,velocity\,of\,separation}}{{\operatorname{Re} letive\,velocity\,of\,separation}} \hfill \\
\,\,\, = \,\frac{{v/4}}{v} = 0.25 \hfill \\
\end{gathered} \)
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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