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A body A, of mass m=0.1\:kg has an initial velocity of 3\: \hat{i}ms^{-1}. It collides elastically with another body, B of the same mass which has an initial velocity of 5\: \hat{j}ms^{-1}. After collision, A moves with a velocity \vec{v}=4(\hat{i}+\hat{j}). The energy of B after collision is written as \frac{x}{10}J. The value of x is __________. 
Option: 1 1
Option: 2 2
Option: 3 3
Option: 4 4
 

Answers (1)

best_answer

 

 

 

 

 

For elastic collision

KEi = KEf

\begin{array}{l}{\frac{1}{2} m \times 25+\frac{1}{2} \times m \times 9=\frac{1}{2} m \times 32+\frac{1}{2} m v^{2}} \\ {34=32+v^{2}\Rightarrow v^2=2}\end{array}

\begin{array}{l}{\mathrm{KE}=\frac{1}{2} \times 0.1 \times 2=0.1 \mathrm{J}=\frac{1}{10}} \\ {x=1}\end{array}

So the answer will be 1

Posted by

vishal kumar

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