Question

A stationary body of mass $3\,kg$ explodes into three equal pieces. Two of the pieces fly off in two mutually perpendicular directions, one with a velocity of $ 3\hat{i}\,m{{s}^{-1}} $ and the other with a velocity of $ 4\hat{j}\,m{{s}^{-1}} $ . If the explosion occurs in $ {{10}^{-4}}s, $ the average force acting on the third piece in newton is

• A. $ (3\hat{i}+4\hat{j})\times {{10}^{-4}} $
• B. $ (3\hat{i}-4\hat{j})\times {{10}^{-4}} $
• C. $ (3\hat{i}+4\hat{j})\times {{10}^{4}} $
• D. $ -(3\hat{i}+4\hat{j})\times {{10}^{4}} $
• E. $ (4\hat{i}-3\hat{j})\times {{10}^{4}} $

Answer 1

Answer: (D)

By law of conservation of linear momentum
$ {{m}_{1}}{{\overrightarrow{v}}_{1}}+{{m}_{2}}{{\overrightarrow{v}}_{2}}+{{m}_{3}}{{\overrightarrow{v}}_{3}}=0 $
Here: $ {{m}_{1}}={{m}_{2}}={{m}_{3}}=1\,kg, $
$ {{\overrightarrow{v}}_{1}}=3\hat{i},{{\overrightarrow{v}}_{2}}=4\hat{j} $
$ \therefore $ $ 3\hat{i}+4\hat{j}+{{\overrightarrow{v}}_{3}}=0 $
The average force acting on the third piece is
$F=\frac{m{{\overrightarrow{v}}_{3}}}{t} $
$ =\frac{1\times (-3\hat{i}+4\hat{j})}{{{10}^{-4}}}N $
$ =-(3\hat{i}+4\hat{j})\times {{10}^{4}}N $