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) \[(4\hat{i}-3\hat{j})\times {{10}^{4}}\]
Correct Answer: D
Solution :
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 \[F=\frac{m{{\overrightarrow{v}}_{3}}}{t}\] \[=\frac{1\times (-3\hat{i}+4\hat{j})}{{{10}^{-4}}}N\] \[=-(3\hat{i}+4\hat{j})\times {{10}^{4}}N\]
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