A) \[45\hat{j}-35\hat{k}\]
B) \[45\hat{i}-35\hat{j}\]
C) \[45\hat{k}-35\hat{j}\]
D) \[-35\hat{i}+45\hat{k}\]
Correct Answer: A
Solution :
[a] : Let the mass of the unexploded bomb be 5m. It explodes into the two pieces of masses m and 4m respectively. Initial momentum of the unexploded bomb \[=5m(40\hat{i}+50\hat{j}-25\hat{k})\] After explosion, momentum of the smaller piece \[=m{{\vec{v}}_{1}}=m(200\hat{i}+70\hat{j}+15\hat{k})\] and momentum of the larger piece \[=4m{{\vec{v}}_{2}}\]where \[{{\vec{v}}_{1}}\]and \[{{\vec{v}}_{2}}\]are the velocities of the two pieces respectively. According to law of conservation of momentum, we get \[5m(40\hat{i}+50\hat{j}-25\hat{k})=m(200\hat{i}+70\hat{j}+15\hat{k})+4m{{\vec{v}}_{2}}\]\[4m{{\vec{v}}_{2}}=5m(40\hat{i}+50\hat{j}-25\hat{k})-m(200\hat{i}+70\hat{j}+15\hat{k})\]\[{{\vec{v}}_{2}}=\frac{1}{4}(180\hat{j}-140\hat{k})=45\hat{j}-35-\hat{k}\]
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