A) 5 kg
B) 7 kg
C) 17 kg
D) 3 kg
Correct Answer: A
Solution :
Key Idea Apply law of conservation of linear momentum. |
Momentum of first part \[=1\times 12=12kg\,m{{s}^{-1}}\] |
Momentum of the second part \[=2\times 8=16\,kg\,m{{s}^{-1}}\] |
\[\xrightarrow[\text{(ii)}{{\text{H}}_{\text{2}}}\text{O,heat}]{\text{(i)}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{room}\,\text{temperature}}Z;\] Resultant momentum |
\[=\sqrt{{{(12)}^{2}}+{{(16)}^{2}}}=20\,kg\,m{{s}^{-1}}\] |
The third part should also have the same momentum. |
Let the mass of the third part be M, then |
\[4\times M=20\] |
\[M=5\,kg\] |
Alternative: |
\[Mv\,\cos \,\theta =12\] |
\[Mv\,\sin \,\theta =16\] |
\[\tan \theta =\frac{16}{12}=\frac{4}{3}\] |
\[M=\frac{12\times 5}{4\times 3}=\frac{60}{12}=5\,kg\] |
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