A) 4 m
B) 1.5 m
C) 3.5 m
D) 2 m
Correct Answer: A
Solution :
| [a] Let M be the mass of the nucleus. |
| Applying conservation of linear momentum, |
| \[mv=m{{v}_{1}}+M{{v}_{2}}\] ...(i) |
| Also, \[\frac{1}{2}mv_{1}^{2}=\frac{36}{100}\frac{1}{2}m{{v}^{2}}\Rightarrow {{v}_{1}}=\frac{6}{10}v\] ...(i) |
| Applying conservation of kinetic energy, |
| \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}mv_{1}^{2}+\frac{1}{2}Mv_{2}^{2}\] |
| \[\Rightarrow \frac{1}{2}Mv_{2}^{2}=\frac{64}{100}\frac{1}{2}m{{v}^{2}}\Rightarrow {{v}_{2}}=\frac{8}{10}v\sqrt{\frac{m}{M}}\] |
| Substituting (ii) and (iii) in eqn. (i) |
| \[mv=-\left( \frac{6}{10}v \right)m+M\left( \frac{8}{10}v\sqrt{\frac{m}{M}} \right)\] |
| \[\Rightarrow \]\[\frac{16}{10}mv=\frac{8}{10}v\sqrt{mM}\] |
| \[\Rightarrow M=4m\] |
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