A) 3 kg
B) 0.6kg
C) 2.4kg
D) 4kg
Correct Answer: B
Solution :
Since, initial velocity\[=u\] Final velocity\[=\frac{u}{4}\] From momentum conservation \[1\times u+0=1\times \frac{u}{4}+m\times {{v}^{2}}\] \[\therefore \] \[m{{v}^{2}}=\frac{3u}{4}\] ? (i) From energy conservation \[\frac{1}{2}\times 1\times {{u}^{2}}+0=\frac{1}{2}\times 1\times {{\left( \frac{u}{4} \right)}^{2}}+\frac{1}{2}m{{v}_{2}}^{2}\] \[\Rightarrow \] \[m{{v}_{2}}^{2}=\frac{15}{16}{{u}^{2}}\] ? (ii) \[\therefore \] \[\frac{{{(m{{v}_{2}})}^{2}}}{m{{v}_{2}}^{2}}=\frac{9/16{{u}^{2}}}{15/16{{u}^{2}}}\] \[\Rightarrow \] \[m=\frac{9}{16}=\mathbf{0}\mathbf{.6}\,\,\mathbf{kg}\]
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