A) 1.0kg and \[\frac{2}{3}v\]
B) 1.2 kg and \[\frac{5}{8}v\]
C) 1.4 and \[\frac{10}{17}v\]
D) 1.5 kg and \[\frac{4}{7}v\]
Correct Answer: B
Solution :
Given that \[{{m}_{1}}=2\,kg\] \[{{m}_{2}}=?\] \[{{u}_{1}}=v\] \[{{v}_{2}}=?\] \[{{v}_{1}}=v/4\] \[{{v}_{cm}}=?\] From conservation of momentum \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] \[2v+0=2\times \frac{v}{4}+{{m}_{2}}{{v}_{2}}\] \[\frac{3}{2}v={{m}_{2}}{{v}_{2}}\] ... (i) Now, from Newtons law of restitution \[{{v}_{1}}-{{v}_{2}}=e({{u}_{2}}-{{u}_{1}})\] (\[\because \,e=1\] for elestic collision) \[\frac{v}{4}-{{v}_{2}}=-v\] \[{{v}_{2}}=\frac{5}{4}v\] From Eq. (i), we get \[{{m}_{2}}=\frac{6}{5}\] = 1.2 kg \[{{v}_{cm}}=\frac{{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}}{{{m}_{1}}+{{m}_{2}}}\] \[=\frac{2\times \frac{v}{4}+1.2+\frac{5}{4}v}{2+12}\] \[=\frac{(0.5+15)v}{32}\] \[=\frac{5}{8}\,v\]
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