A) \[v\]
B) \[2\,\,v\]
C) \[v/2\]
D) \[zero\]
Correct Answer: B
Solution :
From law of conservation of linear momentum \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] \[M\times \varpi \upsilon +m\times 0=M\times {{\upsilon }_{1}}+m{{v}_{2}}\] \[Mv=M{{v}_{1}}+m{{v}_{2}}\] ? (1) \[M(v-{{v}_{1}})=m{{v}_{2}}\] In elastic collision kinetic energy is also conserved \[\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\] \[M{{v}^{2}}=Mv_{1}^{2}+mv_{2}^{2}\] \[M(v+{{v}_{1}})(v-{{v}_{1}})=mv_{2}^{2}\] ... (2) Dividing eq (2) by (1), we get \[v+{{v}_{1}}\simeq {{v}_{2}}\] as \[M>\,\,>m\] so, \[{{v}_{1}}\simeq v\] \[\therefore \] \[{{v}_{2}}=2v\]
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