A) v
B) 2v
C) v/2
D) zero
Correct Answer: B
Solution :
Key Idea: In an elastic collision, linear momentum and kinetic energy are conserved. Initial momentum = Final momentum i.e., \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] \[\therefore \] \[M\times v+m\times 0=M{{v}_{1}}+m{{v}_{2}}\] or \[Mv=M{{v}_{1}}+m{{v}_{2}}\] or \[M(v-{{v}_{1}})=m{{v}_{2}}\] ?. (i) Again kinetic energy is also conserved. \[\frac{1}{2}{{m}_{1}}u_{1}^{2}+\frac{1}{2}{{m}_{2}}u_{2}^{2}=\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}\] \[\therefore \] \[M{{v}^{2}}+m\times 0=Mv_{1}^{2}+mv_{2}^{2}\] or \[M{{v}^{2}}=Mv_{1}^{2}+mv_{2}^{2}\] or \[M\,({{v}^{2}}-v_{1}^{2})=mv_{2}^{2}\] ... (ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{M\,({{v}^{2}}-v_{1}^{2})}{M\,(v-{{v}_{1}})}=\frac{mv_{2}^{2}}{m{{v}_{2}}}\] or \[v+{{v}_{1}}={{v}_{2}}\] As \[M>>m\], so \[{{v}_{1}}\approx v\] \[\therefore \] \[{{v}_{2}}=v+v=2\,v\] Note: The conservation of momentum and the conservation of total energy holds for all the three types of collisions, but KE conservation law holds only for perfectly elastic collision.
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