A) \[\frac{1}{2}\]
B) \[\frac{1}{\sqrt{3}}\]
C) \[\frac{2}{3}\]
D) \[\sqrt{\frac{3}{5}}\]
Correct Answer: B
Solution :
Initial momentum of mass 'm' = mu =5 Final momentum of system\[=(M+m)v=mu=5\] For second collision, mass (m=5, u = 1) coming from right strikes with system of mass 15, both momentum have opposite direction. \[\therefore \] net momentum = zero Similarly for12th collision momentum is zero. For 13th collision, total mass\[=10+12\times 5=70\] Using conservation of momentum \[70\times 0+5\times 1=(70+5)v'\] \[v'=\frac{1}{5}\] Total mass \[=10+13\times 5=75\] Finald KE of system \[=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times 75\times \left[ \frac{1}{15} \right]\left[ \frac{1}{15} \right]\] \[\frac{1}{2}k\,{{A}^{2}}=\frac{1}{2}75\times \frac{1}{15}\times \frac{1}{15}\] \[=\frac{1}{7}\times (1){{A}^{2}}=\frac{1}{2}75\times \frac{1}{15}\times \frac{1}{15}\] \[{{A}^{2}}=\frac{1}{3}\] \[A=\frac{1}{\sqrt{3}}\]
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