A) \[\frac{1}{2}\]
B) \[\frac{1}{\sqrt{3}}\]
C) \[\frac{2}{3}\]
D) \[\sqrt{\frac{3}{5}}\]
Correct Answer: B
Solution :
[b] 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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