A) 0
B) 1.2
C) 2.0
D) 3.2
Correct Answer: A
Solution :
Let \[\Rightarrow \] be the friction exerting between disc surface and ground surface, then for the motion of the disc we can write \[X=\sqrt{\frac{2mgh}{K}}=\sqrt{\frac{2\times 0.04\times 10\times 5}{400}}\] ...... (ii) and \[=\frac{1}{10}m=10cm\approx 9.8cm\] \[V'\] \[mv=2mv'\Rightarrow V'=\frac{V}{2}\] ??... (ii) Here, a = linear acceleration of the disc. Solving Eqs. (i) and (ii) we get, \[\text{e=}\frac{\text{Velocity of separation}}{\text{Velocity of approach}}\]You need to login to perform this action.
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