Mass of a pendulam is 10^-2 kg. It is lifted to 5×10^-2 m and released. At the lowest point of its path an extra mass of 10^{-3 kg} is attached with the pendulam. Find the height travelled by the pendulam At the opposite side.

Dear Subhajit,

Ans:- The velocity at the bottommost point is=√(2*10*0.05) m/s=1m/s

From the conservation of momentum we get the final velocity at the bottommost point after the extra mass is hanged is

=10-²*1/(10-²+10-³)

=0.91 m/s

Let the max height is =H m

Then H=0.91²/(2*10)

=4.1×10-² m  

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