A lead sphere with mass M_pb = 27 kg (ppb = 11.3 g/cm^3) is dropped from height

Question

A lead sphere with mass M_pb = 27 kg (ppb = 11.3 g/cm^3) is dropped from height h_pb = 67 cm onto a long cylindrical piston with radius r_cyl = 9.0 cm and height h_o = 2.5 m. The piston is filled with an ideal gas. Assume the temperature of the gas does not change. Will the lead sphere fit into the cylinder? (Do a geometrical calculation here using the fact that, for a sphere V = 4/3Pir^3.) What is the maximum distance the piston descends after the lead sphere is dropped onto it? Use energy conservation with the work done by the piston given by W_p = -P_oV_o ln (V_f/V_o). The potential energy lost by the ball must be equal to the work done by the gas. Use this fact to show the following: M_pbg(h_pb - Deltah.) = -p_o(Pir^2_cylh_o) In (1 + Deltah/h_o) Complete the calculation by copying and pasting the following text into Wolfram Alpha: 27 times 9.81 times (.67-x)= - 101300 times 3.14 times.09 times.09 times 2.5 times log (1+x/2.5) Assume that, after some complicated oscillations, the system comes to equilibrium with the lead ball upon the stationary piston head. Determine the pressure and the height of the column of gas in the piston

Explanation / Answer

There is smth wrong with the data, or some info is missing: from which reference is measured hPb? From the initial position of the piston (h0) or from the ground (bottom of the piston)? Anyway, the latter option is not possible with the data provided.

Otherwise the calculation cannot be performed since the theorem of the potential (gravitational) energy variation has to be applied. In which case a reference level for measuring the heights is necessary.

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