Suppose a 0.360 kg ball is thrown at 14.0 m/s to a motionless person standing on

Question

Suppose a 0.360 kg ball is thrown at 14.0 m/s to a motionless person standing on ice who catches it with an outstretched arm. (a) Calculate the final linear velocity (in m/s) of the person, given his mass is 68.0 kg. m/s (b) What is his angular velocity (in rad/s) if each arm is 6.00 kg? You may treat the ball as a point mass and treat the person's arms as uniform rods (each has a length of 0.860 m) and the rest of his body as a uniform cylinder of radius 0.170 m. Neglect the effect of the ball on his center of mass so that his center of mass remains in his geometrical center. rad/s (c) Compare the initial and final total kinetic energies. Initial kinetic energy is greater than final kinetic energy. Initial kinetic energy is the same as final kinetic energy. Initial kinetic energy is less than final kinetic energy.

Explanation / Answer

a] using momentum conservation,

(M+m)v = mu

v = mu/(M+m)

v = 0.360*14/(68.360)

= 0.07373 m/s answer

b) By conservation of angular momentum about axis of cylinderical man,

mu (r+L) = i w

0.360*14*(0.17+0.86) = (2*(1/3*6*0.86^2+6*0.17^2) + 0.5*(68-2*6)*0.17^2 + 0.36*(0.17+0.86)^2) w

5.1912 = 4.4963w

w = 5.1912/4.4963

= 1.1545 rad/s answer

c] Initial KE = 0.5mu^2 = 0.5*0.36*14^2 = 35.28 J

Final KE = 0.5(M+m)v^2 + 0.5 iw^2 = 0.5*68.36*0.07373^2 + 0.5*4.4963*1.1545^2 = 3.182 J

initial KE is greater than final KE

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