Two boys are sliding toward each other on a frictionless, ice-covered parking lo

Question

Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45 kg, is gliding to the right at 7.72 m/s, and Ethan, mass 31.0 kg, is gliding to the left at 10.9 m/s along the same line. When they meet, they grab each other and hang on. (a) What is their velocity immediately thereafter? magnitude . m/s direction ---Select--- left right . (b) What fraction of their original kinetic energy is still mechanical energy after their collision? . Your response is off by a multiple of ten.% (c) That was so much fun that the boys repeat the collision with the same original velocities, this time moving along parallel lines 1.06 m apart. At closest approach, they lock arms and start rotating about their common center of mass. Model the boys as particles and their arms as a cord that does not stretch. Find the velocity of their center of mass. magnitude . m/s direction ---Select--- left right . (d) Find their angular speed. . Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. rad/s (e) What fraction of their original kinetic energy is still mechanical energy after they link arms? %

Explanation / Answer

Let the center of mass be 'x' away from Jacob (45 kg) boy. We have {45x-31(1.03 - x)}/(45+31) = 0 or 45x -32.55 +31x = 0 or 76x = 31.93 or x = 31.935/76 = 0.42 Equating KE of translation of both to their rotational KE with 'omega' angular velocity about CM, we have 0.5*[45*(0.42^2)+31*{1.03-0.42)^2}]*om… = 0.5*(45+31)*10.5^2 or 19.478*omega^2 = 8379 or omega = sq rt[8379/19.478] = 20.47 rad/s

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