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? (b) What fraction of their original kinetic energy is still mechanical energy after their collision? % (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. rad/s (e) What fraction of their original kinetic energy is still mechanical energy after they link arms? % (f) Why are the answers to parts (b) and (e) so different?

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

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.