A 52 kg woman anda 70 kg manstand 20.0 m apart onfrictionless ice. (a) How far f

Question

A 52 kg woman anda 70 kg manstand 20.0 m apart onfrictionless ice.
(a) How far from the man is their center of mass(CM)?

(b) If each holds one end of a rope, and the man pulls on therope so that he moves 2.4 m,then how far from the woman will he be now?
(c) How far will the man have moved when he collides with thewoman?
(a) How far from the man is their center of mass(CM)?

(b) If each holds one end of a rope, and the man pulls on therope so that he moves 2.4 m,then how far from the woman will he be now?
(c) How far will the man have moved when he collides with thewoman?

Explanation / Answer

(a) position of cm = ( 1 / total mass ) ( m1x1 + m2 x2 )    Set the man at x = 0   and findthe position of cm   (woman is at x = 18)     position of cm = ( 1 / 122) ( 52 * 20 + 70 * 0) =   7.67     cm is   8.524 meters fromthe man (b) when the man moves 2.4 meters, the womanmoves:    2.4* 70 / 52 =   3.23meters   so he is now:        20 - 3.23 - 2.4 =    14.37 meters fromthe woman

(c) the total of their motion must be 20 meters, becausethat is how far apart they started. The man will move:         20*52 / 122 = 8.524 meters

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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