A 56 kg woman and a 76 kg man stand 20.0 m aparton frictionless ice. (a) How far

Question

A 56 kg woman and a 76 kg man stand 20.0 m aparton frictionless ice. (a) How far from the man is their center ofmass (CM)?
m

(b) If each holds one end of a rope, and the man pulls on the ropeso that he moves 3 m, then how far fromthe woman will he be now?
m

(c) How far will the man have moved when he collides with thewoman?
m (a) How far from the man is their center ofmass (CM)?
m

(b) If each holds one end of a rope, and the man pulls on the ropeso that he moves 3 m, then how far fromthe woman will he be now?
m

(c) How far will the man have moved when he collides with thewoman?
m

Explanation / Answer

a) the position of centre of mass of the two personsassuming that the women at origin ans the men lies on the postivex-axis at a distance 20.0m from the origin                                x = 76 * 20 / (76 + 56 ) b) by the conseravation of momentum                    m1v1 = m2v2              v1 / v2 = m2 / m1             d1 / d2 = m2 /m1 given d2 = 3m        from m1 and m2 we cansolve for d1 where d1 and d2 are the distance moved by the menand women in a time interval t     c) let the distance travelled by the man be x then thedistance travelled by te wom4en = 20-x                       the the ratios of the velocitys must be inverssly related to themasses                      m1 / m2 = v2 /v1              v2 = x /t             v1 = 20-x / t from the above relations we can sovle for x .

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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