A 65 kg person throws a .045 kg snowball forward with a ground speed of 30 m/s.

Question

A 65 kg person throws a .045 kg snowball forward with a ground speed of 30 m/s. A second person, with a mass of 60 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.5 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard friction between the skates and the ice.

Explanation / Answer

considering first person and snow ball conservation of momentum. (m1+m2)u = m1v1 + m2v2 m1v1 is of man and m2v2 is of ball. => velocity of 1st man v1 = 1.85m/s now considering second man and ball. conservation of momentum. m1u1 + m2u2 = (m1+m2)v m1u1 is of man and m2u2, is of ball. => v = 0.022m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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