A 66.0 kg person throws a0.0410 kg snowball forward with aground speed of 27.0 m

Question

A 66.0 kg person throws a0.0410 kg snowball forward with aground speed of 27.0 m/s. A secondperson, with a mass of 56.0 kg,catches the snowball. Both people are on skates. The first personis initially moving forward with a speed of 2.30 m/s, and the second person is initially atrest. What are the velocities of the two people after the snowballis exchanged? Disregard the friction between the skates and theice. So, I need to know the velocity of the thrower and thecatcher. Thanks for any help. A 66.0 kg person throws a0.0410 kg snowball forward with aground speed of 27.0 m/s. A secondperson, with a mass of 56.0 kg,catches the snowball. Both people are on skates. The first personis initially moving forward with a speed of 2.30 m/s, and the second person is initially atrest. What are the velocities of the two people after the snowballis exchanged? Disregard the friction between the skates and theice. So, I need to know the velocity of the thrower and thecatcher. Thanks for any help.

Explanation / Answer

   let us assume that the velocity of the thrower isvt and velocity of the catcher isvc    then applying the conservation of momentumprinciple we get    (66.0 kg) vt + (0.0410 kg) (27.0m / s) = (66.0 kg + 0.0410 kg) (2.30 m / s)    the velocity of the thrower will be    vt = ......... m /s    after the catcher recieves the ball he and theball combindly move with a velocity vc    again by applying the conservation of momentumprinciple we get       (56.0 kg + 0.0410 kg) vc =(0.0410 kg) (27.0 m / s) + (56.0 kg) (0)    the velocity of the catcher willbe    vc = ......... m /s    again by applying the conservation of momentumprinciple we get       (56.0 kg + 0.0410 kg) vc =(0.0410 kg) (27.0 m / s) + (56.0 kg) (0)    the velocity of the catcher willbe    vc = ......... m /s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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