A 65.0 kg ice skater moving forward with a velocity of 1.80 m/s throws a 0.280 kg snowball forward with a velocity of 26.0 m/s.
(a) What is the velocity of the ice skater after throwing the snowball? Disregard the friction between the skates and the ice.
(b) A second skater initially at rest with a mass of 60.0 kg catches the snowball. What is the velocity of the second skater after catching the snowball in a perfectly inelastic collision?
i know u use like MaVai+MbVbi = MaVaf+MbVbf for something..could someone tell me how i would use this formula for this question..? you don't have to solve it for me i just want to know how i would use this.. T_T
2007-11-04 08:45:28 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
a)
65*1.8=65*v+0.28*26
solve for v
v=1.69 m/s
b)
0.28*26=60.28*v
solve for v
v=0.12 m/s
j
2007-11-05 06:32:54 · answer #1 · answered by odu83 7 · 0⤊ 0⤋