A 66.0 kg ice skater moving to the right with a velocity of 2.10 m/s throws a 0.19 kg snowball to the right with a velocity of 27.0 m/s relative to the ground.
(a) What is the velocity of the ice skater after throwing the snowball? Disregard the friction between the skates and the ice.
____m/s to the right
i think i'm supposed to use the conservaton of momentum which is m1v1i + m2v2i = m1v1f + m2v2f but i don't know which one is the v2i or v2f or v1i or v1f.
2006-11-20 15:15:01 · 2 answers · asked by tingerpoo 2 in Science & Mathematics ➔ Physics
before the snowball is thrown, the scater and snowball have the same velocity:
(M1 + M2)Vi = M1V1f + m2V2f
(66 + 0.19)(2.10) = 66V1f + 0.19*27
66V1f = 66.19*2.1 - 5.13
V1f = (138.999 - 5.13)/66
V1f = 133.869/66
V1f = 2.02832
V1f ≈ 2.03 m/s
2006-11-20 15:26:55 · answer #1 · answered by Helmut 7 · 1⤊ 0⤋
You don't seem to understand what i and f stand for: initial and final. You can assign the #s 1 and 2 to the skater and the snowball which ever way you want. Looks like Helmut set the solution up right. Check the math and go with that.
2006-11-21 05:45:56 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋