An 81.1 kg man is standing on a frictionless ice surface when he throws a 3.79 kg book at 18.9 m/s. With what velocity does the man move across the ice?
2006-09-22 08:42:32 · 4 answers · asked by E 1 in Science & Mathematics ➔ Mathematics
The man and the book travel with equal and opposite momentum, so that the total system continues to have zero momentum. The book's momentum is equal to its mass (3.79 kg) times velocity (18.9 m/s). The man's momentum is equal to his mass (81.1 kg) times his velocity (unknown). So 3.79*18.9 = 81.1*v, or v = 3.79*18.9 / 81.1. Don't forget that the man's velocity is in the opposite direction of the book's.
2006-09-22 08:48:11 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋
Zero. There's no such thing as frictionless ice.
2006-09-22 08:52:32 · answer #2 · answered by Krazykraut 3 · 0⤊ 0⤋
well i got it but the answer is already up there
2006-09-22 08:50:50 · answer #3 · answered by jester 2 · 0⤊ 0⤋
.8832429 m/s backwards
2006-09-22 08:47:52 · answer #4 · answered by ishu pal s 1 · 0⤊ 0⤋