A 92 kg man lying on a surface of negligible friction shoves a 71 g stone away from himself, giving it a speed of 4.0 m/s. What speed does the man acquire as a result?
answer should be in meters per second
2007-12-09 17:21:45 · 4 answers · asked by xoxo-oth 2 in Science & Mathematics ➔ Physics
Momentum's conservation: M V + m v = 0
V = - m v/M = - 0.071*4/92 = - 3.1*10^-3 m/s
2007-12-09 17:32:14 · answer #1 · answered by Luigi 74 7 · 2⤊ 1⤋
Use conservation of momentum
Initial momentum = final momentum
0 = MV + mv
MV = -mv
0.071 kg x 4 m/s = - 92 kg x v
v = 0.071 x 4 / -92
v = -0.003 m/s approx.
the negative sign indicates that the velocity of man is in the opposite direction to that of the stone.
2007-12-09 18:00:11 · answer #2 · answered by gauravragtah 4 · 0⤊ 0⤋
The second answer was correct. The person who wrote the first answer forgot to convert the mass of the stone to kg.
2007-12-09 17:52:17 · answer #3 · answered by Miles 2 · 0⤊ 1⤋
momentum is conserved. Since both objects had no momentum at the beginning, their momentums after the push must be equal, but opposite.
momentum is mass times velocity (p = m*v)
so, 92x = 71*4
92x = 284
x = 3.1 m/s
It makes sense that the man would be moving slower since he is more massive than the stone.
2007-12-09 17:30:30 · answer #4 · answered by lhvinny 7 · 0⤊ 2⤋