Two ice-skaters stand next to each other on the ice. One skater, with a mass of 80 kg, pushes the other, who has a mass of only 50 hg. If the pushed skater moves at 4m/s, what is the speed of the heavier skater?
2006-11-29 09:56:40 · 3 answers · asked by James R 1 in Science & Mathematics ➔ Physics
Hmmm. Lessee... two skaters on slick ice. One, mass 80 kg, pushes the other and, according to Newton, is simultaneously pushed back in the opposite direction while the other, mass 50 kg, goes flying off across the ice at 4 m/s. This is a closed system, right? So where did that 200 kilogram meters per second of momentum imparted to the lighter skater come from? Yikes! Could the heavier skater have acquired the SAME momentum in the opposite direction? Is it possible that, after all this time since Newton, momentum is STILL conserved? Does MV = mv, where M is the mass of the heavier skater, V the velocity of the heavier skater, m the mass of the lighter skater, and v the velocity of the lighter skater? Does (80) V = (50) (4)? Does V = 2.5 meters per second? Nah! Lard butt's extra mass caused their skates to dig into the ice, and created enough friction to slow them down to 0.0025 meters per second... probably.
2006-11-29 10:48:04 · answer #1 · answered by hevans1944 5 · 0⤊ 0⤋
Well, momentum is always conserved, so the total momentum before the push is equal to the momentum after the push. Momentum equals mass times velocity. Since before the push they are both motionless, the total momentum is zero. Therefore, the momentum after the collision equals zero. Pa + Pb = 0. As such, you have the equation (80)(v) + (50)(4) = 0, or 80v = 200. 200 divided by 80 equals 2.5 m/s. There's your answer.
2006-11-29 10:50:02 · answer #2 · answered by Where the 'morrow lives 2 · 0⤊ 0⤋
.25m/s
2006-11-29 10:05:02 · answer #3 · answered by zginder 3 · 0⤊ 0⤋