Two ice skaters stand at rest in the center of an ice rink. When they push off against one another the 45 kg skater acquires a speed of 0.73 m/s. If the speed of the other skater is 0.89 m/s, what is this skater's mass?
kg
2007-04-01 06:09:59 · 3 answers · asked by soccerjock 2 in Science & Mathematics ➔ Physics
Momentum is conserved!
Therefore: P1=P2
P1=m1v1=(45 kg)(.73 m/s)=32.85 kg*m/s
Therefore P2=32.85
m2=P2/V2 = (32.85 kg*m/s)/(.89 m/s)=36.9 kg, which with sig figs is 37 kg
2007-04-01 06:20:04 · answer #1 · answered by lukedu186 2 · 0⤊ 0⤋
Momentum conservation is to be used. The momentum of the 45 kg skater is 45 X 0.73 and that must be equal to m X 0.89
So, m = 45 X 0.73 / 0.89 = 36.9 kg (approx.)
2007-04-01 13:16:06 · answer #2 · answered by Swamy 7 · 0⤊ 0⤋
m1v1 = m2v2
45kg*0.73m/s = xkg*0.89m/s
36.9kg = x
2007-04-01 13:20:52 · answer #3 · answered by neuro 2 · 0⤊ 0⤋