Initial x momentum = 820 kg m/s
Initial y momentum = 55*7 = 385 kg m/s
total mass in the end clinging together = 82+55 = 137 kg
so
137 Vx = 820
Vx = 5.99 m/s
and
137 Vy = 385
Vy = 2.81 m/s
magnitude of velocity squared= Vx^2 + Vy^2 = 36 + 7.9 = 43.9 m^2/s^2
so
Ke = (1/2) m v^2 = .5*137 * 43.9 = 3007 Joules = 3*10^3
so a
Two skaters glide toward each other and then hold on to each other, gliding together as one mass. Skater A has a mass of 82.0 kg and an initial velocity of 10 m/s [0°], skater B has a mass of 55.0 kg and an initial velocity of 7.00 m/s [90°].
What is the kinetic energy of the two skaters when they are skating together?
Select one:
a.
2.99 x 10^3 J
b.
1.20 x 10^4 J
c.
5.30 x 10^3 J
d.
537 J
1 answer
1 answer