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anonymous
 one year ago
A 50kg ice skater moving at 6m/s collides with a second stationary skater with mass 70kg. The skaters cling together after the collision and move without friction. Compute their speed after the collision.
anonymous
 one year ago
A 50kg ice skater moving at 6m/s collides with a second stationary skater with mass 70kg. The skaters cling together after the collision and move without friction. Compute their speed after the collision.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0M1V1 + M2V2 = (M1 + M2)V3 (50)(6) + (70)(0) = (120)(V3) 300 = 120V3 V3 = 2.5 m/s

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did you get that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is a completely inelastic collision, in which kinetic energy is not conserved. Therefore, you must use conservation of momentum. First calculate the momentum of the first ice skater: P= mv = (50 kg)(6 m/s) = 300 kg*m/s = 300 N*s Then calculate the momentum of the second ice skater: P= mv = (70kg) (0 m/s) = 0 N*s (The second skater has no momentum because he/she is stationary.) Then calculate their final momentum: We will leave v as a variable standing for their velocity: P(final)= (total mass)(v) = (50 kg + 70kg) (v) = 120v N*s The momentum at the beginning must be equal to the final momentum: (300) + (0) = (120v) 300 = 120 v v = (300/120) = 5/2 m/s = 2.5 m/s
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