Question

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.

Answer 1

M1V1 + M2V2 = (M1 + M2)V3
(50)(6) + (70)(0) = (120)(V3)
300 = 120V3
V3 = 2.5 m/s

Answer 2

how did you get that

Answer 3

This 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

Answer 4

your welcome